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. |
Answer:
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.
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