The minimum number of times a fair coin needs | vedclass

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

Question

(D) Let $n$ be the number of tosses. The probability of getting $k$ heads in $n$ tosses is given by the binomial distribution $P(X=k) = {^nC_k} (\frac

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Let $n$ be the number of tosses. The probability of getting $k$ heads in $n$ tosses is given by the binomial distribution $P(X=k) = {^nC_k} (\frac{1}{2})^n$.The probability of getting at least two heads is $P(X \ge 2) = 1 - P(X=0) - P(X=1)$.$P(X=0) = {^nC_0} (\frac{1}{2})^n = \frac{1}{2^n}$.$P(X=1) = {^nC_1} (\frac{1}{2})^n = \frac{n}{2^n}$.So,$P(X \ge 2) = 1 - \frac{1+n}{2^n}$.We want $1 - \frac{1+n}{2^n} \ge 0.96$,which implies $\frac{1+n}{2^n} \le 0.04 = \frac{1}{25}$.Testing values for $n$:For $n=7$: $\frac{1+7}{2^7} = \frac{8}{128} = \frac{1}{16} = 0.0625 > 0.04$.For $n=8$: $\frac{1+8}{2^8} = \frac{9}{256} \approx 0.03515 Thus,the minimum number of tosses required is $8$.

The minimum number of times a fair coin needs to be tossed so that the probability of getting at least two heads is at least $0.96$ is:

Similar Questions

The mean and variance of a binomial distribution are $x$ and $5$ respectively. If $x$ is an integer,then the possible values for $x$ are

If the mean and variance of a binomial variable $X$ are $2$ and $1$ respectively,then what is the probability that $X \geq 1$?

$A$ random variable $X$ follows a binomial distribution in which the difference between its mean and variance is $1$. If $2 P(X=2)=3 P(X=1)$,then $n^2 P(X>1)=$

What is the probability of getting at least $1$ tail when a coin is tossed $4$ times?

$A$ random variable $X \sim B(n, p)$. If the values of the mean and variance of $X$ are $18$ and $12$ respectively,then the total number of possible values of $X$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module