Question

The guidance department has reported that of the senior class, 2.3% are members of key club, K, 8.6% are enrolled in AP physics, P, and 1.9% are enrolled in both. Determine the probability of P given K, to the nearest tenth of a percent.

Answer 1

P(P|K) = 82.6%.

P(P|K) = P(K and P)/P(K) = 1.9%/2.3% = 0.019/0.023 = 0.8261 = 82.6%

Answer 2

The question is related to conditional probability .

It is given in the question that, the guidance department has reported that of the senior class, 2.3% are members of key club, K, 8.6% are enrolled in AP physics, P, and 1.9% are enrolled in both.

So we have ,

$P(K) = 2.3 % = 0.023 \\ P(p) = 8.6% = 0.086 \\ P(K \cap p) = 0.019$

We have to use the following formula

$P(p/K) = \frac{ P(p \cap K) }{P(K)}$

Substituting the values, we will get

$P(p/K) = \frac{0.019}{0.023} = 0.826= 82.6%$