SOLUTION: 1. Randomly pick out 40 mixed-color M&M’s from a small bag of M&M’s. Record the number of each color below. Yellow- 5; Green- 10; Blue - 8; Brown - 4; Orange - 6; Red -

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Question 627606: 1. Randomly pick out 40 mixed-color M&M’s from a small bag of M&M’s. Record the number of each color below. Yellow- 5; Green- 10; Blue - 8; Brown - 4; Orange - 6; Red - 7;

2. Construct two tree diagrams (1 for with replacement and 1 for without replacement) showing the drawing of two M&Ms, one at a time, from the 40 M&Ms above to complete the theoretical probability questions below.
With Replacement Without Replacement
P(BL1 and BL2):
P(BL1 and BR2 or BR1 and BL2):
P(BL1 and O2 ):
P(O2 |BL1):
P(no yellows on either draw):
P(doubles):
P(no doubles):
Note: O2 = orange on second pick; BL1 = blue on first pick; BL2 = blue on second pick; doubles = both picks are the same color. BR1= brown on first pick; BR2 = brown on second pick.
1. Why are the “With Replacement” and “Without Replacement” probabilities different?
2. Explain the differences in what P(BL1 and O2) and P(O2 | BL1) represent.



Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 

Hi, Re TY, unable to do tree diagemas on this site.  Perhaps You can find

an example in your materials to follow. 

 40 M&Ms: 5 Yellow, 10 Green, 8 Blue, 4 Brown, 6 Orange and 7 Red 

P(BL1 and BL2):  or w/o replacement  

P(BL1 and BR2 or BR1 and BL2): P(BL1 and BR2) + P( BR1 and BL2): as done above

P(BL1 and O2 ):  or w/o replacement 

P(O2 |BL1): 

P(no yellows on either draw): 1 - P(both Yellow)

                                  P(both Yellow) or w/o replacement  

P(doubles): Adding All 2 of a kind (with or without replacement

P(no doubles):  1 - P(doubles) 

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