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A bag contains 9 red marbles, 4 blue marbles, and 7 yellow marbles. You randomly select three marbles from the bag. a. What is the probability that all three marbles are red when you replace each marble before selecting the next marble? Round your answer to the nearest tenth. about % b. What is the probability that all three marbles are red when you do not replace each marble before selecting the next marble? Round your answer to the nearest tenth. about %

User Wpnpeiris
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2 Answers

4 votes

Answer:

9.1%

7.4%

Explanation:

User Jeppe Christensen
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7.5k points
4 votes

Answer:

A) 9.1%

B) 7.4%

Explanation:

Part A

Because you are replacing each marble before selecting the next, the probability of choosing one red marble remains the same, so choosing 3 red marbles with replacement would mean the probability gets multiplied by itself 3 times:


\displaystyle P(\text{Red+Replace})=\biggr((9)/(20)\biggr)^3=(9)/(20)*(9)/(20)*(9)/(20)=(729)/(8000)\approx9.1\%

Part B


\displaystyle P(\text{Red+No Replace})=(C(9,3)C(4,0)C(7,0))/(C(20,3))=((9!)/(3!(9-3)!))/((20!)/(3!(20-3)!))=((9*8*7)/(3*2*1))/((20*19*18)/(3*2*1))=(9*8*7)/(20*19*18)=(7)/(95)\approx7.4\%This can also be calculated with
(P(9,3))/(P(20,3)) because the order does matter in which you pick the 3 red marbles with no replacement.

User DarkDeny
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