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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-> 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.
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)