Conditional Probability and Multiplication Theorem

M and N are two events such that $\mathrm{P}\left(\mathrm{M}\right|\mathrm{N}) = 0.3, \mathrm{P}(\mathrm{M})= 0.2 \mathrm{and} \mathrm{P}(\mathrm{N}) = 0.4$. Which of the following is the value of $\mathrm{P}(\mathrm{M} \cap \mathrm{N}\text{'})?$

If $A\text{ and }B$ are two events, such that $P\left(A\right)=\frac{3}{8},P\left(B\right)=\frac{5}{8}\text{ and }P(A\cup B)=\frac{3}{4}$, then find $P\left(\frac{A}{B}\right)$.

A coin is tossed three times, where event $E$: head on third toss, event $F$: heads on first two tosses. Determine $P\left(E|F\right)$.

Given that $\mathrm{E}$ and $\mathrm{F}$ are events such that $\mathrm{P}\left(\mathrm{E}\right)=0.6,\mathrm{P}\left(\mathrm{F}\right)=0.3$ and $\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=0.2$, find $\mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)$and $\mathrm{P}\left(\mathrm{F}|\mathrm{E}\right)$.

$A$ speaks truth in $75\%$ of the cases, while $B$ in $90\%$ of the cases. In what percent of cases, are they likely to contradict each other in stating the same fact?

A fair die is rolled. Consider events $A=\left\{1,3,5\right\}, B=\left\{2,3\right\}$ and $C=\left\{2,3,4,5\right\}$, then find $P\left(\frac{A\cup B}{C}\right)$ and $P\left(\frac{A\cap B}{C}\right)$.

Two urns contains$2$ red, $4$ white and $3$ red, $7$ white balls. One of the urn is chosen at a random and a ball is drawn from it. If the probability of a white ball drawn is $\frac{k}{60}$, then find the value of $k$.

Two urns contains$2$ red, $4$ white and $3$ red, $7$ white balls respectively. One of the urn is chosen at a random and a ball is drawn from it. If the probability of a red ball drawn is $\frac{k}{60}$, then find the value of $k$.

If $P\left(A\right)=0.8, P\left(B\right)=0.5$ and $P(B / A)=0.4$, the value of $100\times P(A\cup B)$  is 

If $P\left(B\right)=0.5$ and $P\left(\frac{A}{B}\right)=0.4$, then the value of $10P(A\cap B)$ is

Two integers are selected at random from the set $\left\{1, 2, .......,11\right\}$. Given that the sum of selected number is even, the conditional probability that both the numbers are even is:

A dice is rolled three times. Let $E_1$ denotes the event of getting a number larger than the previous number each time and $E_2$ denotes the event that the numbers (in order) form an increasing AP then -

A committee of three person is to be randomly selected from a group of three men and two women and the chairperson is to be randomly selected form the committee. The probability that the committee will have exactly two women and one men, and that the chairperson will be a women, is

There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines are identified. Then the probability that only two tests are needed is

$A$ and $B$ toss a fair coin each simultaneously $50$ times. The probability that both of them will not get tail at the same toss is

A and B draw two cards each, one after another, from a pack of well shuffled pack of 52 cards. The probability that all the four cards drawn are of the same suit is

If two events $A$and $B$are such that $P\left(A^{\prime }\right)=0.3, P\left(B\right)=0.4$ and $\left(A\cap B^{\prime }\right)=0.5,$ then $P\left(\frac{B}{A\cup B^{\prime }}\right)=$

If $A \& B$ are two events such that $P\left(A\right)>0,$ and $P\left(B\right)\ne 1$ , then $P\left(\frac{\overline{A}}{\overline{B}}\right)$ is equal to

A pack of cards contains $4$aces, $4$ kings, $4$ queens and $4$ jacks. Two cards are drawn at random from this pack without replacement. The probability that at least one of them will be an ace, is

A box contains$15$ transistors, $5$ of which are defective. An inspector takes out one transistor at random, examines it for defects and replaces it. After it has replaced another inspector does the same thing and then so does a third inspector. The probability that atleast one of the inspectors finds a defective transistor, is equal to