Consider 5 independent Bernoulli trials,each | vedclass

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

Question

(A) Let $X$ be the number of successes in $5$ independent Bernoulli trials. $X$ follows a binomial distribution $B(n, p)$ with $n = 5$. The probabilit

Vedclass · Accepted Answer

$[0, 1/2]$

Vedclass · Answer

(A) Let $X$ be the number of successes in $5$ independent Bernoulli trials. $X$ follows a binomial distribution $B(n, p)$ with $n = 5$. The probability of success is $p$ and the probability of failure is $q = 1 - p$. The probability of at least one failure is $P(X Since $P(X = 5) = p^5$,the probability of at least one failure is $1 - p^5$. We are given that $1 - p^5 \ge 31/32$. Rearranging the inequality: $1 - 31/32 \ge p^5$,which simplifies to $1/32 \ge p^5$. Taking the fifth root on both sides: $(1/2)^5 \ge p^5$,which implies $p \le 1/2$. Since $p$ is a probability,$p \ge 0$. Thus,$p \in [0, 1/2]$.

Consider $5$ independent Bernoulli trials,each with a probability of success $p$. If the probability of at least one failure is greater than or equal to $31/32$,then in which interval does $p$ lie?

Similar Questions

In a family with $4$ children,the probability that there are at least two girls is:

From a lot containing $10$ defective and $90$ non-defective bulbs,$8$ bulbs are selected one by one with replacement. Then the probability of getting at least $7$ defective bulbs is:

$A$ man makes attempts to hit the target. The probability of hitting the target is $\frac{3}{5}$. Then the probability that he hits the target exactly $2$ times in $5$ attempts is:

It is given that the discrete random variable is $X \sim B(n, p)$ and $P(X=2)=P(X=3)$. Find the mean of the distribution.

Suppose $X$ follows a binomial distribution with parameters $n$ and $p$,where $0 < p < 1$. If $\frac{P(X=r)}{P(X=n-r)}$ is independent of $n$ for every $r$,then $p$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module