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There are 98 balls in a box. 25 of them are red, 19 of them are green, 30 of them are purple,and 24 of them are blue. Suppose Hao draws 22 balls from the box with replacement (hedraws the ball, records its color, and then puts it back into the box). Find the probabilitythat he draws 2 red balls, 5 green balls, 10 purple balls, and 5 blue balls.

User Genekogan
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2 Answers

5 votes

Answer:

Since this problem related to the replacement problem thus

Pr(2 red)=(25/98)(25/98)=0.0651

Pr(5 green)=(19/98)(19/98)(19/98)(19/98)(19/98)

Pr(5 green)=2.73*10^-4

Pr(10 purple)=((30/98))^10=7.22*10^-6

Pr(5 blue)=((24/98))^5=8.8*10^-4

User Kthompson
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4.7k points
5 votes

Answer:

Since the balls are drawn with replacement, it means the individual probability remain constant,

I have solved this problem on paper (Figures Attached).

Thanks.

There are 98 balls in a box. 25 of them are red, 19 of them are green, 30 of them-example-1
There are 98 balls in a box. 25 of them are red, 19 of them are green, 30 of them-example-2
User KIMA
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