1 Answer

Tinesha · Answer 1 · 2024-09-25T20:03:26+0000

The probability of the first toss being a prime and the second toss resulting in a 5 is
((1)/(12)) .

The requested probability is
$(P(A \cap B))$ , where
$(A = {\text{first toss is a prime}})$ and
$(B = {\text{second toss is a 5}})$ .

To find the probability of the intersection of two events, we can use the formula:

$[P(A \cap B) = P(A) * P(B)]$

The probability of the first toss being a prime is
((3)/(6)) since there are 3 prime numbers (2, 3, 5) on a standard die, and the die has 6 faces.

The probability of the second toss resulting in a 5 is
((1)/(6)) since there is 1 face with a 5 out of 6 faces.

Therefore, the probability of both events occurring is:

$[P(A \cap B)$ =
(3)/(6) * (1)/(6) = (1)/(12)]

So, the probability of the first toss being a prime and the second toss resulting in a 5 is
((1)/(12)) .

Complete question:

Suppose a die is rolled twice and let

A = {first toss is a prime} C = {second toss is a 5}

Find the requested probability. (Enter the probability as a fraction.) P(A n B)

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