Probability: MCQs with Answers

Probability is the ratio of favourable outcomes to all equally likely outcomes.

The favourable count lies between 0 and the total count.

Therefore, 0 ≤ P(A) ≤ 1.

Every probability lies between 0 and 1, inclusive.

Here 4% = 0.04 and 0.04% = 0.0004.

The value 4 is greater than 1, so it cannot be a probability.

An event becomes less likely as its probability approaches 0.

Among the given values, 0.0001 is the smallest and closest to 0.

Therefore, 0.0001 represents the most unlikely event among these choices.

An event and its complementary event together cover all possible outcomes.

P(event) + P(not event) = 1.

Substituting the given probabilities gives p + q = 1.

P(not event) = 1 − P(event).

Since P(event) = p, the required probability is 1 − p.

P(event) = 57% = 0.57.

P(not event) = 1 − P(event).

Therefore, P(not event) = 1 − 0.57 = 0.43.

Let E denote the given event. For equally likely outcomes, P(E) = favourable outcomes ÷ total outcomes.

Favourable outcomes = P(E) × total outcomes.

Substituting gives 0.098 × 500 = 49.

Each of the 800 tickets has an equal chance of winning the first prize.

Let x be the number of tickets bought by Meena. Then P(Meena wins) = $\frac{\text{x}}{\text{800}}$.

Thus $\frac{\text{x}}{\text{800}}$ = 0.08, so x = 0.08 × 800 = 64.

Let x be the number of chocolate-flavoured ice creams.

P(chocolate) = $\frac{\text{x}}{\text{600}}$ = 0.055.

Therefore, x = 600 × 0.055 = 33.

The equally likely outcomes are HH, HT, TH and TT, where H denotes a head and T a tail.

Exactly one tail occurs in HT and TH: 2 favourable outcomes out of 4.

P(exactly one tail) = $\frac{\text{2}}{\text{4}}$ = $\frac{\text{1}}{\text{2}}$.

The 8 equally likely outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH and TTT.

Exactly two tails occur in HTT, THT and TTH: 3 favourable outcomes.

P(exactly two tails) = $\frac{\text{3}}{\text{8}}$.

Three coins have 2 × 2 × 2 = 8 equally likely outcomes.

The only outcome with no head is TTT, so P(no head) = $\frac{\text{1}}{\text{8}}$.

P(at least one head) = 1 − P(no head) = 1 − $\frac{\text{1}}{\text{8}}$ = $\frac{\text{7}}{\text{8}}$.

There are 8 equally likely outcomes for three coins.

At most one tail means either no tail or exactly one tail.

The favourable outcomes are HHH, HHT, HTH and THH: 4 outcomes.

P(at most one tail) = $\frac{\text{4}}{\text{8}}$ = $\frac{\text{1}}{\text{2}}$.

The 6 equally likely outcomes are 1, 2, 3, 4, 5 and 6.

The only even number less than 3 is 2.

P(even number less than 3) = $\frac{\text{1}}{\text{6}}$.

A die has 6 equally likely outcomes.

The prime numbers less than 5 on the die are 2 and 3.

P(prime number less than 5) = $\frac{\text{2}}{\text{6}}$ = $\frac{\text{1}}{\text{3}}$.

A die has 6 equally likely outcomes.

The composite numbers among 1, 2, 3, 4, 5 and 6 are 4 and 6.

P(composite number) = $\frac{\text{2}}{\text{6}}$ = $\frac{\text{1}}{\text{3}}$.

The possible outcomes are 1, 2, 3, 4, 5 and 6.

Among these, 1, 2, 3, 4 and 6 divide 36; only 5 does not.

P(not a factor of 36) = $\frac{\text{1}}{\text{6}}$.

A die has 6 equally likely outcomes.

The even outcomes are 2, 4 and 6, but only 2 is less than 4.

P(even number less than 4) = $\frac{\text{1}}{\text{6}}$.

The two dice give 6 × 6 = 36 equally likely ordered outcomes.

The favourable outcomes are (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6).

P(same number) = $\frac{\text{6}}{\text{36}}$ = $\frac{\text{1}}{\text{6}}$.

Two dice have 36 equally likely ordered outcomes.

There are 6 outcomes with equal numbers, one for each doublet.

