Probability 2-Mark Questions Class 10: CBSE PYQs with Answers

Given: 52 cards; all 13 diamonds removed; 3 kings remain.

To find: The probability of a king.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “a king”.

Counting: There are 52 − 13 = 39 cards left. Removing the diamond suit removes one of the four kings, leaving 3.

n(S) = 39; n(E) = 3.

Substitution: P(E) = $\frac{\text{3}}{\text{39}}$.

Calculation: P(E) = $\frac{\text{1}}{\text{13}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: a king: $\frac{\text{1}}{\text{13}}$

Given: 4 kings and 4 queens removed from 52 cards.

To find: The probability of (i) a red card; (ii) an ace.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) a red card”.

Counting: 8 cards are removed, leaving 44. Four of the removed cards are red, so 26 − 4 = 22 red cards remain.

n(S) = 44; n(E1) = 22.

Substitution: P(E1) = $\frac{\text{22}}{\text{44}}$.

Calculation: P(E1) = $\frac{\text{1}}{\text{2}}$.

Event: Let E2 be the event “(ii) an ace”.

Counting: None of the four aces is a king or queen; all 4 aces remain among the 44 cards.

n(S) = 44; n(E2) = 4.

Substitution: P(E2) = $\frac{\text{4}}{\text{44}}$.

Calculation: P(E2) = $\frac{\text{1}}{\text{11}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) a red card: $\frac{\text{1}}{\text{2}}$; (ii) an ace: $\frac{\text{1}}{\text{11}}$

Related method: Question 1.

Given: 52 cards originally; one black card lost; queen of hearts is red.

To find: The probability of the queen of hearts.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “the queen of hearts”.

Counting: 51 cards remain. The lost card is black, so the one queen of hearts is still present.

n(S) = 51; n(E) = 1.

Substitution: P(E) = $\frac{\text{1}}{\text{51}}$.

Calculation: P(E) = $\frac{\text{1}}{\text{51}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: the queen of hearts: $\frac{\text{1}}{\text{51}}$

Related method: Question 1.

Given: A standard deck of 52 cards, containing 1 queen of hearts and 4 jacks.

To find: The probability of (i) the queen of hearts; (ii) not a jack.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) the queen of hearts”.

Counting: There is exactly one queen of hearts.

n(S) = 52; n(E1) = 1.

Substitution: P(E1) = $\frac{\text{1}}{\text{52}}$.

Calculation: P(E1) = $\frac{\text{1}}{\text{52}}$.

Event: Let E2 be the event “(ii) not a jack”.

Counting: 4 cards are jacks, so 52 − 4 = 48 are not jacks.

n(S) = 52; n(E2) = 48.

Related formula: P(E2) = 1 − P(E2̅), where E2̅ is the complementary event.

Substitution: P(E2) = 1 − $\frac{\text{4}}{\text{52}}$.

Calculation: P(E2) = $\frac{\text{52 − 4}}{\text{52}}$ = $\frac{\text{12}}{\text{13}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) the queen of hearts: $\frac{\text{1}}{\text{52}}$; (ii) not a jack: $\frac{\text{12}}{\text{13}}$

Related method: Question 1.

Given: The king, queen and ace from each of clubs and diamonds are removed: 6 cards in total.

To find: The probability of (i) a club; (ii) a red card.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) a club”.

Counting: 52 − 6 = 46 cards remain. Three clubs were removed, so 13 − 3 = 10 clubs remain.

n(S) = 46; n(E1) = 10.

Substitution: P(E1) = $\frac{\text{10}}{\text{46}}$.

Calculation: P(E1) = $\frac{\text{5}}{\text{23}}$.

Event: Let E2 be the event “(ii) a red card”.

Counting: Only three diamonds were removed from the 26 red cards; 26 − 3 = 23 red cards remain.

n(S) = 46; n(E2) = 23.

Substitution: P(E2) = $\frac{\text{23}}{\text{46}}$.

Calculation: P(E2) = $\frac{\text{1}}{\text{2}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) a club: $\frac{\text{5}}{\text{23}}$; (ii) a red card: $\frac{\text{1}}{\text{2}}$

Related method: Question 1.

Given: 52 cards; 12 face cards (jacks, queens and kings); 2 black kings.

To find: The probability of (i) not a face card; (ii) a black king.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) not a face card”.

Counting: 52 − 12 = 40 cards are not face cards.

n(S) = 52; n(E1) = 40.

Related formula: P(E1) = 1 − P(E1̅), where E1̅ is the complementary event.

Substitution: P(E1) = 1 − $\frac{\text{12}}{\text{52}}$.

Calculation: P(E1) = $\frac{\text{52 − 12}}{\text{52}}$ = $\frac{\text{10}}{\text{13}}$.

Event: Let E2 be the event “(ii) a black king”.

Counting: The kings of clubs and spades are the two black kings.

n(S) = 52; n(E2) = 2.

Substitution: P(E2) = $\frac{\text{2}}{\text{52}}$.

