Can you assist with working this out

Question

A bag contains 24 pieces of candy, of which 12 are peppermints and 12 are butterscotch. Let X be the number of peppermints in a sample of 5 pieces of candy selected at random and without replacement from the bag. Find the following:

1. Pr(X=2)

2. Pr(X≤2)

3.E(X),σ= VarX

Answer 1

You will need to use the binomial probability formula:

[n! / x!*(n-x)!]*p^x*q^n-x , where

n = the number of pieces of candy

x = the number of peppermints

p = the success probability (picking peppermints) = 12/24 = 1/2

q = the failure probability (picking butterscotch) = 12/24 = 1/2

P(X = 2) = (5 choose 2) = (5! / [2!*(5-2)!])*(1/2)²*(1/2)³ = 0.3125

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= (5! / [0!*(5-0)!])*(1/2)⁰*(1/2)⁵ + (5! / [1!*(5-1)!])*(1/2)¹*(1/2)⁴ + (5! / [2!*(5-2)!])*(1/2)²*(1/2)³

= 0.03125 + 0.15625 + 0.3125 = 0.5

The probability of picking 2 peppermints in the set of 5 pieces of candy is 0.3125. The probability of picking up to 2 peppermints in the set of 5 pieces of candy is 0.5.

E[X] = np = 5*(1/2) = 2.5

Var[X] = npq = 5*(1/2)*(1/2) = 1.25

σ = sqrt(1.25) = 1.118 ≈ 1.12