Probability — Assertion-Reason Type Questions

Assertion (A): The probability that a leap year has 53 Sundays is $\dfrac{2}{7}$.

Reason (R): The probability that a non-leap year has 53 Sundays is $\dfrac{5}{7}$.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

In a leap year, there are 366 days.

366 days = 52 weeks + 2 days

These 2 days can be (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), and (Sun, Mon).

Total number of possible outcomes = 7

Number of favourable outcomes (Getting Sunday as one of the extra days) = 2 (i.e., (Sat, Sun), (Sun, Mon)).

P(Getting Sunday as one of the extra days) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{7}$

∴ Assertion (A) is true.

In a non - leap year, there are 365 days.

365 days = 52 weeks + 1 days

These 1 days can be Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.

Total number of possible outcomes = 7

Number of favourable outcomes (Getting Sunday as one of the extra days) = 1

P(Getting Sunday as one of the extra days) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{7}$

∴ Reason (R) is false.

∴ Assertion (A) is true, but Reason (R) is false.

Hence, option 1 is the correct option.

Assertion (A): Two players Sania and Ashma play a tennis match. If the probability of Sania winning the match is 0.79, then the probability of Ashma winning the match is 0.21.

Reason (R): The sum of probabilities of two complementary events is 1.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given, the probability of Sania winning the match = 0.79.

As we know that the sum of probabilities of two complementary event is 1.

⇒ P(Sania wins) + P(Ashma wins) = 1

∴ Reason (R) is true.

⇒ 0.79 + P(Ashma wins) = 1

⇒ P(Ashma wins) = 1 - P(Sania wins)

⇒ P(Ashma wins) = 1 - 0.79

⇒ P(Ashma wins) = 0.21

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Assertion (A): If the probability of occurrence of an event E is $\dfrac{5}{11}$, then the probability of non-occurrence of the event E is $\dfrac{7}{11}$.

Reason (R): If E is an event, then P(E) + P($\overline{E}$) = 1.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given, the probability of occurrence of an event E = $\dfrac{5}{11}$

As we know that the sum of probabilities of two complementary event is 1.

⇒ P(E) + P($\overline{E}$) = 1

∴ Reason (R) is true.

$\Rightarrow \dfrac{5}{11} + P(\overline{E}) = 1\\[1em] \Rightarrow P(\overline{E}) = 1 - \dfrac{5}{11}\\[1em] \Rightarrow P(\overline{E}) = \dfrac{11 - 5}{11}\\[1em] \Rightarrow P(\overline{E}) = \dfrac{6}{11}$

∴ Assertion (A) is false.

∴ Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Assertion (A): In a throw of two fair points once, the probability of getting one head is $\dfrac{1}{2}$.

Reason (R): In a throw of two fair coin, the sample space is HH, HT, TH, TT.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

When two coins are tossed together, the total number of possible outcomes = 4 (i.e. HH, HT, TH and TT)

∴ Reason (R) is true.

Number of favourable outcomes (Getting one head) = 2 (HT and TH)

P(getting one head) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{4} = \dfrac{1}{2}$

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Assertion (A): The probability of getting a prime number, when a die is thrown once, is $\dfrac{2}{3}$.

Reason (R): On the faces of a die, prime numbers are 2, 3, 5.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

On the faces of a die, numbers are 1, 2, 3, 4, 5 and 6. Out of which 2, 3 and 5 are only prime numbers.

∴ Reason (R) is true.

Number of favourable outcomes (of getting prime number) = 3

Number of possible outcomes = 6

P(getting a prime number) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}$

∴ Assertion (A) is false.

∴ Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Assertion (A): A die is thrown once and the probability of getting an even number is $\dfrac{2}{3}$.

Reason (R): The sample space for even number on a die is {2, 4, 6}.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

On the faces of a die, numbers are 1, 2, 3, 4, 5 and 6. Out of which 2, 4 and 6 are even numbers.

So, the sample space for even number on a die = {2, 4, 6}

∴ Reason (R) is true.

Number of favourable outcomes = 3 (for getting even numbers)

The total number of possible outcomes = 6

P(getting a even number) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}$.

∴ Assertion (A) is false.

∴ Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.