$A$ shopkeeper buys a particular type of electric bulbs from three manufacturers $M_1, M_2$,and $M_3$. He buys $25 \%$ of his requirement from $M_1$,$45 \%$ from $M_2$,and $30 \%$ from $M_3$. Based on past experience,he found that $2 \%$ of type $M_3$ bulbs are defective,whereas only $1 \%$ of type $M_1$ and type $M_2$ bulbs are defective. If a bulb chosen by him at random is defective,then the probability that it was of type $M_3$ is
- A$\frac{5}{13}$
- B$\frac{6}{13}$
- C$\frac{7}{13}$
- D$\frac{8}{13}$
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DifficultJEE MAIN 2019
View SolutionThree boxes $B_1$,$B_2$ and $B_3$ contain balls with different colors as follows:
Box White,Black,Red $B_1$ $2, 1, 2$ $B_2$ $3, 2, 4$ $B_3$ $4, 3, 2$
$A$ die is thrown. Box $B_1$ is chosen if either $1$ or $2$ turns up. Box $B_2$ is chosen if $3$ or $4$ turns up and box $B_3$ is chosen if $5$ or $6$ turns up. Having chosen a box in this way,a ball is drawn at random from that box. If the ball drawn is found to be Red,then the probability that it is drawn from box $B_2$ is
| Box | White,Black,Red |
|---|---|
| $B_1$ | $2, 1, 2$ |
| $B_2$ | $3, 2, 4$ |
| $B_3$ | $4, 3, 2$ |
$A$ die is thrown. Box $B_1$ is chosen if either $1$ or $2$ turns up. Box $B_2$ is chosen if $3$ or $4$ turns up and box $B_3$ is chosen if $5$ or $6$ turns up. Having chosen a box in this way,a ball is drawn at random from that box. If the ball drawn is found to be Red,then the probability that it is drawn from box $B_2$ is
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