Calculation: P(E2) = $\frac{\text{1}}{\text{26}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) not a face card: $\frac{\text{10}}{\text{13}}$; (ii) a black king: $\frac{\text{1}}{\text{26}}$

Related method: Question 1.

Given: All 12 face cards removed; 40 cards remain.

To find: The probability of (i) a spade; (ii) not an ace.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) a spade”.

Counting: Each suit loses its jack, queen and king. Thus 13 − 3 = 10 spades remain.

n(S) = 40; n(E1) = 10.

Substitution: P(E1) = $\frac{\text{10}}{\text{40}}$.

Calculation: P(E1) = $\frac{\text{1}}{\text{4}}$.

Event: Let E2 be the event “(ii) not an ace”.

Counting: All 4 aces remain, so 40 − 4 = 36 remaining cards are not aces.

n(S) = 40; n(E2) = 36.

Related formula: P(E2) = 1 − P(E2̅), where E2̅ is the complementary event.

Substitution: P(E2) = 1 − $\frac{\text{4}}{\text{40}}$.

Calculation: P(E2) = $\frac{\text{40 − 4}}{\text{40}}$ = $\frac{\text{9}}{\text{10}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) a spade: $\frac{\text{1}}{\text{4}}$; (ii) not an ace: $\frac{\text{9}}{\text{10}}$

Related method: Question 1.

Given: 4 aces and 4 kings removed; 44 cards remain.

To find: The probability of (i) a face card; (ii) a red card.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) a face card”.

Counting: Of the 12 face cards, the 4 kings are removed. The 4 jacks and 4 queens remain: 8 face cards.

n(S) = 44; n(E1) = 8.

Substitution: P(E1) = $\frac{\text{8}}{\text{44}}$.

Calculation: P(E1) = $\frac{\text{2}}{\text{11}}$.

Event: Let E2 be the event “(ii) a red card”.

Counting: Two red aces and two red kings are removed, so 26 − 4 = 22 red cards remain.

n(S) = 44; n(E2) = 22.

Substitution: P(E2) = $\frac{\text{22}}{\text{44}}$.

Calculation: P(E2) = $\frac{\text{1}}{\text{2}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) a face card: $\frac{\text{2}}{\text{11}}$; (ii) a red card: $\frac{\text{1}}{\text{2}}$

Related method: Question 1.

Given: 6 red face cards removed; 46 cards remain.

To find: The probability of (i) a red card; (ii) a king or queen.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) a red card”.

Counting: 26 − 6 = 20 red cards remain.

n(S) = 46; n(E1) = 20.

Substitution: P(E1) = $\frac{\text{20}}{\text{46}}$.

Calculation: P(E1) = $\frac{\text{10}}{\text{23}}$.

Event: Let E2 be the event “(ii) a king or queen”.

Counting: Only the black kings and black queens remain: 2 + 2 = 4. A card cannot be both a king and a queen.

n(S) = 46; n(E2) = 4.

Substitution: P(E2) = $\frac{\text{4}}{\text{46}}$.

Calculation: P(E2) = $\frac{\text{2}}{\text{23}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) a red card: $\frac{\text{10}}{\text{23}}$; (ii) a king or queen: $\frac{\text{2}}{\text{23}}$

Related method: Question 1.

Given: 90 discs numbered 1–90, inclusive.

To find: The probability of a multiple of 15.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “a multiple of 15”.

Counting: The favourable numbers are 15, 30, 45, 60, 75 and 90: 6 numbers.

n(S) = 90; n(E) = 6.

Substitution: P(E) = $\frac{\text{6}}{\text{90}}$.

Calculation: P(E) = $\frac{\text{1}}{\text{15}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: a multiple of 15: $\frac{\text{1}}{\text{15}}$

Given: 20 equally selectable discs numbered 1–20, inclusive.

To find: The probability of (i) a two-digit number; (ii) a number less than 10.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) a two-digit number”.

Counting: The two-digit numbers are 10–20, giving 20 − 10 + 1 = 11.

n(S) = 20; n(E1) = 11.

Substitution: P(E1) = $\frac{\text{11}}{\text{20}}$.

Calculation: P(E1) = $\frac{\text{11}}{\text{20}}$.

Event: Let E2 be the event “(ii) a number less than 10”.

Counting: The favourable numbers are 1–9: 9 numbers.

n(S) = 20; n(E2) = 9.

Substitution: P(E2) = $\frac{\text{9}}{\text{20}}$.

Calculation: P(E2) = $\frac{\text{9}}{\text{20}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) a two-digit number: $\frac{\text{11}}{\text{20}}$; (ii) a number less than 10: $\frac{\text{9}}{\text{20}}$

Related method: Question 10.

Given: 30 discs numbered 1–30, inclusive.

To find: The probability of (a) divisible by 6; (b) greater than 25.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(a) divisible by 6”.

Counting: The multiples are 6, 12, 18, 24 and 30: 5 numbers.

n(S) = 30; n(E1) = 5.

Substitution: P(E1) = $\frac{\text{5}}{\text{30}}$.