Outcomes with different numbers = 36 − 6 = 30.

P(different numbers) = $\frac{\text{30}}{\text{36}}$ = $\frac{\text{5}}{\text{6}}$.

There are 36 equally likely ordered outcomes.

A sum more than 10 is either 11 or 12.

The favourable outcomes are (5,6), (6,5) and (6,6).

P(sum more than 10) = $\frac{\text{3}}{\text{36}}$ = $\frac{\text{1}}{\text{12}}$.

There are 36 equally likely ordered outcomes.

If neither die shows 6, each die can show one of 1, 2, 3, 4 and 5, giving 5 × 5 = 25 outcomes.

Outcomes with at least one 6 = 36 − 25 = 11.

P(at least one 6) = $\frac{\text{11}}{\text{36}}$.

There are 36 equally likely ordered outcomes.

If the first die is 4, the second has 5 choices other than 4.

If the second die is 4, the first has 5 choices other than 4.

Favourable outcomes = 5 + 5 = 10.

P(exactly one 4) = $\frac{\text{10}}{\text{36}}$ = $\frac{\text{5}}{\text{18}}$.

There are 36 equally likely ordered outcomes.

Each die must show one of 1, 2 or 3.

Favourable outcomes = 3 × 3 = 9.

P(both numbers less than 4) = $\frac{\text{9}}{\text{36}}$ = $\frac{\text{1}}{\text{4}}$.

There are 36 equally likely ordered outcomes.

For first-die values 1, 2, 3, 4 and 5, the allowed second-die values are respectively {1,2,3,4,5}, {1,2}, {1}, {1} and {1}.

There is no allowed second-die value when the first die shows 6.

Favourable outcomes = 5 + 2 + 1 + 1 + 1 = 10.

P(product less than 6) = $\frac{\text{10}}{\text{36}}$ = $\frac{\text{5}}{\text{18}}$.

There are 36 equally likely ordered outcomes.

The product is 12 for (2,6), (3,4), (4,3) and (6,2).

P(product 12) = $\frac{\text{4}}{\text{36}}$ = $\frac{\text{1}}{\text{9}}$.

There are 36 equally likely ordered outcomes.

A difference of 3 means the larger number minus the smaller number is 3.

The favourable outcomes are (1,4), (4,1), (2,5), (5,2), (3,6) and (6,3).

P(difference 3) = $\frac{\text{6}}{\text{36}}$ = $\frac{\text{1}}{\text{6}}$.

There are 36 equally likely ordered outcomes, with x on the first die and y on the second.

For x = 1, 2, 3, 4, 5, 6, the numbers of possible y values smaller than x are 0, 1, 2, 3, 4, 5.

Favourable outcomes = 0 + 1 + 2 + 3 + 4 + 5 = 15.

P(x > y) = $\frac{\text{15}}{\text{36}}$ = $\frac{\text{5}}{\text{12}}$.

There are 36 equally likely ordered outcomes.

The ordered pairs satisfying b = 2a are (1,2), (2,4) and (3,6).

P(b = 2a) = $\frac{\text{3}}{\text{36}}$ = $\frac{\text{1}}{\text{12}}$.

There are 36 equally likely ordered outcomes.

The sum is 8 for (2,6), (3,5), (4,4), (5,3) and (6,2): 5 outcomes.

P(sum 8) = $\frac{\text{5}}{\text{36}}$.

P(sum not 8) = 1 − $\frac{\text{5}}{\text{36}}$ = $\frac{\text{31}}{\text{36}}$.

There are 52 equally likely cards.

Hearts and diamonds are red suits. Each has 3 face cards: jack, queen and king.

Red face cards = 2 × 3 = 6.

P(red face card) = $\frac{\text{6}}{\text{52}}$ = $\frac{\text{3}}{\text{26}}$.

There are 6 red face cards: jack, queen and king in each of hearts and diamonds.

Cards that are not red face cards = 52 − 6 = 46.

P(not a red face card) = $\frac{\text{46}}{\text{52}}$ = $\frac{\text{23}}{\text{26}}$.

There are 4 aces and 2 black kings in the deck.

No ace is a king, so favourable cards = 4 + 2 = 6.

P(ace or black king) = $\frac{\text{6}}{\text{52}}$ = $\frac{\text{3}}{\text{26}}$.

