If every student in a large Statistics class selects peanut M&M’s at random until they get a red candy, on average how many M&M’s will the students need to select? (Round your answer to two decimal places.)

Question:

According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. You randomly select peanut M&M's from an extra-large bag looking for a red candy.

If every student in a large Statistics class selects peanut M&M’s at random until they get a red candy, on average how many M&M’s will the students need to select? (Round your answer to two decimal places.).

If every student in the large Statistics class selects peanut M&M’s at random until they get a red candy the expected value of the number of  M&M’s the students need to select is 8.33 M&M's.

Step-by-step explanation:

To solve the question, we note that the statistical data presents a geometric mean. That is the probability of success is of the form.

The amount of repeated Bernoulli trials required before n eventual success outcome or

The probability of having a given number of failures before the first success is recorded.

In geometric distribution, the probability of having an eventual successful outcome depends on the the completion of a certain number of attempts with each having the same probability of success.

If the probability of each of the preceding trials is p and the kth trial is the  first successful trial, then the probability of having k is given by

Pr(X=k) =  

The number of expected independent trials to arrive at the first success for a variable Xis 1/p where p is the expected success of each trial hence p is the probability for the red and the expected value of the number of trials is 1/p or where p = 12 % which is 0.12

1/p = 1/0.12 or 25/3 or 8.33.

14.3

step-by-step explanation:

add all the numbers together.

9 + 12 + 15 + 16 + 17 + 17 = 86

now, divide 86 by how many numbers we have: 6.

86 / 6 = 14.3