Question

While trick or treating, each child gets to randomly pick two candy items from a paper bag. The paper bag is filled 15% full of peanut M&Ms and 85% full of candies without peanuts. Create a probability distribution for a child that is allergic to peanuts. Let the random variable, x, represent the number of candies picked that do not contain peanuts (hint: first consider sample space)

Answer 1

Given that,

While trick or treating, each child gets to randomly pick two candy items from a paper bag. The paper bag is filled 15% full of peanut M&Ms and 85% full of candies without peanuts.

Let X be the random variable that the number of candies picked that do not contain peanuts.

Here X can take values 0,1 and 2.

That means here X follows Binomial distribution with n = 2 and p = 85% = 0.85

The pmf of X=x is,

P(X = x) = (2 C x) * 0.85^x * (1 - 0.85)^(2-x)

P(X = 0) = (2 C 0 ) * 0.85⁰ * (1 - 0.85)^(2-0) = 0.0225

P(X = 1) = (2 C 1 ) * 0.85¹ * (1 - 0.85)^(2-1) = 0.2550

P(X = 2) = (2 C 2) * 0.85² * (1 - 0.85)^(2-2) = 0.7225

Now we have to find probability for x = 0,1 and 2

The probability distribution of X=x is,

x	p
0	0.0225
1	0.2550
2	0.7225
total	1