An urn contains 7 white balls and 3 red balls

Asked by admin @ 06/08/2021 in Math viewed by 362 People

an urn contains 7 white balls and 3 red balls. two balls are drawn together at random from this urn. compute probability that neither of them is white. hence compute the expeted number of white balls drawn

Answered by admin @ 06/08/2021

1.4 is expected number of white balls drawn

0.09 is the probability that neither of them is white

Step-by-step explanation:

an urn contains 7 white balls and 3 red balls

Total Balls = 7 + 3 = 10

Probability of White = 7/10 = 0.7

Probability of Red = 3/10 = 0.3

Probability of not white = 0.3

two balls are drawn together at random from this urn.

Neither White = 0 White

P(0) =²C₀(0.7)⁰(0.3)² = 0.09

P(1) =²C₁(0.7)¹(0.3)¹ = 0.42

P(2) =²C₂(0.7)²(0.3)⁰ = 0.49

0.09 is the probability that neither of them is white

expected number of white balls drawn = 0 * P(0) + 1 * P(1) + 2 * P(2)

= 0 + 0.42 + 0.98

= 1.4

1.4 is expected number of white balls drawn

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