Question

A candy company claims that 20% of the candies in its bags are colored green. Steve buys 30 bags of 30 candies, randomly selects one candy from each, and counts the number of green candies. If there are 5, 6, or 7 green candies, Steve will conclude that the company’s claim is correct. What is the probability of Steve agreeing with the company’s claim? Use a TI-83, TI-83 plus, or TI-84 calculator to find the probability.

Answer 1

Answer:

The probability of Steve agreeing with the company’s claim is 0.50502.

Step-by-step explanation:

Let X denote the number of green candies.

The probability of green candies is, p = 0.20.

Steve buys 30 bags of 30 candies, randomly selects one candy from each, and counts the number of green candies.

So, n = 30 candies are randomly selected.

All the candies are independent of each other.

The random variable X follows a binomial distribution with parameter n = 30 and p = 0.20.

It is provided that if there are 5, 6, or 7 green candies, Steve will conclude that the company’s claim is correct.

Compute the probability of 5, 6 and 7 green candies as follows:

$P(X=5)={30\choose 5}(0.20)^{5}(1-0.20)^{30-5}=0.17228\\\\P(X=6)={30\choose 6}(0.20)^{6}(1-0.20)^{30-6}=0.17946\\\\P(X=7)={30\choose 7}(0.20)^{7}(1-0.20)^{30-7}=0.15328$

Then the probability of Steve agreeing with the company’s claim is:

P (Accepting the claim) = P (X = 5) + P (X = 6) + P (X = 7)

                                       = 0.17228 + 0.17946 + 0.15328

                                       = 0.50502

Thus, the probability of Steve agreeing with the company’s claim is 0.50502.