$A$ car manufacturing factory has two plants $X$ and $Y$. Plant $X$ manufactures $70 \%$ of cars and plant $Y$ manufactures $30 \%$ of cars. $80 \%$ of cars at plant $X$ and $90 \%$ of cars at plant $Y$ are rated as standard quality. $A$ car is chosen at random and is found to be standard quality. The probability that it has come from plant $X$ is

Two coins $A$ and $B$ are kept in an urn. When coin $A$ is flipped,the probability of getting a head is $1/4$,while for coin $B$ it is $3/4$. One coin is randomly chosen from this bag,tossed twice,and it falls heads on both occasions. The probability that it is coin $A$ is:

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.

Let $H_1, H_2, \ldots, H_{n}$ be mutually exclusive and exhaustive events with $P(H_i) > 0, i = 1, 2, \ldots, n$. Let $E$ be any other event with $0 < P(E) < 1$.
$STATEMENT-1$: $P(H_i \mid E) > P(E \mid H_i) \cdot P(H_i)$ for $i = 1, 2, \ldots, n$.
$STATEMENT-2$: $\sum_{i=1}^{n} P(H_i) = 1$.

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:

If $E_1$ and $E_2$ are equally likely, mutually exclusive and exhaustive events and $P(A|E_1) = 0.2$, $P(A|E_2) = 0.3$, then $P(E_1|A)$ is equal to...