Probability Assertion–Reason PYQs Class 10 CBSE (With Answers)

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
A standard die has outcomes 1, 2, 3, 4, 5, 6.
There is no outcome 8.
P(getting 8) = $\frac{\text{0}}{\text{6}}$ = 0.

Reason: True.
An impossible event has no favourable outcome, so its probability is 0.

Link: Getting 8 on a standard die is impossible. R gives exactly the principle used in A, so R explains A. Answer: A.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
For a fair die, the six outcomes 1, 2, 3, 4, 5, 6 are equally likely.
The favourable outcomes are 2, 4, 6.
P(even number) = $\frac{\text{3}}{\text{6}}$ = $\frac{\text{1}}{\text{2}}$.

Reason: True.
The die has six possible outcomes, of which three are even.

Link: R identifies both counts needed for A. R explains A. Answer: A.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: False.
For a fair die, the prime outcomes are 2, 3, 5.
P(prime number) = $\frac{\text{3}}{\text{6}}$ = $\frac{\text{1}}{\text{2}}$, not $\frac{\text{2}}{\text{3}}$.
The number 1 is not prime.

Reason: True.
Among 1, 2, 3, 4, 5, 6, exactly 2, 3 and 5 are prime.

Link: R correctly identifies the favourable outcomes, but their probability disproves A. A is false and R is true. Answer: D.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
The prime outcomes are 2, 3, 5.
P(prime number) = $\frac{\text{3}}{\text{6}}$ = $\frac{\text{1}}{\text{2}}$.

Reason: True.
Here factors means positive factors. A prime number has exactly two: 1 and the number itself.

Link: R identifies which die outcomes are prime. Counting those outcomes gives A, so R explains A. Answer: A.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
Using the usual model of two fair dice, there are 6 × 6 = 36 equally likely ordered outcomes.
The doublets are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
P(doublet) = $\frac{\text{6}}{\text{36}}$ = $\frac{\text{1}}{\text{6}}$.

Reason: True.
Every probability lies between 0 and 1, including the endpoints.

Link: R gives the possible range of a probability. It does not count the six doublets among 36 outcomes, so it does not explain A. Answer: B.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
For the usual fair-coin model, the equally likely outcomes are HH, HT, TH and TT.
No tail means HH only.
P(no tail) = $\frac{\text{1}}{\text{4}}$.

Reason: True.
A probability is at least 0 and at most 1.

Link: R states a range. It does not identify the four outcomes or the single favourable outcome, so it does not explain A. Answer: B.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: False.
For two fair coins, the equally likely outcomes are HH, HT, TH and TT.
The category “one head and one tail” contains two outcomes: HT and TH.
The three categories printed in A are therefore not equally likely.
P(two heads) = $\frac{\text{1}}{\text{4}}$, not $\frac{\text{1}}{\text{3}}$.

Reason: True.
Equally likely outcomes have equal probabilities.

Link: R is correct, but A wrongly treats its three categories as equally likely. A is false and R is true. Answer: D.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
There are 100 equally likely numbered cards.
The even numbers are 2, 4, 6, …, 100, giving 50 favourable cards.
P(even number) = $\frac{\text{50}}{\text{100}}$ = $\frac{\text{1}}{\text{2}}$.

Reason: True.
For equally likely outcomes, divide the favourable count by the total count.

Link: R gives the probability formula used with 50 favourable cards out of 100. R explains A. Answer: A.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: False.
The multiples of 4 from 1 to 15 are 4, 8 and 12.
P(multiple of 4) = $\frac{\text{3}}{\text{15}}$ = $\frac{\text{1}}{\text{5}}$, not $\frac{\text{1}}{\text{3}}$.

Reason: True under the usual fair-coin model.
The equally likely outcomes are HH, HT, TH and TT.
At least one head occurs in HH, HT and TH.
P(at least one head) = $\frac{\text{3}}{\text{4}}$.

Link: A gives the wrong probability for the number selection. R gives the correct probability for a separate coin experiment. A is false and R is true. Answer: D.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
Face cards are the jack, queen and king in each suit.
There are two red suits, hearts and diamonds.
Red face cards = 2 × 3 = 6.
P(red face card) = $\frac{\text{6}}{\text{52}}$ = $\frac{\text{3}}{\text{26}}$.

Reason: False.
There are 4 × 3 = 12 face cards, not 16.
Aces are not face cards.

Link: A has the correct probability, while R gives an incorrect count. A is true and R is false. Answer: C.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True in the intended completed-match model, where one of the two players wins.
Take P(Sania wins) = 0.79 as given.
P(Ashnam wins) = 1 − P(Sania wins).
= 1 − 0.79 = 0.21.

Reason: True.
Complementary events exhaust all possibilities and cannot happen together.
Their probabilities add to 1.

