You have a bag of M&Ms.; Your friend has a Snickers bar. After opening your bag

Question

You have a bag of M&Ms.; Your friend has a Snickers bar. After opening your bag of M&Ms; and inspecting its contents you discover that the bag contains only 7 red M&Ms; and 5 blue M&Ms.; You are hungry so you propose the following game to your friend. He will draw, without replacement, 3 M&Ms; from your bag. If he selects 2 or more blue M&Ms; he wins and you will give him the entire bag. Otherwise, you win and he will give you his Snickers bar. Define the experiment for this game and the pertinent random variable (X). Fully specify the PMF for the RV defined in question 1. Fully specify the CDF for the RV defined in question 1. Calculate the expected value, variance, and standard deviation of the RV defined in question 1. What is the probability that you win your friend's Snickers bar?

Explanation / Answer

1)Experiment:To draw more than 2 Blue M&M in a draw of 3 without replacement

Hence the random variable is the Blue M&M

2)PMF contains events with cases Blue M&M draws =0,1,2,3

Hence PMF = prob(red=3 and blue=0)=7*6*5/(12*11*10) =7/44    for X=0

                  =prob(red=2 and blue=1)=3*7*6*5/(12*11*10)=21/44     for X=1

                  =prob(red=1 and blue=2)=3*7*5*4/(12*11*10)=7/22 for X=2

                 =prob(red=0 and blue=3)=5*4*3/(12*11*10)=1/22 for X=3

3)CDF for the RV CDF(X)=sum of all PMF's <=X

CDF= 7/44    for X=0

      =7/11     for X<=1

      =21/22    for X<=2

      =1          for X<=3

4)Expected Value =0*7/44 +1*21/44+2*7/22+3*1/22 =1.25

Variance=(((7/44)^2)*(0-1.25)^2 +((21/44)^2)*(1-1.25)^2 +((7/22)^2)*(2-1.25)^2 +((1/22)^2)*(3-1.25)^2)/(4-1) =0.039019

Hence std dev=0.1975

5)Prob that frnd wins is P(X>=2) =P(X=2)+P(X=3)=7/22 +1/22 =8/22 =4/11=0.3636

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