If $a$ and $b$ are chosen randomly from the set consisting of numbers $1, 2, 3, 4, 5, 6$ with replacement. If the probability that $\lim_{x \to 0} \left(\frac{a^x + b^x}{2}\right)^{\frac{1}{x}} = 6$ is $\frac{p}{q}$ (where $H.C.F(p, q) = 1$) then $q - p = \ldots\ldots\ldots\ldots\ldots\ldots$
If \(a^2 + 4b^2 + 4c^2 - 2ab - 4bc - 2ac = 0\), then the number of ordered triplets \((a, b, c)\) with \(a, b, c \in \{1, 2, 3, 4, 5, 6\}\) satisfying the equation is 3, i.e., \((2,1,1), (4,2,2), (6,3,3)\). Two points \((2,1,1)\) and \((4,2,2)\) lie inside a given tetrahedron. If the required probability is \(\dfrac{2}{3} = \dfrac{5}{\lambda}\), then \(\lambda =\)
Four candidates A, B, C and D have applied for the post in government office. If A is twice as likely to be selected as B, and B and C are given about the same chances of being selected, while C is twice as likely to be selected as D, what are the probabilities that(i) C will be selected?(ii) A will not be selected?
A random variable $X$ takes values $0,1,2,3$ with probabilities $\dfrac{2a+1}{30}$, $\dfrac{8a-1}{30}$, $\dfrac{4a+1}{30}$, $b$ respectively, where $a,b\in\mathbb{R}$. Let $\mu$ and $\sigma$ respectively be the mean and standard deviation of $X$ such that $\sigma^2+\mu^2=2$. Then $\dfrac{a}{b}$ is equal to:
Consider the following assignments of probabilities for outcomes of sample space \(S = \{1, 2, 3, 4, 5, 6, 7, 8\}\).Number (X)12345678Probability P(X)0.150.230.120.100.200.080.070.05Find the probability that (b) \(X\) is a number greater than 4.
Consider the following assignments of probabilities for outcomes of sample space \(S = \{1, 2, 3, 4, 5, 6, 7, 8\}\).Number (X)12345678Probability P(X)0.150.230.120.100.200.080.070.05Find the probability that (a) \(X\) is a prime number.