JEE Main 2021 — Mathematics Probability & Statistics

Given that

An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is $0.9$0.9 and that of the second unit is $0.8.$0.8. The instrument is switched on and it fails to operate. If the probability that only the first unit failed and second unit is functioning is $p.$p.

Let ${E}_{1}=$E1​= Event of first unit is working

${E}_{2}=$E2​= Event of second unit is working

$$P({E}_{1})=0.9P({E}_{2})=0.8$$P(E1​)=0.9P(E2​)=0.8

We know that $P(E)=1-P(E)$P(E)=1−P(E)

$$P({\bar{E}}_{1})=0.1P({\bar{E}}_{2})=0.2$$P(Eˉ1​)=0.1P(Eˉ2​)=0.2

By using Baye's theorem

Required probability

$$p=\frac{P({E}_{2})\times P({E}_{1})}{P({E}_{2})\times P({E}_{1})+P({E}_{1})\times P({E}_{2})+P({E}_{2})\times P({E}_{1})}$$p=P(E2​)×P(E1​)+P(E1​)×P(E2​)+P(E2​)×P(E1​)P(E2​)×P(E1​)​

$$p=\frac{0.8\times 0.1}{0.1\times 0.2+0.9\times 0.2+0.1\times 0.8}=\frac{(\frac{8}{100})}{\frac{(2+18+8)}{100}}=\frac{8}{28}$$p=0.1×0.2+0.9×0.2+0.1×0.80.8×0.1​=100(2+18+8)​(1008​)​=288​

So, $98P=98\times \frac{8}{28}=28$98P=98×288​=28