The following is the number of each color of M&M found in a sample bag of 280 M&
ID: 2934509 • Letter: T
Question
The following is the number of each color of M&M found in a sample bag of 280 M&M’s.
Red-39
Orange-58
Yellow-55
Green-38
Blue-27
Brown-63
Suppose you reach into the bag and randomly select one M&M. Calculate the following
probabilities. Round your answers to 4 decimals.
1.
P (Red) =
2.
P (Yellow) =
3.
P (Blue) =
Suppose you reach into the sample bag and randomly select THREE M&M’s.
Calculate the following probabilities (with and without replacement).
Show your calculations and round your final answers to 4 decimals.
4.
The probability that the first M&M is Red, the second M&M is Yellow, and the third M&M is Blue.
(with replacement)
5.
The probability that the first M&M is Red, the second M&M is Yellow, and the third M&M is Blue.
(without replacement)
6.
The probability that all three M&M’s are Blue.
(with replacement)
7.
The probability that all three M&M’s are Blue.
(without replacement)
8.
Would it be unusual for all three M&M’s to be blue if the sampling is done without replacement?
Justify your answer using a complete sentence and proper grammar.
Explanation / Answer
(1) P(Red) = 39/280 = 0.1393
(2) P(Yellow) = 55/280 = 0.1964
(3) P(Blue) = 27/280 = 0.0964
(4) With Replacement:
P(RYB) = (39/280) X (55/280) X (27/280)
REASON: Since Replacement, Total remains as 280 throughout)
= 0.1393 X 0.1964 X 0.0964 = 0.002637
(5) Without Replacement - 280 becomes 279 & 278 after each selection)
P(RYB) = (39/280) X (55/279) X (27/278) = 0.1393 X 0.1971 X 0.0971 = 0.002666
(6) With Replacement:
P(BBB) := (27/280) X (27/280) X (27/280) = 0.09643 = 0.0008958
(7) Without Replacement:
P(BBB) = (27/280) X (26/279) X (25/278) = 0.0964 X 0.0932 X 0.0899 = 0.0008708
(8) Since the probability of all three M&M's to be blue if the sampling is done without replacement = 0.0008708 is extremely small, it would be unusual.
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