In an experiment, a person gets success alpha | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The probability of success in a single trial is $p = \frac{\alpha}{\beta}$.The probability of failure in a single trial is $q = 1 - p = \frac{\bet

Vedclass · Accepted Answer

$\frac{(\beta-\alpha)^{n-1}}{\beta^n}(n \alpha+\beta-\alpha)$

Vedclass · Answer

(B) The probability of success in a single trial is $p = \frac{\alpha}{\beta}$.The probability of failure in a single trial is $q = 1 - p = \frac{\beta - \alpha}{\beta}$.We need the probability of failing at least $(n-1)$ times in $n$ trials, which means failing $(n-1)$ times or failing $n$ times.This is equivalent to succeeding at most $1$ time.Using the binomial distribution formula $P(X = k) = {}^{n}C_{k} p^k q^{n-k}$:$P(X \le 1) = P(X = 0) + P(X = 1)$$P(X = 0) = {}^{n}C_{0} p^0 q^n = 1 \cdot 1 \cdot \left(\frac{\beta - \alpha}{\beta}\right)^n = \frac{(\beta - \alpha)^n}{\beta^n}$$P(X = 1) = {}^{n}C_{1} p^1 q^{n-1} = n \cdot \left(\frac{\alpha}{\beta}\right) \cdot \left(\frac{\beta - \alpha}{\beta}\right)^{n-1} = \frac{n \alpha (\beta - \alpha)^{n-1}}{\beta^n}$Adding these probabilities:$P = \frac{(\beta - \alpha)^n + n \alpha (\beta - \alpha)^{n-1}}{\beta^n}$$P = \frac{(\beta - \alpha)^{n-1} [(\beta - \alpha) + n \alpha]}{\beta^n}$$P = \frac{(\beta - \alpha)^{n-1} (n \alpha + \beta - \alpha)}{\beta^n}$Thus, option $B$ is correct.

In an experiment, a person gets success $\alpha$ times out of $\beta$ trials. If the experiment consists of $n$ trials, then the probability that he fails at least $(n-1)$ times is

Similar Questions

Let $X \sim B(6, 1/2)$,then $P[|X-4| \leq 2]$ is

If $X$ is a binomial variate with the range $\{0, 1, 2, 3, 4, 5, 6\}$ and $P(X=2) = 4 P(X=4)$,then the parameter $p$ of $X$ is

The items produced by a firm are supposed to contain $5\%$ defective items. The probability that a sample of $8$ items will contain less than $2$ defective items is:

Let a random variable $X$ have a Binomial distribution with mean $8$ and variance $4$. If $P(X \leqslant 2) = \frac{k}{2^{16}}$,then $k$ is equal to:

Let $\lim _{t \rightarrow 0}(1+5 t)^{\frac{1}{t}}=K$ and $X$ be the random variable representing the number of successes in $100$ independent trials. If the probability of success in each trial is $0.05$, then the probability of getting at least one success is

Vedclass Test Series

Exam Paper Generator

Online Exam Module