• # question_answer A company has two plants for manufacturing   scooters.   Plant I manufactures 70% of the scooters and plant II manufactures 30% of the scooters. At plant I, 30% of the scooters are maintaining pollution norms and plant II, 90% of the scooters are maintaining pollution norms. A scooter chosen at random and is found to be fit oh pollution norms. Find the probability that it has come from plant II. OR Events A and B are such that $P(A)=\frac{1}{2},$ $P(B)=\frac{7}{12}$ and P (not A or not B) $=\frac{1}{4}.$ State whether A and B are independent.

 Let ${{E}_{1}}$be the event, when plant I is chosen ${{E}_{2}}$ be the event when plant II is chosen A be the event, when chosen scooter is fit. Then, $P({{E}_{1}})=70%=\frac{70}{100},$ $P({{E}_{2}})=30%=\frac{30}{100},$ $P\left( \frac{A}{{{E}_{1}}} \right)=30%=\frac{30}{100},$ $P\left( \frac{A}{{{E}_{2}}} \right)=90%=\frac{90}{100},$ By using Baye?s theorem, we get $P\left( \frac{{{E}_{2}}}{A} \right)=\frac{P({{E}_{2}})P\left( \frac{A}{{{E}_{2}}} \right)}{P({{E}_{1}})P\left( \frac{A}{{{E}_{1}}} \right)+P({{E}_{2}})P\left( \frac{A}{{{E}_{2}}} \right)}$ $=\frac{\left( \frac{300}{100}\times \frac{90}{100} \right)}{\left( \frac{70}{100}\times \frac{30}{100} \right)+\left( \frac{30}{100}\times \frac{90}{100} \right)}$ $=\frac{27}{21+27}=\frac{27}{48}=\frac{9}{16}$ Thus, probability that he fit scooter has come from plant 2 is$\frac{9}{16}$. OR Give, P (not A or not B) =$\frac{1}{4}$ $\Rightarrow$   $P(A'\,\,or\,\,B')=\frac{1}{4}$ $\Rightarrow$   $P\,(A'\cup B')=\frac{1}{4}$ $\Rightarrow$   $P\,[(A\cap B)']=\frac{1}{4}$     $[\because P\,(A'\cup B')=P\,(A\cap B)']$ $\Rightarrow$   $1-P\,(A\cap B)=\frac{1}{4}$ $\Rightarrow$   $P\,(A\cap B)=1-\frac{1}{4}=\frac{3}{4}$                        ?(i) Also,     $P\,(A)\times P\,(B)=\frac{1}{2}\cdot \frac{7}{12}$ $\left[ given,\,\,P\,(A)=\frac{1}{2},\,\,P\,(B)=\frac{7}{12} \right]$ $=\frac{7}{24}\ne \frac{3}{4}$ $\Rightarrow$   $P\,(A)\cdot P\,(B)\ne P\,(A\cap B)$ Hence, A and B are not independent events.