Visual SolutionPYQ 2024 · Jan Shift 2Standard

Question

Two dice are thrown simultaneously. The probability that the sum is a prime number, given that the sum is odd, is:

(A)

\frac{5}{11}

(B)

\frac{5}{18}

(C)

\frac{7}{9}

(D)

\frac{7}{18}

Solution Path

Odd sums:

18/36

. Odd prime sums

(3,5,7,11)

14

outcomes.

P(\text{prime}|\text{odd}) = 14/18 = 7/9

01Question Setup

1/4

Two dice are thrown. Find the probability that the sum is prime, given that the sum is odd.

P(\text{prime} \mid \text{odd}) = \,?

02Odd Sums

2/4

An odd sum requires one die even and one odd. There are 18 such outcomes out of 36.

P(\text{odd}) = \frac{18}{36} = \frac{1}{2}

03Prime and Odd SumsKEY INSIGHT

3/4

Odd prime sums:

3, 5, 7, 11

. Count outcomes:

2 + 4 + 6 + 2 = 14

outcomes that are both prime and odd.

P(\text{prime} \mid \text{odd}) = \frac{14}{18} = \frac{7}{9}

04Final Answer

4/4

By conditional probability:

P(\text{prime} \mid \text{odd}) = P(\text{prime} \cap \text{odd}) / P(\text{odd}) = (14/36) / (18/36) = 14/18

\boxed{\dfrac{7}{9}}

Concepts from this question1 concepts unlocked

★

Conditional Probability

EASY

P(B|A) = P(A ∩ B) / P(A). The probability of B given that A has already occurred.

P(B \mid A) = \frac{P(A \cap B)}{P(A)}

Without-replacement problems are conditional by nature -the sample space changes after each draw. Misidentifying the reduced sample space is the #1 error.

Without replacementBayes theoremIndependent events test

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