There are 13 spades and 4 aces.

The ace of spades belongs to both groups and is counted only once.

Favourable cards = 13 + 4 − 1 = 16.

P(spade or ace) = $\frac{\text{16}}{\text{52}}$ = $\frac{\text{4}}{\text{13}}$.

There are 4 aces and 4 tens, with no overlap.

Favourable cards = 4 + 4 = 8.

P(ace or ten) = $\frac{\text{8}}{\text{52}}$ = $\frac{\text{2}}{\text{13}}$.

Removed cards = 4 queens + 4 jacks + 4 aces = 12.

Remaining cards = 52 − 12 = 40.

All 4 kings remain.

P(king) = $\frac{\text{4}}{\text{40}}$ = $\frac{\text{1}}{\text{10}}$.

After one card is lost, 51 cards remain.

The king of hearts is red, so it cannot be the lost black card.

Exactly 1 of the 51 remaining cards is the king of hearts.

P(king of hearts) = $\frac{\text{1}}{\text{51}}$.

Remaining cards = 52 − 2 − 4 = 46.

A full deck has 26 black cards; the 4 missing spades are black.

Remaining black cards = 26 − 4 = 22.

P(black card) = $\frac{\text{22}}{\text{46}}$ = $\frac{\text{11}}{\text{23}}$.

Remaining cards = 52 − 2 − 4 = 46.

No heart is missing, so all 13 hearts remain.

P(heart) = $\frac{\text{13}}{\text{46}}$.

Total cards = 40 − 7 + 1 = 34.

The multiples of 7 in this range are 7, 14, 21, 28 and 35: 5 numbers.

P(multiple of 7) = $\frac{\text{5}}{\text{34}}$.

Total cards = 30 − 10 + 1 = 21.

Multiples of 4 are 12, 16, 20, 24 and 28.

Multiples of 5 are 10, 15, 20, 25 and 30.

Counting 20 once gives 9 favourable numbers.

P(multiple of 4 or 5) = $\frac{\text{9}}{\text{21}}$ = $\frac{\text{3}}{\text{7}}$.

There are 7 equally likely choices.

The numbers whose squares are at most 1 are −1, 0 and 1.

P(square ≤ 1) = $\frac{\text{3}}{\text{7}}$.

The choices are the 30 numbers from 1 to 30.

The prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29: 10 numbers.

P(prime number) = $\frac{\text{10}}{\text{30}}$ = $\frac{\text{1}}{\text{3}}$.

There are 30 equally likely numbers.

The only even prime number is 2.

P(even prime number) = $\frac{\text{1}}{\text{30}}$.

Total cards = 50 − 6 + 1 = 45.

The perfect squares in this range are 9, 16, 25, 36 and 49.

P(perfect square) = $\frac{\text{5}}{\text{45}}$ = $\frac{\text{1}}{\text{9}}$.

There are 100 equally likely cards.

The perfect cubes from 1 to 100 are 1, 8, 27 and 64.

P(perfect cube) = $\frac{\text{4}}{\text{100}}$ = $\frac{\text{1}}{\text{25}}$.

Total T-shirts = 99 − 4 + 1 = 96, with one for each number.

Perfect squares are 4, 9, 16, 25, 36, 49, 64 and 81: 8 numbers.

Perfect cubes are 8, 27 and 64: 3 numbers.

The common number 64 is counted once, so favourable numbers = 8 + 3 − 1 = 10.

P(square or cube) = $\frac{\text{10}}{\text{96}}$ = $\frac{\text{5}}{\text{48}}$.

The word MOBILE contains 6 letters.

Its vowels are O, I and E: 3 letters.

P(vowel) = $\frac{\text{3}}{\text{6}}$ = $\frac{\text{1}}{\text{2}}$.

The word MATHEMATICS contains 11 letter positions.

The vowel positions are A, E, A and I: 4 positions.

P(vowel) = $\frac{\text{4}}{\text{11}}$.

There are 26 equally likely letters.

The 5 vowels are A, E, I, O and U, so consonants = 26 − 5 = 21.

P(consonant) = $\frac{\text{21}}{\text{26}}$.

MATHEMATICS has 11 letter positions, of which 4 are vowels.

