Question

Consider the following binomial experiment. The probability that a green jelly bean is chosen at random from a large package of jelly beans is 2/5. If Sally chooses 13 jelly beans, what is the probability that at most two will be green jelly beans?
a) 0.0349
b) 0.3100
c) 0.0579
d) 0.9421
e) 0.5000

Answer 1

Answer:

There is a 5.79% probability that at most 2 jelly beans are chosen.

Step-by-step explanation:

For each jelly bean chosen, there are only two possible outcomes. Either it is green, or it is not. This is why we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

There are 13 jelly beans, so $n = 13$.

The probability that a green jelly bean is chosen at random from a large package of jelly beans is 2/5. This means that $p = 0.4$.

If Sally chooses 13 jelly beans, what is the probability that at most two will be green jelly beans?

This is

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$.

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{13,0}.(0.4)^{0}.(0.6)^{13} = 0.0013$

$P(X = 1) = C_{13,1}.(0.4)^{1}.(0.6)^{12} = 0.0113$

$P(X = 1) = C_{13,2}.(0.4)^{2}.(0.6)^{111} = 0.0453$

So

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0013 + 0.0113 + 0.0453 = 0.0579$.

There is a 5.79% probability that at most 2 jelly beans are chosen.