In a carnival game, there are six identical boxes, one of which contains a prize. A contestant wins the prize by selecting the box containing it. Before each game, the old prize is removed and another prize is placed at random in one of the six boxes. Is it appropriate to use the binomial probability distribution to find the probability that a contestant who plays the game five times wins exactly twice? A) Yes. The five trials are independent, have only two outcomes, and have the same P(success); n = 5, r = 2, p = 1/5.
B) Yes. The five trials are independent, have only two outcomes, and have the same P(success); n = 2, r = 5, p = 1/6.
C) No. The five trials are independent, but have more than two outcomes.
D) Yes. The five trials are independent, have only two outcomes, and have the same P(success); n = 5, r = 2, p = 1/6.

D) Yes. The five trials are independent, have only two outcomes, and have the same P(success); n = 5, r = 2, p = 1/6.

Step-by-step explanation:

The Number of boxes = 6

Box containing a prize = 1

Probability of success, p = box containing a price / number of boxes = 1 /6

Number of trials = 5

Probability of success on exactly 2 trials, r = 2

Hence,

P(r = 2) = nCr * p^r * (1-p)^(n-r)

n = 5 ; r = 2 ; p = 1/6

Using a binomial probability calculator :

P(r = 2) = 0.1608

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!