JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

In a bolt factory, machines A, B and C manufacture respectively 20%, 30% and 50% of
the total bolts. Of their output 3, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random form the product. If the bolt drawn is found the defective, then the probability that it is manufactured by the machine C is.

Choose the correct answer:

Explanation

\begin{aligned}
& \mathrm{Given:~P(A)}=\frac{20}{100}=\frac{2}{10} \\
& \mathrm{P(B)}=\frac{30}{100}=\frac{3}{10};\mathrm{P(C)}=\frac{50}{100}=\frac{5}{10}
\end{aligned}

Let E $\to$ Event that the bolt is defective.

So, P( E/ A) = $\frac{3}{100}$ , P $(\frac{E}{B})$ = $\frac{4}{100}$ , P $(\frac{E}{C})$ = $\frac{2}{100}$ .

$am p; S6, P (C/E) am p; = \frac{P ( \frac{E}{C} ) \times P ( C )}{P ( \frac{E}{A} ) \times P ( A ) + P ( \frac{E}{B} ) \times P ( B ) + P ( \frac{E}{C} ) \times P ( C )}$

$= \frac{\frac{5}{10} \times \frac{2}{100}}{\frac{3}{100} \times \frac{2}{10} + \frac{4}{100} \times \frac{3}{10} + \frac{2}{100} \times \frac{5}{10}}$

$= \frac{10}{6 + 12 + 10} = \frac{10}{28} = \frac{5}{14}$