The following is the number of each color of M&M; found in a sample bag of 280 M
ID: 3358215 • 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 Yellow55 Green 38 27 Blue 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 grammarExplanation / Answer
1. P(Red) = 39/280 = 0.1393
2. P(Yellow) = 55/280 = 0.1964
3. P(Blue) = 27/280 = 0.0964
4. P(RYB) With Replacement
= (39/280) X (55/280) X (27/280) = 0.1393 X 0.1964 X 0.0964 = 0.0026
5. P(RYB) Without Replacement
= (39/280) X (55/279) X (27/278) = 0.1393 X 0.1971 X 0.0971 = 0.0027
(SINCE WITHOUT REPLACEMENT, TOTAL REDUCES BY 1 AFTER EACH DRAW)
6. P(BBB) With replacement
= (27/280) X (27/280) X (27/280) = 0.09643 = 0.0009
7. P(BBB) Without replacement
= (27/280) X (26/279) X (25/278) = 0.0964 X 0.0932 X 0.0899 = 0.0008
8. Since the probability of occurrence of all 3 M&M's to be blue if the sampling is done without replacement is only 0.08 %, it would be unusual.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.