P(vowel) = $\frac{\text{4}}{\text{11}}$, so $\frac{\text{2}}{\text{2x + 1}}$ = $\frac{\text{4}}{\text{11}}$.

Cross-multiplying gives 22 = 4(2x + 1).

Thus 22 = 8x + 4, so 8x = 18 and x = $\frac{\text{18}}{\text{8}}$ = $\frac{\text{9}}{\text{4}}$.

INDEPENDENCE has 12 letter positions.

Among its vowels, I occurs once and E occurs 4 times. Therefore, E is the most frequent vowel.

P(selecting E) = $\frac{\text{4}}{\text{12}}$ = $\frac{\text{1}}{\text{3}}$.

Total balls = 4 + 2 = 6.

There are 2 yellow balls.

P(yellow ball) = $\frac{\text{2}}{\text{6}}$ = $\frac{\text{1}}{\text{3}}$.

Total balls = 3 + 5 + 7 = 15.

A ball that is neither red nor black must be white, giving 5 favourable balls.

P(neither red nor black) = $\frac{\text{5}}{\text{15}}$ = $\frac{\text{1}}{\text{3}}$.

Total balls = 5 + 8 + 7 = 20.

A ball that is neither blue nor pink must be yellow.

P(yellow ball) = $\frac{\text{7}}{\text{20}}$.

White balls = 24 − 4 − 11 = 9.

There are 24 equally likely balls.

P(white ball) = $\frac{\text{9}}{\text{24}}$ = $\frac{\text{3}}{\text{8}}$.

Let the numbers of red and black balls be 3k and 7k, respectively, where k is positive.

Total balls = 3k + 7k = 10k.

P(black ball) = $\frac{\text{7k}}{\text{10k}}$ = $\frac{\text{7}}{\text{10}}$ = 0.7.

Let the numbers of ₹1 and ₹2 coins be 9k and 11k, respectively.

Total number of coins = 9k + 11k = 20k.

P(₹2 coin) = $\frac{\text{11k}}{\text{20k}}$ = $\frac{\text{11}}{\text{20}}$ = 0.55.

Total balls = n + 5.

P(green) = $\frac{\text{n}}{\text{n + 5}}$ and P(red) = $\frac{\text{5}}{\text{n + 5}}$.

The given condition gives $\frac{\text{n}}{\text{n + 5}}$ = 3 × $\frac{\text{5}}{\text{n + 5}}$.

Multiplying by n + 5 gives n = 3 × 5 = 15.

Let the number of black balls be x. Then the number of red balls is x + 10.

Total balls = x + x + 10 = 2x + 10.

P(red) = $\frac{\text{x + 10}}{\text{2x + 10}}$ = $\frac{\text{3}}{\text{5}}$.

Cross-multiplying gives 5x + 50 = 6x + 30, so x = 20.

Total balls = 2(20) + 10 = 50.

For each box, P(green) = number of green balls ÷ total balls.

For A, P(green) = $\frac{\text{4}}{\text{9}}$ = $\frac{\text{60}}{\text{135}}$.

For B, P(green) = $\frac{\text{7}}{\text{15}}$ = $\frac{\text{63}}{\text{135}}$.

For C, P(green) = $\frac{\text{8}}{\text{18}}$ = $\frac{\text{4}}{\text{9}}$ = $\frac{\text{60}}{\text{135}}$.

Since $\frac{\text{63}}{\text{135}}$ > $\frac{\text{60}}{\text{135}}$, box B gives the greatest probability.

Use the area model: assume the sapling’s position is selected uniformly at random within the circular park.

Area of park = πr² = π × 20² = 400π m².

Area of lawn = side² = 8² = 64 m².

Area outside lawn = 400π − 64 m².

P(outside lawn) = $\frac{\text{area outside lawn}}{\text{area of park}}$ = $\frac{\text{400π − 64}}{\text{400π}}$.

Assume the dot is placed uniformly at random over the square board.

The circle is inscribed in the square, so its diameter is 2a and its radius is a.

Area of board = (2a)² = 4a².

Shaded area = area of square − area of circle = 4a² − πa² = (4 − π)a².

P(shaded region) = $\frac{\text{shaded area}}{\text{area of square}}$ = $\frac{\text{(4 − π)a²}}{\text{4a²}}$ = $\frac{\text{4 − π}}{\text{4}}$.