If a random variable X follows the Binomial distribution B(33, p) such that $$3P(X=0)=P(X=1)$$, then the value of {{P(X = 15)} \over {P(X = 18)}} - {{P(X = 16)} \over {P(X = 17)}} is equal to :

If a random variable X follows the Binomial distribution B(5, p) such that P(X = 0) = P(X = 1), then {{P(X = 2)} \over {P(X = 3)}} is equal to :

If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :

If the numbers appeared on the two throws of a fair six faced die are $$\alpha$$ and $$\beta$$, then the probability that $$x^2+\alpha x+\beta >0$$, for all $$x\in \mathbf{R}$$, is :

If $$A$$ and $$B$$ are two events such that $$P(A)=\frac{1}{3},P(B)=\frac{1}{5}$$ and $$P(A\cup B)=\frac{1}{2}$$, then $$P\left(A\mid B^{\prime }\right)+P\left(B\mid A^{\prime }\right)$$ is equal to :

The mean and variance of a binomial distribution are $$\alpha$$ and $$\frac{\alpha }{3}$$ respectively. If $$\mathrm{P}(X=1)=\frac{4}{243}$$, then $$\mathrm{P}(X=4$$ or 5$$)$$ is equal to :

Let $${\mathrm{E}}_1,{\mathrm{E}}_2,{\mathrm{E}}_3$$ be three mutually exclusive events such that $$\mathrm{P}\left({\mathrm{E}}_1\right)=\frac{2+3\mathrm{p}}{6},\mathrm{P}\left({\mathrm{E}}_2\right)=\frac{2-\mathrm{p}}{8}$$ and $$\mathrm{P}\left({\mathrm{E}}_3\right)=\frac{1-\mathrm{p}}{2}$$. If the maximum and minimum values of $$\mathrm{p}$$ are $${\mathrm{p}}_1$$ and $${\mathrm{p}}_2$$, then $$\left({\mathrm{p}}_1+{\mathrm{p}}_2\right)$$ is equal to :

Let $$X$$ be a binomially distributed random variable with mean 4 and variance $$\frac{4}{3}$$. Then, $$54P(X\le 2)$$ is equal to :

Let $$S$$ be the sample space of all five digit numbers. It $$p$$ is the probability that a randomly selected number from $$S$$, is a multiple of 7 but not divisible by 5 , then $$9p$$ is equal to :

Let X have a binomial distribution B(n, p) such that the sum and the product of the mean and variance of X are 24 and 128 respectively. If $$P(X>n-3)=\frac{k}{2^n}$$, then k is equal to :

A six faced die is biased such that $$3\times \mathrm{P}($$a prime number$$)=6\times \mathrm{P}($$a composite number$$)=2\times \mathrm{P}(1)$$. Let X be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of X is :

Out of $$60\%$$ female and $$40\%$$ male candidates appearing in an exam, $$60\%$$ candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is :

Let $$\mathrm{A}$$ and $$\mathrm{B}$$ be two events such that $$P(B\mid A)=\frac{2}{5},P(A\mid B)=\frac{1}{7}$$ and $$P(A\cap B)=\frac{1}{9}\cdot$$ Consider (S1) $$P\left(A^{\prime }\cup B\right)=\frac{5}{6}$$, (S2) $$P\left(A^{\prime }\cap B^{\prime }\right)=\frac{1}{18}$$ Then :

Let $$S=\{1,2,3,\dots ,2022\}$$. Then the probability, that a randomly chosen number n from the set S such that $$\mathrm{HCF}(\mathrm{n},2022)=1$$, is :

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is :

If $${\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5},\,|y|

Let f : R $$\to$$ R be defined as $$f(x)=x^3+x-5$$. If g(x) is a function such that $$f(g(x))=x,{\forall }^{\prime }{x}^{\prime }\in R$$, then g'(63) is equal to ________________.

If $$y = {\tan ^{ - 1}}\left( {\sec {x^3} - \tan {x^3}} \right),{$\pi$ \over 2}

The value of $${\log }_{e}2\frac{d}{dx}\left({\log }_{\cos x}\mathrm{cosec}x\right)$$ at $$x=\frac{\pi }{4}$$ is

Let $$x(t)=2\sqrt{2}\cos t\sqrt{\sin 2t}$$ and $$y(t)=2\sqrt{2}\sin t\sqrt{\sin 2t},t\in \left(0,\frac{\pi }{2}\right)$$. Then $$\frac{1+{\left(\frac{dy}{dx}\right)}^2}{\frac{d^{2}y}{d{x}^2}}$$ at $$t=\frac{\pi }{4}$$ is equal to :