An 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?

Assume that the probability of a patient having a heart attack is $40 \%$. It is also assumed that a meditation and yoga course reduces the risk of a heart attack by $30 \%$ and the prescription of a certain drug reduces its risk by $25 \%$. $A$ patient can choose any one of the two options with equal probability. Given that a patient selected at random suffers a heart attack after choosing one of the two options,find the probability that the patient followed the course of meditation and yoga.

Recent studies suggest that $12\%$ of the world population is left-handed. Depending on parents' hand usage, the chances of having left-handed children are as follows: $A$. Both parents are left-handed, chances of having left-handed children = $24\%$. $B$. Both parents are right-handed, chance of having left-handed children = $9\%$. $C$. Father left-handed and mother right-handed, chances of having left-handed children = $17\%$. $D$. Father right-handed and mother left-handed, chances of having left-handed children = $22\%$. Given $P(A) = P(B) = P(C) = P(D) = 1/4$ and $L$ denotes the event that the child is left-handed. What is the probability $P(A|L)$?

$A$ 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

$A$ bag contains $(N+1)$ coins: $N$ fair coins, and one coin with 'Head' on both sides. $A$ coin is selected at random and tossed. If the probability of getting 'Head' is $\frac{9}{16}$, then $N$ is equal to:

$A$ and $B$ are two events of a random experiment such that $P(B)=0.4$, $P(A \cap \bar{B})=0.5$, and $P(A \cup B) + P\left(\frac{B}{A \cup \bar{B}}\right) = 1.15$. Then $P(A) = $