Calculation: P(E1) = $\frac{\text{1}}{\text{6}}$.

Event: Let E2 be the event “(b) greater than 25”.

Counting: The numbers are 26, 27, 28, 29 and 30: 5 numbers.

n(S) = 30; n(E2) = 5.

Substitution: P(E2) = $\frac{\text{5}}{\text{30}}$.

Calculation: P(E2) = $\frac{\text{1}}{\text{6}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (a) divisible by 6: $\frac{\text{1}}{\text{6}}$; (b) greater than 25: $\frac{\text{1}}{\text{6}}$

Related method: Question 10.

Given: The integers 1–50, inclusive, are equally likely.

To find: The probability of divisible by 6.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “divisible by 6”.

Counting: The multiples are 6, 12, 18, 24, 30, 36, 42 and 48: 8 integers.

n(S) = 50; n(E) = 8.

Substitution: P(E) = $\frac{\text{8}}{\text{50}}$.

Calculation: P(E) = $\frac{\text{4}}{\text{25}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: divisible by 6: $\frac{\text{4}}{\text{25}}$

Related method: Question 10.

Given: 80 cards numbered 1–80, inclusive.

To find: The probability of not divisible by 8.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “not divisible by 8”.

Counting: The multiples 8, 16, …, 80 number 80 ÷ 8 = 10. Thus 80 − 10 = 70 cards are favourable.

n(S) = 80; n(E) = 70.

Related formula: P(E) = 1 − P(E̅), where E̅ is the complementary event.

Substitution: P(E) = 1 − $\frac{\text{10}}{\text{80}}$.

Calculation: P(E) = $\frac{\text{80 − 10}}{\text{80}}$ = $\frac{\text{7}}{\text{8}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: not divisible by 8: $\frac{\text{7}}{\text{8}}$

Related method: Question 10.

Given: Cards numbered 5–100, inclusive; total = 100 − 5 + 1 = 96.

To find: The probability of (i) a perfect square; (ii) a two-digit number.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) a perfect square”.

Counting: The squares in the range are 9, 16, 25, 36, 49, 64, 81 and 100: 8 numbers.

n(S) = 96; n(E1) = 8.

Substitution: P(E1) = $\frac{\text{8}}{\text{96}}$.

Calculation: P(E1) = $\frac{\text{1}}{\text{12}}$.

Event: Let E2 be the event “(ii) a two-digit number”.

Counting: The two-digit numbers run from 10 to 99, giving 99 − 10 + 1 = 90.

n(S) = 96; n(E2) = 90.

Substitution: P(E2) = $\frac{\text{90}}{\text{96}}$.

Calculation: P(E2) = $\frac{\text{15}}{\text{16}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) a perfect square: $\frac{\text{1}}{\text{12}}$; (ii) a two-digit number: $\frac{\text{15}}{\text{16}}$

Related method: Question 10.

Given: Balls numbered 2–91, inclusive; total = 91 − 2 + 1 = 90.

To find: The probability of (i) a two-digit number; (ii) a multiple of 1.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) a two-digit number”.

Counting: The numbers are 10–91, giving 91 − 10 + 1 = 82.

n(S) = 90; n(E1) = 82.

Substitution: P(E1) = $\frac{\text{82}}{\text{90}}$.

Calculation: P(E1) = $\frac{\text{41}}{\text{45}}$.

Event: Let E2 be the event “(ii) a multiple of 1”.

Counting: Every integer is a multiple of 1. All 90 balls are favourable, so this is a certain event.

n(S) = 90; n(E2) = 90.

Substitution: P(E2) = $\frac{\text{90}}{\text{90}}$.

Calculation: P(E2) = 1.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) a two-digit number: $\frac{\text{41}}{\text{45}}$; (ii) a multiple of 1: 1

Related method: Question 10.

Given: 120 discs numbered 1–120, inclusive.

To find: The probability of (i) a two-digit number; (ii) a perfect square.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) a two-digit number”.

Counting: The two-digit numbers 10–99 give 99 − 10 + 1 = 90.

n(S) = 120; n(E1) = 90.

Substitution: P(E1) = $\frac{\text{90}}{\text{120}}$.

Calculation: P(E1) = $\frac{\text{3}}{\text{4}}$.

Event: Let E2 be the event “(ii) a perfect square”.

Counting: The squares are 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100: 10 numbers. The next square, 121, exceeds 120.

n(S) = 120; n(E2) = 10.

Substitution: P(E2) = $\frac{\text{10}}{\text{120}}$.

Calculation: P(E2) = $\frac{\text{1}}{\text{12}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) a two-digit number: $\frac{\text{3}}{\text{4}}$; (ii) a perfect square: $\frac{\text{1}}{\text{12}}$

Related method: Question 10.

Given: 40 balls numbered 21–60, inclusive.

To find: The probability of (i) a perfect square; (ii) a multiple of 8.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) a perfect square”.

Counting: The squares in 21–60 are 25, 36 and 49. The next square, 64, is outside the range.

n(S) = 40; n(E1) = 3.

