hemanthp wrote:

When a random experiment is conducted, the probability that event A occurs is \(\frac{1}{3}\). If the random experiment is conducted 5 independent times, what is the probability that event A occurs exactly twice?

5/243

25/243

64/243

80/243

16/17

took more time than I hoped for in the tests (

Kaplan). Don't forget to KUDOS if you like the question.

Tx.

The probability that event A occurs is \(\frac{1}{3}\);

The probability that event A will not occur is \(1-\frac{1}{3}=\frac{2}{3}\).

We want to calculate the probability of event YYNNN (Y-occurs, N-does not occur).

\(P(YYNNN)=\frac{5!}{2!3!}*(\frac{1}{3})^2*(\frac{2}{3})^3=\frac{80}{243}\), we are multiplying by \(\frac{5!}{2!3!}\) as event YYNNN can happen in # of times: YYNNN, YNYNN, NNNYY, ... basically the # of permutations of 5 letters YYNNN out of which 2 Y's and 3 N's are identical (\(\frac{5!}{2!3!}\)).

Answer: D.

Hope it's clear.