Probability

The probability that, on the examination day, the student $S_1$ gets the previously allotted seat $R_1$, and NONE of the remaining students gets the seat previously allotted to him/her is

For $i = 1, 2, 3, 4$, let $T_i$ denote the event that the students $S_i$ and $S_{i+1}$ do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event $T_1 \cap T_2 \cap T_3 \cap T_4$ is

Let $X$ and $Y$ be two events such that $P(X) = \frac{1}{3}$, $P(X|Y) = \frac{1}{2}$ and $P(Y|X) = \frac{2}{5}$. Then

Three randomly chosen nonnegative integers $x$, $y$ and $z$ are found to satisfy the equation $x + y + z = 10$. Then the probability that $z$ is even, is

For three events A, B and C,

P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = $\dfrac14$ and P(All the three events occur simultaneously) = $\dfrac1{16}$.

Then the probability that at least one of the events occurs, is :

If two different numbers are taken from the set {0, 1, 2, 3, ......, 10}; then the probability that their sum as well as absolute difference are both multiple of 4, is :

Three persons P, Q and R independently try to hit a target. If the probabilities of their hitting the target are $\dfrac34$, $\dfrac12$ and $\dfrac58$ respectively, then the probability that the target is hit by P or Q but not by R is :

An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is :

From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men, is :

Let E and F be two independent events. The probability that both E and F happen is $\dfrac1{12}$ and the probability that neither E nor F happens is $\dfrac12$, then a value of $\dfrac{P(E)}{P(F)}$ is :

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

$P$(computer turns out to be defective given that it is produced in plant $T_1$) $=10P$(computer turns out to be defective given that it is produced in plant $T_2$),

where $P(E)$ denotes the probability of an event $E$. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $T_2$ is

$P(X>Y)$ is

$P(X=Y)$ is

Let two fair six-faced dice A and B be thrown simultaneously. If $E_1$ is the event that die A shows up four, $E_2$ is the event that die B shows up two and $E_3$ is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ?

If A and B are any two events such that $P(A)=\frac25$ and $P(A\cap B)=\frac3{20}$, then the conditional probability, $P(A\mid(A'\cup B'))$, where A′ denotes the complement of A, is equal to :

The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is

One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is $\dfrac13$, then the correct option(s) with the possible values of $n_1$, $n_2$, $n_3$ and $n_4$ is(are)

A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is $\dfrac13$, then the correct option(s) with the possible values of $n_1$ and $n_2$ is(are)

If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is :

Let X be a set containing 10 elements and P(X) be its power set. If A and B are picked up at random from P(X), with replacement, then the probability that A and B have equal number of elements, is :