Substitution: P(E1) = $\frac{\text{3}}{\text{40}}$.

Calculation: P(E1) = $\frac{\text{3}}{\text{40}}$.

Event: Let E2 be the event “(ii) a multiple of 8”.

Counting: The multiples are 24, 32, 40, 48 and 56: 5 numbers.

n(S) = 40; n(E2) = 5.

Substitution: P(E2) = $\frac{\text{5}}{\text{40}}$.

Calculation: P(E2) = $\frac{\text{1}}{\text{8}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) a perfect square: $\frac{\text{3}}{\text{40}}$; (ii) a multiple of 8: $\frac{\text{1}}{\text{8}}$

Related method: Question 10.

Given: 32 slips numbered 20–51, inclusive.

To find: The probability of (i) a prime number; (ii) divisible by 5.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) a prime number”.

Counting: The primes are 23, 29, 31, 37, 41, 43 and 47: 7 numbers.

n(S) = 32; n(E1) = 7.

Substitution: P(E1) = $\frac{\text{7}}{\text{32}}$.

Calculation: P(E1) = $\frac{\text{7}}{\text{32}}$.

Event: Let E2 be the event “(ii) divisible by 5”.

Counting: The numbers are 20, 25, 30, 35, 40, 45 and 50: 7 numbers.

n(S) = 32; n(E2) = 7.

Substitution: P(E2) = $\frac{\text{7}}{\text{32}}$.

Calculation: P(E2) = $\frac{\text{7}}{\text{32}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) a prime number: $\frac{\text{7}}{\text{32}}$; (ii) divisible by 5: $\frac{\text{7}}{\text{32}}$

Related method: Question 10.

Given: 31 integers from 30 to 60, inclusive.

To find: The probability of (i) prime; (ii) a multiple of 6.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) prime”.

Counting: The primes are 31, 37, 41, 43, 47, 53 and 59: 7 numbers.

n(S) = 31; n(E1) = 7.

Substitution: P(E1) = $\frac{\text{7}}{\text{31}}$.

Calculation: P(E1) = $\frac{\text{7}}{\text{31}}$.

Event: Let E2 be the event “(ii) a multiple of 6”.

Counting: The multiples are 30, 36, 42, 48, 54 and 60: 6 numbers.

n(S) = 31; n(E2) = 6.

Substitution: P(E2) = $\frac{\text{6}}{\text{31}}$.

Calculation: P(E2) = $\frac{\text{6}}{\text{31}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) prime: $\frac{\text{7}}{\text{31}}$; (ii) a multiple of 6: $\frac{\text{6}}{\text{31}}$

Related method: Question 10.

Given: BACKGROUND contains 10 letters, each occurring once: B, A, C, K, G, R, O, U, N, D.

To find: The probability of (i) a vowel; (ii) present in BALL.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) a vowel”.

Counting: The vowels are A, O and U: 3 slips.

n(S) = 10; n(E1) = 3.

Substitution: P(E1) = $\frac{\text{3}}{\text{10}}$.

Calculation: P(E1) = $\frac{\text{3}}{\text{10}}$.

Event: Let E2 be the event “(ii) present in BALL”.

Counting: BALL contains the letter types B, A and L. BACKGROUND has B and A but no L, so 2 slips are favourable.

n(S) = 10; n(E2) = 2.

Substitution: P(E2) = $\frac{\text{2}}{\text{10}}$.

Calculation: P(E2) = $\frac{\text{1}}{\text{5}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) a vowel: $\frac{\text{3}}{\text{10}}$; (ii) present in BALL: $\frac{\text{1}}{\text{5}}$

Related method: Question 10.

Given: 4 red, 3 blue and 2 yellow balls; 9 balls in total.

To find: The probability of (i) red; (ii) yellow.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) red”.

Counting: 4 + 3 + 2 = 9 balls are possible; 4 are red.

n(S) = 9; n(E1) = 4.

Substitution: P(E1) = $\frac{\text{4}}{\text{9}}$.

Calculation: P(E1) = $\frac{\text{4}}{\text{9}}$.

Event: Let E2 be the event “(ii) yellow”.

Counting: The same 9 balls are possible; 2 are yellow.

n(S) = 9; n(E2) = 2.

Substitution: P(E2) = $\frac{\text{2}}{\text{9}}$.

Calculation: P(E2) = $\frac{\text{2}}{\text{9}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) red: $\frac{\text{4}}{\text{9}}$; (ii) yellow: $\frac{\text{2}}{\text{9}}$

Given: 145 good pens and 15 defective pens.

To find: The probability of a good pen.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “a good pen”.

Counting: Total pens = 145 + 15 = 160; 145 are good.

n(S) = 160; n(E) = 145.

Substitution: P(E) = $\frac{\text{145}}{\text{160}}$.

Calculation: P(E) = $\frac{\text{29}}{\text{32}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: a good pen: $\frac{\text{29}}{\text{32}}$

Related method: Question 21.

Given: 165 pens in total; 30 defective; only good pens are acceptable.

