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  • The probability that certain electronic component fails when first used is 0.10 . If it does not fail immediately, the probability that it lasts for one year is 0.99 . If the probability that a new

The probability that certain electronic component fails when first used is 0.10 . If it does not fail immediately, the probability that it lasts for one year is 0.99 . If the probability that a new component will last for one year is \lambda then, value of 1000 \lambda must be

Option: 1

891


Option: 2

0.891


Option: 3

8910


Option: 4

.0891


Answers (1)

best_answer

Given probability of electronic component fails when first used =0.10

\mathrm{\text { ie, } P(F)=0.10 }

\mathrm{\therefore P(\bar{F})=1-P(F)=0.90}

and let \mathrm{P(Y)=}probability of new component to last for one year

\mathrm{ \because \quad P(F)+P(\bar{F})=1 }

Obviously the two events are mutually exclusive and exhaustive

\mathrm{\therefore P\left(\frac{Y}{F}\right)=0 \text { and } P\left(\frac{Y}{\bar{F}}\right) =0.99 }

\mathrm{\therefore \quad P(Y) =P(F) \cdot P\left(\frac{Y}{F}\right)+P(\bar{F}) \cdot P\left(\frac{Y}{\bar{F}}\right) }

\mathrm{=0.10 \times 0+0.90 \times 0.99 }

\mathrm{ =0.891=\lambda \text { (given) } }

\mathrm{\therefore \quad 1000 \lambda =891}

Hence option 1 is correct.






 


 

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Ritika Harsh

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