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An urn contains nine blue balls and seven yellow balls. If three balls are selected randomly without being​ replaced, what is the probability that of the balls​ selected, one of them will be blue and two of them will be​ yellow?

User Brian Dant
by
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1 Answer

1 vote

Answer:


Probability = (27)/(80)

Explanation:

Given


Blue =9


Yellow = 7

Required

Determine the probability that one of the balls is blue while others is yellow

The order of selection may be any of these three:

Yellow, Yellow, Blue or Yellow, Blue, Yellow or Blue, Yellow, Yellow

Solving Yellow, Yellow, Blue


Probability = P(Y_1) * P(Y_2) * P(B)


Probability = (7)/(16) * (6)/(15) * (9)/(14)


Probability = (1)/(8) * (1)/(5) * (9)/(2)


Probability = (9)/(80)

Solving Yellow, Blue, Yellow


Probability = P(Y_1) * P(B) * P(Y_2)


Probability = (7)/(16) * (9)/(15) * (6)/(14)


Probability = (9)/(80)

Solving Blue, Yellow, Yellow


Probability = P(B) *P(Y_1) * P(Y_2)


Probability = (9)/(16) * (7)/(15) * (6)/(14)


Probability = (9)/(80)

Hence:

The required probability is:


Probability = (9)/(80) + (9)/(80) + (9)/(80)


Probability = (27)/(80)

User Samveen
by
8.3k points

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