To find: The probability of a pen Rakshita will buy.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “a pen Rakshita will buy”.

Counting: Good pens = 165 − 30 = 135.

n(S) = 165; n(E) = 135.

Substitution: P(E) = $\frac{\text{135}}{\text{165}}$.

Calculation: P(E) = $\frac{\text{9}}{\text{11}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: a pen Rakshita will buy: $\frac{\text{9}}{\text{11}}$

Related method: Question 21.

Given: 48 good shirts, 8 with major defects and 4 with minor defects; Anmol rejects only major defects.

To find: The probability of acceptable to Anmol.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “acceptable to Anmol”.

Counting: Anmol accepts good shirts and shirts with minor defects: 48 + 4 = 52. Nigam’s acceptance rule is not used for this event.

n(S) = 60; n(E) = 52.

Substitution: P(E) = $\frac{\text{52}}{\text{60}}$.

Calculation: P(E) = $\frac{\text{13}}{\text{15}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: acceptable to Anmol: $\frac{\text{13}}{\text{15}}$

Related method: Question 21.

Given: 50 cards numbered 1–50, inclusive.

To find: The probability of a perfect square.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “a perfect square”.

Counting: The squares are 1, 4, 9, 16, 25, 36 and 49: 7 numbers.

n(S) = 50; n(E) = 7.

Substitution: P(E) = $\frac{\text{7}}{\text{50}}$.

Calculation: P(E) = $\frac{\text{7}}{\text{50}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: a perfect square: $\frac{\text{7}}{\text{50}}$

Related method: Question 21.

Given: 50 coins of ₹1, 100 coins of ₹2 and 150 coins of ₹5; each individual coin is equally likely.

To find: The probability of not a ₹2 coin.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “not a ₹2 coin”.

Counting: Total coins = 50 + 100 + 150 = 300. Favourable coins = 50 + 150 = 200. Count coins, not their monetary values.

n(S) = 300; n(E) = 200.

Related formula: P(E) = 1 − P(E̅), where E̅ is the complementary event.

Substitution: P(E) = 1 − $\frac{\text{100}}{\text{300}}$.

Calculation: P(E) = $\frac{\text{300 − 100}}{\text{300}}$ = $\frac{\text{2}}{\text{3}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: not a ₹2 coin: $\frac{\text{2}}{\text{3}}$

Related method: Question 21.

Given: Two independent tosses of a fair coin; S = {HH, HT, TH, TT}.

To find: The probability of at most one head.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “at most one head”.

Counting: At most one means zero or one head. The favourable outcomes are TT, HT and TH.

n(S) = 4; n(E) = 3.

Substitution: P(E) = $\frac{\text{3}}{\text{4}}$.

Calculation: P(E) = $\frac{\text{3}}{\text{4}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: at most one head: $\frac{\text{3}}{\text{4}}$

Given: Use the standard model of three fair, independently tossed coins.

To find: The probability of (ii) at most two heads.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Sample space: (i) S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.

Event: Let E be the event “(ii) at most two heads”.

Counting: The eight outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH and TTT. Only HHH has more than two heads.

n(S) = 8; n(E) = 7.

Related formula: P(E) = 1 − P(E̅), where E̅ is the complementary event.

Substitution: P(E) = 1 − $\frac{\text{1}}{\text{8}}$.

Calculation: P(E) = $\frac{\text{8 − 1}}{\text{8}}$ = $\frac{\text{7}}{\text{8}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}; (ii) at most two heads: $\frac{\text{7}}{\text{8}}$

Related method: Question 26.

Given: A fair die with outcomes S = {1, 2, 3, 4, 5, 6}.

To find: The probability of (i) greater than 3; (ii) an even prime number.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) greater than 3”.

Counting: The favourable outcomes are 4, 5 and 6.

n(S) = 6; n(E1) = 3.

Substitution: P(E1) = $\frac{\text{3}}{\text{6}}$.

Calculation: P(E1) = $\frac{\text{1}}{\text{2}}$.

Event: Let E2 be the event “(ii) an even prime number”.

Counting: 2 is the only even prime number.

n(S) = 6; n(E2) = 1.

Substitution: P(E2) = $\frac{\text{1}}{\text{6}}$.

Calculation: P(E2) = $\frac{\text{1}}{\text{6}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) greater than 3: $\frac{\text{1}}{\text{2}}$; (ii) an even prime number: $\frac{\text{1}}{\text{6}}$

Related method: Question 26.

Given: 5 red, 8 white and 4 green marbles; total 17.

To find: The probability of (i) white; (ii) not green.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) white”.

Counting: 5 + 8 + 4 = 17 marbles; 8 are white.

n(S) = 17; n(E1) = 8.

Substitution: P(E1) = $\frac{\text{8}}{\text{17}}$.

Calculation: P(E1) = $\frac{\text{8}}{\text{17}}$.

Event: Let E2 be the event “(ii) not green”.

Counting: 5 + 8 = 13 marbles are not green.

n(S) = 17; n(E2) = 13.

Related formula: P(E2) = 1 − P(E2̅), where E2̅ is the complementary event.

