Algebra of Probability

Two students come late to the class and give reasons why they are late. The probability that the teacher believes the excuse of each student is $0.4.$ Let $p$ be the probability that the teacher believes both the students and $q$ be the probability that the teacher believes the second student but not the first. Then

Two players flip three fair coins each. The chance that they get the same number of heads is

Thousand tickets are sold in a lottery in which there is one top prize of Rs.$500$, four prizes of Rs.$100$ each and five prizes of Rs.$10$ each. A ticket costs Rs.$1$. The expected gain when you buy a ticket is

If $P\left(A\right)=0.65, P\left(B\right)=0.80$, then $P\left(A\cap B\right)$ lies in the interval

Events $A, B, C$ are mutually exclusive events such that $P\left(A\right)=\frac{3x+1}{3}, P\left(B\right)=\frac{1-x}{4}$ and $P\left(C\right)=\frac{1-2x}{2}.$ The set of possible values of $x$ are in the interval

Let $A, B, C$ be three events associated with sample space $S$. $A, B, C$ are pair wise independent and $P\left(A\right)=P\left(B\right)=P\left(C\right)=P.$ If all of them cannot occur simultaneously then $P(A\cup B\cup C)=$

The probability of solving a problem in Mathematics by two students $X$ and $Y$ are $\frac{1}{4}$ and $\frac{1}{5}$, respectively. If both of them try to solve the problem independently, then the probability that the problem will be solved by exactly one of them is equal to

If $P\left(A\right)=0.35, P\left(B\right)=0.2$ and $P(A\cap B)=0.42$, find the probability that neither $A$ nor $B$ occurs.

If $P\left(E\right)=\frac{4}{5}, P(E\cap F)=\frac{1}{5}$ and $P(D\cap E\cap \overline{F})=\frac{2}{5}$, then the value of $P(\overline{D}\cap E\cap \overline{F})$ is

There are $8$ laddus, $5$ mangoes and $4$ samoshas in a sweets box. If $1$ Item is picked at random, What is the probability of having either a laddu or a mango ?

If $X~B(n,p)$ and $n=10,E\left(X\right)=5,$ then the value of $p$ is

$A$ speaks truth in $75\%$ of the cases and $B$ in $80\%$, cases. What is the probability that their statements about an incident do not match?

If $P\left(A\right)=\frac{6}{11}, P\left(B\right)=\frac{5}{11}$ and $P(A\cup B)=\frac{7}{11}$, then value of $P\left(A\cap B\right)$ is 

If $A, B$ and $C$ are exhaustive events satisfying $P\left(\left(A\cup B\right)\cap \overline{C}\right)=\frac{1}{5}$, $P\left(B\cap C\right)-P\left(A\cap B\cap C\right)=\frac{1}{15}$ and $P\left(A\cap C\right)=\frac{1}{10}$ then $P\left(C\cap \left(\overline{A\cup B}\right)\right)$ is equal to

A fair coin is tossed until one of the $2$ sides occurs twice in a row. Probability that even number of tosses required is

Let $P_1, P_2$ and $P_3$ are the probabilities of a student passing three independent exams $A, B$ and $C$ respectively. If $P_1, P_2$ and $P_3$ are the roots of equation $20x^3-27x^2+14x-2=0$, then the probability that the student passes in exactly one of $A, B$ and $C$ is

A die is thrown. Let $A$ be the event that the number obtained is greater than $3 .$ Let $B$ be the event that the number obtained is less than $5 .$Then $P(A\cup B)$ is

Face is just like moon. This sentence has which figure of speech?

Let $A$ and $B$ are two events such that $P\left(\overline{A\cup B}\right)=\frac{1}{6}$, $P\left(A\cap B\right)=\frac{1}{4}$ and $P\left(\overline{A}\right)=\frac{1}{4},$ where $\overline{A}$ represents the complement of event $A$. Then, events $A$ and $B$ are

The probability that a problem is solved correctly by at least person, if four persons independently can solve the problem with probabilities $\frac{1}{2}\text{, }\frac{3}{4}\text{, }\frac{1}{4}\text{, }\frac{1}{8}$, is