Link: Winning by Ashnam is the complement of winning by Sania in this model. R explains the calculation in A. Answer: A.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: False.
E₁ contains 1 and 2.
E₂ contains 4, 5 and 6.
The outcome 3 belongs to neither event, so the events do not cover all outcomes.
For a fair die, P(E₁) + P(E₂) = $\frac{\text{2}}{\text{6}}$ + $\frac{\text{3}}{\text{6}}$ = $\frac{\text{5}}{\text{6}}$, not 1.

Reason: True.
An event and its complement cover all outcomes without overlap, so their probabilities add to 1.

Link: R states a correct property, but E₁ and E₂ in A are not complementary. A is false and R is true. Answer: D.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: False.
P(not E) = 1 − P(E).
= 1 − $\frac{\text{1}}{\text{999}}$
= $\frac{\text{998}}{\text{999}}$, which is approximately 0.998999, not 0.001.

Reason: True.
The probabilities of an event and its complement add to 1.

Link: R is the correct identity, and applying it disproves A. A is false and R is true. Answer: D.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True using the observed relative frequency intended in the question.
Let E be hitting a boundary.
The observed estimate P(E) = $\frac{\text{9}}{\text{45}}$ = $\frac{\text{1}}{\text{5}}$.
P(not E) = 1 − P(E).
= 1 − $\frac{\text{1}}{\text{5}}$ = $\frac{\text{4}}{\text{5}}$.
This is the estimate from the given record, not a guarantee for future balls.

Reason: True.
An event and its complement have probabilities adding to 1.

Link: R gives the complement rule used to obtain the non-boundary estimate in A. R explains A. Answer: A.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
The event is selecting any number from the stated set, not selecting one particular number.
Every one of the 20 outcomes satisfies the event.
P(selecting a number from 1 to 20) = $\frac{\text{20}}{\text{20}}$ = 1.

Reason: True.
A sure event occurs for every possible outcome and has probability 1.

Link: The selection must produce a number from the stated set, so it is a sure event. R identifies the certainty used in A and is the explanation accepted by the official key. Answer: A.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
0.9999 is very close to 1.
The event is therefore very likely, although it is not certain.

Reason: True.
A sure event has probability 1.

Link: R states the probability of certainty. It does not state the interpretation of probabilities close to 1, so it does not explain A. Answer: B.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
$\frac{\text{1}}{\text{0.9}}$ = $\frac{\text{10}}{\text{9}}$ > 1.
A probability cannot exceed 1.

Reason: True.
Every probability lies in the interval from 0 to 1, including both endpoints.

Link: R rules out the value in A because it exceeds 1. R explains A. Answer: A.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
A probability cannot exceed 1.
0.2p ≤ 1.
p ≤ $\frac{\text{1}}{\text{0.2}}$ = 5.

Reason: True.
The probability of the complementary event is 1 minus the probability of the event.

Link: Both statements are true. The direct justification for the bound is P(E) ≤ 1; R states the complementary-event identity. The official marking scheme treats R as not explaining A, so the answer is B. Answer: B.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True under the standard model in which the starting weekday is equally likely.
366 = 52 × 7 + 2.
The two extra days can be Mon–Tue, Tue–Wed, Wed–Thu, Thu–Fri, Fri–Sat, Sat–Sun or Sun–Mon.
Two of these seven pairs contain Sunday.
P(53 Sundays in a leap year) = $\frac{\text{2}}{\text{7}}$.

Reason: True under the same model.
365 = 52 × 7 + 1.
A non-leap year has 53 Sundays only when its extra day is Sunday.
P(53 Sundays in a non-leap year) = $\frac{\text{1}}{\text{7}}$.

Link: R concerns a non-leap year with one extra day. A requires the separate two-extra-day count for a leap year. Both are true, but R does not explain A. Answer: B.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True under the standard model of equally likely starting weekdays.
A leap year has 52 full weeks and two extra consecutive days.
Of the seven possible extra-day pairs, Sat–Sun and Sun–Mon contain Sunday.
P(53 Sundays) = $\frac{\text{2}}{\text{7}}$.

Reason: False.
A non-leap year has 52 full weeks and one extra day.
Only Sunday among the seven possible extra weekdays gives a 53rd Sunday.
The probability is $\frac{\text{1}}{\text{7}}$, not $\frac{\text{5}}{\text{7}}$.

Link: A has the correct leap-year probability. R gives the wrong non-leap-year probability. A is true and R is false. Answer: C.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True under the standard model of equally likely starting weekdays.
366 = 52 × 7 + 2.
The extra pairs Sun–Mon and Mon–Tue contain Monday.
P(53 Mondays in a leap year) = $\frac{\text{2}}{\text{7}}$.

Reason: False.
A non-leap year has 52 full weeks and one extra day.
There is a 53rd Monday only if that extra day is Monday.
The probability is $\frac{\text{1}}{\text{7}}$, not $\frac{\text{5}}{\text{7}}$.

Link: A gives the correct leap-year count, whereas R gives an incorrect non-leap-year probability. A is true and R is false. Answer: C.