Substitution: P(E2) = 1 − $\frac{\text{4}}{\text{17}}$.

Calculation: P(E2) = $\frac{\text{17 − 4}}{\text{17}}$ = $\frac{\text{13}}{\text{17}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) white: $\frac{\text{8}}{\text{17}}$; (ii) not green: $\frac{\text{13}}{\text{17}}$

Related method: Question 26.

Given: Each fair die has six labels: 4, 6, 7, 9, 11, 12. There are 36 ordered outcomes.

To find: The probability of an odd product.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “an odd product”.

Counting: Both labels must be odd. The odd labels are 7, 9 and 11, giving 3 × 3 = 9 favourable ordered pairs.

n(S) = 36; n(E) = 9.

Substitution: P(E) = $\frac{\text{9}}{\text{36}}$.

Calculation: P(E) = $\frac{\text{1}}{\text{4}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: an odd product: $\frac{\text{1}}{\text{4}}$

Related method: Question 26.

Given: All 900 integers from 100 to 999 are equally likely; required form is 8_5.

To find: The probability of hundreds digit 8 and units digit 5.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Required count: Number of such integers = 10.

Event: Let E be the event “hundreds digit 8 and units digit 5”.

Counting: The tens digit has 10 choices, giving 805, 815, 825, …, 895. Thus the required count is 10.

n(S) = 900; n(E) = 10.

Substitution: P(E) = $\frac{\text{10}}{\text{900}}$.

Calculation: P(E) = $\frac{\text{1}}{\text{90}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: Number of such integers = 10. hundreds digit 8 and units digit 5: $\frac{\text{1}}{\text{90}}$

Related method: Question 26.

Given: Two independent throws of a fair six-sided die; 6 × 6 = 36 ordered outcomes.

To find: The probability of 5 at least once.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “5 at least once”.

Counting: For no 5, each throw has 5 choices: 5 × 5 = 25 outcomes. Thus 36 − 25 = 11 outcomes have at least one 5.

n(S) = 36; n(E) = 11.

Related formula: P(E) = 1 − P(E̅), where E̅ is the complementary event.

Substitution: P(E) = 1 − $\frac{\text{25}}{\text{36}}$.

Calculation: P(E) = $\frac{\text{36 − 25}}{\text{36}}$ = $\frac{\text{11}}{\text{36}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: 5 at least once: $\frac{\text{11}}{\text{36}}$

Related method: Question 26.

Given: 3 diamond face cards removed; 49 cards remain.

To find: The probability of a face card or a spade.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “a face card or a spade”.

Related formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B), where A is the set of remaining face cards and B is the set of remaining spades.

Counting: 9 face cards and 13 spades remain. The 3 spade face cards belong to both sets, so favourable cards = 9 + 13 − 3 = 19.

n(S) = 49; n(E) = 19.

Substitution: P(E) = $\frac{\text{19}}{\text{49}}$.

Calculation: P(E) = $\frac{\text{19}}{\text{49}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: a face card or a spade: $\frac{\text{19}}{\text{49}}$

Related method: Question 26.

Given: Three fair, independently tossed coins; 8 equally likely outcomes.

To find: The probability of at least two tails.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “at least two tails”.

Counting: The favourable outcomes are HTT, THT, TTH and TTT: 4 outcomes.

n(S) = 8; n(E) = 4.

Substitution: P(E) = $\frac{\text{4}}{\text{8}}$.

Calculation: P(E) = $\frac{\text{1}}{\text{2}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: at least two tails: $\frac{\text{1}}{\text{2}}$

Related method: Question 26.

Given: Two fair six-sided dice; 36 equally likely ordered outcomes.

To find: The probability of a prime product.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E be the event “a prime product”.

Counting: For a prime product, one label is 1 and the other is 2, 3 or 5. The favourable pairs are (1,2), (2,1), (1,3), (3,1), (1,5), (5,1).

n(S) = 36; n(E) = 6.

Substitution: P(E) = $\frac{\text{6}}{\text{36}}$.

Calculation: P(E) = $\frac{\text{1}}{\text{6}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: a prime product: $\frac{\text{1}}{\text{6}}$

Related method: Question 26.

Given: 4 red and 5 white balls; P(red) = $\frac{\text{1}}{\text{5}}$; number of yellow balls unknown.

To find: The probability of drawing a yellow ball.

Formula: P(red) = $\frac{\text{number of red balls}}{\text{total number of balls}}$; P(yellow) = $\frac{\text{number of yellow balls}}{\text{total number of balls}}$.

Substitution: Let y be the number of yellow balls. Total = 4 + 5 + y = 9 + y.

$\frac{\text{4}}{\text{9 + y}}$ = $\frac{\text{1}}{\text{5}}$.

Calculation: 9 + y = 4 × 5 = 20.

y = 20 − 9 = 11.

P(yellow) = $\frac{\text{11}}{\text{20}}$.

Check: 4 + 5 + 11 = 20, and $\frac{\text{4}}{\text{20}}$ = 1/5.

