Embibe Experts Solutions for Chapter: Probability, Exercise 1: VITEEE 2017

If $A$ and $B$ are events such that $P\left(A\right)=0.42, P\left(B\right)=0.48$ and $P \left(A \mathrm{and} B\right)=0.16$. Then,

$\mathrm{I}.$ $P \left(\mathrm{not} A\right)=0.58$

$\mathrm{II}.$ $P \left(\mathrm{not} B\right)=0.52$

$\mathrm{III}.$ $P \left(A \mathrm{or} B\right)=0.47$

$A$ and $B$ are events such that $P\left(A\cup B\right)=\frac{3}{4}, P\left(A\cap B\right)=\frac{1}{4}, P\left(\overline{A}\right)=\frac{2}{3}$, then $P\left(\overline{A}\cap B\right)$ is

It is given that the events $A$ and $B$ are such that $P\left(A\right)=\frac{1}{4}, P\left(A\mid B\right)=\frac{1}{2}$ and $P\left(B\mid A\right)=\frac{2}{3}$. Then $P\left(B\right)$ is

A signal which can be green or red with probability $\frac{4}{5}$ and $\frac{1}{5}$ respectively, is received by station $\text{A}$ and then transmitted to station $\text{B}$ . The probability of each station receiving the signal correctly is $\frac{3}{4}$ . If the signal received at station $\text{B}$ is green, then the probability that the original signal is green, is

$A$ and $B$ are two independent witnesses (i.e., there is no collision between them) in a case. The probability that $A$ will speak the truth is $x$ and the probability that $B$ will speak the truth is $y.$ $A$ and $B$ agree in a certain statement. The probability that the statement is true is

A random variable $X$ has the probability distribution :

The mean and variance of a random variable $X$ in binomial distribution are $4$ and $2$ respectively, then $P\left(X=1\right)$ is