Question

a candy machine holds 500 pieces of candy, 20%, percent of which are blue. customers get an srs of 4 pieces of candy per purchase from this machine. let b, equals the number of blue candies a customer gets in a purchase from a full machine. which of the following would find p(b=1)?

Answer 1

Answer:

p(b = 1) = 0.4096 = 40.96%

Step-by-step explanation:

For each candy, there are only two possible outcomes. Either it is blue, or it is not. The probability of a candy being blue is independent of any other candy. This means that the binomial probability distribution is used to solve this question.

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 combinations 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.

20% are blue

This means that $p = 0.2$

Sample of 4

This means that $n = 4$.

Which of the following would find p(b=1)?

P(X = 1). So

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

$P(X = 1) = C_{4,1}.(0.2)^{1}.(0.8)^{3} = 0.4096$

So, p(b = 1) = 0.4096 = 40.96%