Answer: P(yellow) = $\frac{\text{11}}{\text{20}}$

Given: 40 marbles; P(black) = 3/5.

To find: The number of white marbles.

Formula: P(black) = $\frac{\text{number of black marbles}}{\text{total number of marbles}}$.

Substitution: Let b be the number of black marbles. Then $\frac{\text{b}}{\text{40}}$ = $\frac{\text{3}}{\text{5}}$.

Calculation: 5b = 3 × 40.

5b = 120.

b = 24.

Number of white marbles = 40 − 24 = 16.

Check: $\frac{\text{24}}{\text{40}}$ = $\frac{\text{3}}{\text{5}}$, and 24 + 16 = 40.

Answer: 16 white marbles

Related method: Question 33.

Given: 40 cards numbered 20–59, inclusive.

To find: The probability of (i) prime; (ii) a perfect square.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) prime”.

Counting: The primes are 23, 29, 31, 37, 41, 43, 47, 53 and 59: 9 numbers.

n(S) = 40; n(E1) = 9.

Substitution: P(E1) = $\frac{\text{9}}{\text{40}}$.

Calculation: P(E1) = $\frac{\text{9}}{\text{40}}$.

Event: Let E2 be the event “(ii) a perfect square”.

Counting: The squares in the range are 25, 36 and 49: 3 numbers.

n(S) = 40; n(E2) = 3.

Substitution: P(E2) = $\frac{\text{3}}{\text{40}}$.

Calculation: P(E2) = $\frac{\text{3}}{\text{40}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) prime: $\frac{\text{9}}{\text{40}}$; (ii) a perfect square: $\frac{\text{3}}{\text{40}}$

Related method: Question 33.

Given: P(correct) = $\frac{\text{x}}{\text{12}}$; P(not correct) = 5/6.

To find: The value of x.

Formula: P(E) + P(E̅) = 1, where E is guessing correctly and E̅ is not guessing correctly.

Substitution: $\frac{\text{x}}{\text{12}}$ + $\frac{\text{5}}{\text{6}}$ = 1.

Calculation: $\frac{\text{x}}{\text{12}}$ = 1 − $\frac{\text{5}}{\text{6}}$ = $\frac{\text{1}}{\text{6}}$.

x = 12 × $\frac{\text{1}}{\text{6}}$ = 2.

Check: $\frac{\text{2}}{\text{12}}$ + $\frac{\text{5}}{\text{6}}$ = $\frac{\text{1}}{\text{6}}$ + $\frac{\text{5}}{\text{6}}$ = 1.

Answer: x = 2

Related method: Question 33.

Given: Only red and black balls; red count exceeds black count by 3; P(red) = 12/23.

To find: The total number of balls.

Formula: P(red) = $\frac{\text{number of red balls}}{\text{total number of balls}}$.

Substitution: Let the number of black balls be x. Red balls = x + 3; total balls = 2x + 3.

$\frac{\text{x + 3}}{\text{2x + 3}}$ = $\frac{\text{12}}{\text{23}}$.

Calculation: 23(x + 3) = 12(2x + 3).

23x + 69 = 24x + 36.

x = 33.

Total number of balls = 2 × 33 + 3 = 69.

Check: Red balls = 36; black balls = 33. Thus $\frac{\text{36}}{\text{69}}$ = $\frac{\text{12}}{\text{23}}$ and 36 − 33 = 3.

Answer: 69 balls

Related method: Question 33.

Given: 25 balls; only yellow and green; P(green) = 3/5.

To find: The number of yellow balls.

Formula: P(green) = $\frac{\text{number of green balls}}{\text{total number of balls}}$; yellow balls = total balls − green balls.

Substitution: Let g be the number of green balls. Then $\frac{\text{g}}{\text{25}}$ = $\frac{\text{3}}{\text{5}}$.

Calculation: g = 25 × $\frac{\text{3}}{\text{5}}$ = 15.

Yellow balls = 25 − 15 = 10.

Check: $\frac{\text{15}}{\text{25}}$ = $\frac{\text{3}}{\text{5}}$, and 15 + 10 = 25.

Answer: 10 yellow balls

Related method: Question 33.

Given: There are 31 possible dates. Use the textbook model of independent birthdays with all dates in the stated month/year equally likely.

To find: The probability of (i) the same birthday; (ii) different birthdays.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) the same birthday”.

Counting: Fix the first birthday. Exactly 1 of the 31 equally likely dates matches it.

n(S) = 31; n(E1) = 1.

Substitution: P(E1) = $\frac{\text{1}}{\text{31}}$.

Calculation: P(E1) = $\frac{\text{1}}{\text{31}}$.

Event: Let E2 be the event “(ii) different birthdays”.

Counting: The remaining 31 − 1 = 30 dates differ.

n(S) = 31; n(E2) = 30.

Related formula: P(E2) = 1 − P(E2̅), where E2̅ is the complementary event.

Substitution: P(E2) = 1 − $\frac{\text{1}}{\text{31}}$.

