On every evening,a student either watches $TV$ or reads a book. The probability of watching $TV$ is $\frac{4}{5}$. If he watches $TV$,the probability that he will fall asleep is $\frac{3}{4}$ and it is $\frac{1}{4}$ when he reads a book. If the student is found to be asleep on an evening,the probability that he watched the $TV$ is
- A$\frac{11}{13}$
- B$\frac{12}{13}$
- C$\frac{2}{13}$
- D$\frac{4}{13}$
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Similar Questions
Suppose we have four boxes $A, B, C$ and $D$ containing coloured marbles as given below:
Box Red White Black $A$ $1$ $6$ $3$ $B$ $6$ $2$ $2$ $C$ $8$ $1$ $1$ $D$ $0$ $6$ $4$
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red,what is the probability that it was drawn from box $A$,box $B$,or box $C$?
| Box | Red | White | Black |
| $A$ | $1$ | $6$ | $3$ |
| $B$ | $6$ | $2$ | $2$ |
| $C$ | $8$ | $1$ | $1$ |
| $D$ | $0$ | $6$ | $4$ |
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red,what is the probability that it was drawn from box $A$,box $B$,or box $C$?
Difficult
View SolutionThere are $3$ bags,each containing $5$ white balls and $3$ black balls. Also,there are $2$ bags,each containing $2$ white balls and $4$ black balls. $A$ white ball is drawn at random. Find the probability that this white ball is from a bag of the first group.
$A$ box $A$ contains $2$ white,$3$ red and $2$ black balls. Another box $B$ contains $4$ white,$2$ red and $3$ black balls. If two balls are drawn at random,without replacement,from a randomly selected box and one ball turns out to be white while the other ball turns out to be red,then the probability that both balls are drawn from box $B$ is
DifficultJEE MAIN 2018
View SolutionAn insurance company insured $2000$ scooter drivers,$4000$ car drivers,and $6000$ truck drivers. The probabilities of an accident are $0.01, 0.03,$ and $0.15$ respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Difficult
View Solution$A$ bag contains $2n+1$ coins. It is known that $n$ of these coins have a head on both sides, whereas the remaining $n+1$ coins are fair. $A$ coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$, then $n$ is equal to
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