The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs (i) none (ii) not more than one (iii) more than one (iv) at least one will fuse after 150 days of use.

Answer

N/a

Explanation

Let `X` represents the number of bulbs fused in 150 days in an experiment of 5 bulbs. It is the Bernaulli's experiment.
`p=0.05` and `q=1-p=1-0.05=0.95`
`X` is a binomial distribution where `n=5, p=0.05` and `q=0.95`.
`:.P(X=r)=.^(5)C_(r).(0.05)^(r)(0.95)^(5-r)` (i) Required Probability
`=P(X=0)=.^(5)C_(0)p^(0)q^(5)=q^(5)=(0.95)^(5)`
(ii) Required Probability
`=P(Xle1)=P(0)+P(1)=.^(5)_(0)p^(0)q^(5)+.^(5)C_(1)p^(1)q^(4)`
`=q^(5)+5pq^(4)=q^(4)(q+5p)`
`=(0.95)^(4)[0.95+5xx(0.05)]`
`=(0.95)^(4)(0.95+0.25)-(0.95)^(4)xx1.2`
(iii) Required Probability
`=P(Xgt1)=1p{P(0)+P(1)}`
`=1-(0.95)^(4)xx1.2`
(iv) Required Probability
`=P` (at least one bulb will fuse)
`=P(Xge1)=1P(0)`
`=1-(0.95)^(5)`