Math, asked by sweetyng7129, 1 year ago

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

Answers

Answered by Bhanu11118
8
probability of neither of them is white= 3/10

probability of getting white ball= 7/10
Answered by amitnrw
4

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

Learn more:

Q6) Calculate the Expected number of candies for a randomly ...

https://brainly.in/question/12858893

Onit two urns contain 5 white 3 black and 2 white 3 black balls ...

https://brainly.in/question/13795488

Similar questions