Calculation: P(E2) = $\frac{\text{31 − 1}}{\text{31}}$ = $\frac{\text{30}}{\text{31}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) the same birthday: $\frac{\text{1}}{\text{31}}$; (ii) different birthdays: $\frac{\text{30}}{\text{31}}$

Given: There are 30 possible dates. Use the textbook model of independent birthdays with all dates in the stated month/year equally likely.

To find: The probability of (i) different birthdays; (ii) the same birthday.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) different birthdays”.

Counting: The remaining 30 − 1 = 29 dates differ.

n(S) = 30; n(E1) = 29.

Related formula: P(E1) = 1 − P(E1̅), where E1̅ is the complementary event.

Substitution: P(E1) = 1 − $\frac{\text{1}}{\text{30}}$.

Calculation: P(E1) = $\frac{\text{30 − 1}}{\text{30}}$ = $\frac{\text{29}}{\text{30}}$.

Event: Let E2 be the event “(ii) the same birthday”.

Counting: Fix the first birthday. Exactly 1 of the 30 equally likely dates matches it.

n(S) = 30; n(E2) = 1.

Substitution: P(E2) = $\frac{\text{1}}{\text{30}}$.

Calculation: P(E2) = $\frac{\text{1}}{\text{30}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) different birthdays: $\frac{\text{29}}{\text{30}}$; (ii) the same birthday: $\frac{\text{1}}{\text{30}}$

Related method: Question 38.

Given: There are 366 possible dates. Use the textbook model of independent birthdays with all dates in the stated month/year equally likely.

To find: The probability of (i) the same birthday; (ii) different birthdays.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) the same birthday”.

Counting: Fix the first birthday. Exactly 1 of the 366 equally likely dates matches it.

n(S) = 366; n(E1) = 1.

Substitution: P(E1) = $\frac{\text{1}}{\text{366}}$.

Calculation: P(E1) = $\frac{\text{1}}{\text{366}}$.

Event: Let E2 be the event “(ii) different birthdays”.

Counting: The remaining 366 − 1 = 365 dates differ.

n(S) = 366; n(E2) = 365.

Related formula: P(E2) = 1 − P(E2̅), where E2̅ is the complementary event.

Substitution: P(E2) = 1 − $\frac{\text{1}}{\text{366}}$.

Calculation: P(E2) = $\frac{\text{366 − 1}}{\text{366}}$ = $\frac{\text{365}}{\text{366}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) the same birthday: $\frac{\text{1}}{\text{366}}$; (ii) different birthdays: $\frac{\text{365}}{\text{366}}$

Related method: Question 38.

Given: Rectangle 3 m × 2 m; circle radius 0.7 m; use the uniform landing-point model.

To find: The probability of landing inside the circle.

Model / formula: For a uniformly distributed landing point, P(inside circle) = $\frac{\text{area of circle}}{\text{area of rectangle}}$. Area of circle = πr²; area of rectangle = length × breadth. Take π = $\frac{\text{22}}{\text{7}}$, as in the marking scheme.

Substitution: Area of circle = $\frac{\text{22}}{\text{7}}$ × $\frac{\text{7}}{\text{10}}$ × $\frac{\text{7}}{\text{10}}$ m².

Calculation: Area of circle = $\frac{\text{77}}{\text{50}}$ m².

Area of rectangle = 3 × 2 = 6 m².

P(inside circle) = $\frac{\text{77}}{\text{50}}$ ÷ 6 = $\frac{\text{77}}{\text{300}}$.

Check: The probability lies between 0 and 1. Area units cancel in the ratio.

Answer: P(inside circle) = $\frac{\text{77}}{\text{300}}$

Related method: Question 38.

Given: Two independent throws of a fair six-sided die; 36 ordered outcomes.

To find: The probability of (i) a difference of 3; (ii) a sum of 8.

Formula: P(E) = $\frac{\text{n(E)}}{\text{n(S)}}$, where n(E) is the number of favourable outcomes and n(S) is the number of equally likely possible outcomes.

Event: Let E1 be the event “(i) a difference of 3”.

Counting: The favourable pairs are (1,4), (2,5), (3,6), (4,1), (5,2) and (6,3). “Difference” is taken as the non-negative difference.

n(S) = 36; n(E1) = 6.

Substitution: P(E1) = $\frac{\text{6}}{\text{36}}$.

Calculation: P(E1) = $\frac{\text{1}}{\text{6}}$.

Event: Let E2 be the event “(ii) a sum of 8”.

Counting: The favourable pairs are (2,6), (3,5), (4,4), (5,3) and (6,2).

n(S) = 36; n(E2) = 5.

Substitution: P(E2) = $\frac{\text{5}}{\text{36}}$.

Calculation: P(E2) = $\frac{\text{5}}{\text{36}}$.

Check: Each probability is between 0 and 1. Probabilities have no physical unit.

Answer: (i) a difference of 3: $\frac{\text{1}}{\text{6}}$; (ii) a sum of 8: $\frac{\text{5}}{\text{36}}$

Related method: Question 38.