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For an initial screening of an admission...

For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any proglem is `(4)/(5)`, then the probability that he is unable to solve less than two problem is

A

`(201)/(5)((1)/(5))^(49)`

B

`(316)/(25)((4)/(5))^(48)`

C

`(54)/(5)((4)/(5))^(49)`

D

`(164)/(25)((1)/(5))^(48)`

Text Solution

Verified by Experts

The correct Answer is:
C
Given that, there are 50 problems to solve in an admission test and probability that the candidate can solve any problem is `(4)/(5)`=q(say). So, probability that the candidate cannot solve a problem is p=1-q=1-(4)/(5)=(1)/(5)`.
Now, let X be a random variable which denotes the number of problems that the candidate is unabel to solve. Then, X follows binomial distribution with parameters n=50 and `p=(1)/(5)`.
Now, according to binomial probability distribution concept
`P(X=r)=overset(50)""C_(r )((1)/(5))^(r )((4)/(5))^(50-r),r=0,1,.......,50`
`:. ` Required probability
`=P(X lt 2)=P(X=0)+P(X=1)`
`=overset(50)""C_(0)((4)/(5))^(50)+overset(50)""C_(1)(4^(49))/((5)^(50))=((4)/(5))^(49)((4)/(5)+(50)/(5))=(54)/(5)((4)/(5))^(49)`
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