$A$ 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

Urn $A$ contains $6$ red and $4$ black balls and urn $B$ contains $4$ red and $6$ black balls. One ball is drawn at random from urn $A$ and placed in urn $B$. Then one ball is drawn at random from urn $B$ and placed in urn $A$. If one ball is now drawn at random from urn $A$,the probability that it is found to be red,is

$A$ student has to write the words $ABILITY$, $PROBABILITY$, $FACILITY$, $MOBILITY$. He wrote one word and erased all the letters in it except two consecutive letters. If '$LI$' is left after erasing, then the probability that the boy wrote the word $PROBABILITY$ is

The probability that a person goes to college by car is $\frac{1}{5}$,by bus is $\frac{2}{5}$,and by train is $\frac{3}{5}$. The probabilities that he reaches the college late if he takes a car,bus,or train are $\frac{2}{7}$,$\frac{4}{7}$,and $\frac{1}{7}$ respectively. If he reaches the college in time,the probability that he traveled by car is:

Bag $A$ contains $2$ white,$1$ black and $3$ red balls and bag $B$ contains $3$ black,$2$ red and $n$ white balls. One bag is chosen at random and $2$ balls drawn from it at random,are found to be $1$ red and $1$ black. If the probability that both balls come from Bag $A$ is $\frac{6}{11}$,then $n$ is equal to

$A$ computer producing factory has only two plants $T_1$ and $T_2$. Plant $T_1$ produces $20 \%$ and plant $T_2$ produces $80 \%$ of the total computers produced. $7 \%$ of computers produced in the factory turn out to be defective. It is known that $P(\text{defective} | T_1) = 10 P(\text{defective} | T_2)$. $A$ computer produced in the factory is randomly selected and it is not defective. Then the probability that it is produced in plant $T_2$ is