an urn contains 1 white, 1 green, and 2 red balls. draw 3 balls with replacement. find probability we did not see all the colours

Chardonnay Felix

Chardonnay Felix

Answered question

2021-05-21

an urn contains 1 white, 1 green, and 2 red balls. draw 3 balls with replacement. find probability we did not see all the colours

Answer & Explanation

Luvottoq

Luvottoq

Skilled2021-05-22Added 95 answers

Given: Urn contains 1 white, 1 green and 2 red balls
All colors were seen
First we determine the probability of seeing all three colors among the 3 balls.
1 of the 4 balls are white, 1 of the 4 balls are green and 2 of the 4 balls are red. The probability is the number of favorable outcomes divided by the number of possible outcomes:
P(white)= # of favorable outcomes # of possible outcomes=14
P(green)= # of favorable outcomes # of possible outcomes=14
P(red)= # of favorable outcomes # of possible outcomes=24=12
Since the balls are drawn with replacement, the selection of the different balls are independent.
Thus it is then appropriate to use the Multiplication rule for independent events: P(AB)=P(AandB)=P(A)P(B).
P(all different colors)=P(Red, white and green)=P(red)P(white)P(green)=1/21/41/4=1/(244)=1/32
2.Not all colors were seen Complement rule:
P(Ac)=P(¬A)=1P(A)
P(not all different colors)=1P(All different colors)=1(1/32)=31/32=0.06875=96.875

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