In a bag of Thanksgiving M&M;\'s, there are 100 total M&M;\'s. Of those 100 M&M;
ID: 3362723 • Letter: I
Question
In a bag of Thanksgiving M&M;'s, there are 100 total M&M;'s. Of those 100 M&M;'s, 46 are Red M&M;'s, 31 are Orange M&M;'s, and 23 are Brown M&M;'s. 1. If randomly selecting 1 M&M;, what is the probability that the M&M; is either Red or Brown? 2. If randomly selecting 2 M&M;'s, with replacement, what is the probability that both M&M;'s selected are Orange? 3. If randomly selecting 3 M&M;'s, with replacement, what is the probability that at least one M&M;'s selected will be Red?Explanation / Answer
2.
Total number of M&Ms =100
Number of Red M&Ms = 46
Number of Orange M&Ms = 31
Number of Brown M&Ms = 23
2 M&Ms with replacement , Probability that both the M&M's selected are Orange:
Probability of first selected M&M is orange = Number of Orange M&Ms/Total number of M&Ms = 31/100 = 0.31
With replacement,i.e. the selected one is put back into the bag, therefore all the M&Ms are intact.
Probability of second selected M&M is Orange = 31/100 =0.31
Probability that both the M&M's selected are Orange = Probability of first selected M&M is orange x Probability of second selected M&M is Orange = 0.31 x 0.31 = 0.0961
3. If randomly selected 3 M&M with replacement, Probability that atleast one M&M's sellected will be Red = 1- Probability that None of the M&M's selected will be Red(Either Orange or Brown) .
Proabability of first selected M&M is not red (Orange or Brown) = Number of non-Red(Orange,Brown) M&Ms / Total number of M&Ms = (31+23)/ 100 = 54/100 = 0.54
Proabability of second selected M&M is not red (Orange or Brown) = 0.54
Proabability of third selected M&M is not red (Orange or Brown) = 0.54
Probability that None of the M&M's selected will be Red(Either Orange or Brown) (when three M&M are selected with replacement = 0.54x0.54x0.54= 0.157464
If randomly selected 3 M&M with replacement, Probability that atleast one M&M's sellected will be Red = 1- Probability that None of the M&M's selected will be Red(Either Orange or Brown) = 1-0.157464=0.842536
If randomly selected 3 M&M with replacement, Probability that atleast one M&M's sellected will be Red = 0.842536