Question

I’m stuck w/ problem B. Please help!

Answer 1

We know that the probability of picking a nervous person is 0.2. This implies that the probability of picking someone who is not nervous is 0.8.

When we pick a sample of two people, we multiply the probabilities of each single person. So, the possible outcomes and their respective probabilities are:

• Both people not nervous: $0.8\cdot 0.8 = 0.64$
• First person nervous, second person not nervous: $0.2\cdot 0.8 = 0.16$
• First person not nervous, second person nervous: $0.8\cdot 0.2 = 0.16$
• Both people nervous: $0.2\cdot 0.2 = 0.04$

Out of these four scenarios, the last three correspond to the description "at least one of the two is nervous". So, we have to sum their probability:

$0.16+0.16+0.04=0.36$

Alternatively, and probably more easily, we could solve this problem by complementary probability: we know that if event $E$ has probability $p$, then the negation of $E$ has probability $1-p$.

So, the negation of "at least one person is nervous" is "none of the two is nervous". We know that the probability of "no nervous people" is 0.64 (see above), so the probability of its negation (at least one nervous) is

$1-0.64=0.36$