A student appeared in an examination consisti | vedclass

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

Question

(A) Let $X$ be the number of correct answers. Since the student guesses,$X$ follows a binomial distribution $B(n=8, p=1/2)$.We want to find the smalle

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Let $X$ be the number of correct answers. Since the student guesses,$X$ follows a binomial distribution $B(n=8, p=1/2)$.We want to find the smallest $n$ such that $P(X \geq n) $P(X \geq n) = \sum_{r=n}^{8} {}^{8}C_{r} (\frac{1}{2})^{r} (\frac{1}{2})^{8-r} = \frac{1}{2^{8}} \sum_{r=n}^{8} {}^{8}C_{r} $\sum_{r=n}^{8} {}^{8}C_{r} We know that $\sum_{r=0}^{8} {}^{8}C_{r} = 2^{8} = 256$.So,$\sum_{r=n}^{8} {}^{8}C_{r} = 256 - \sum_{r=0}^{n-1} {}^{8}C_{r} $\sum_{r=0}^{n-1} {}^{8}C_{r} > 128$.For $n=5$,$\sum_{r=0}^{4} {}^{8}C_{r} = {}^{8}C_{0} + {}^{8}C_{1} + {}^{8}C_{2} + {}^{8}C_{3} + {}^{8}C_{4} = 1 + 8 + 28 + 56 + 70 = 163$.Since $163 > 128$,the condition holds for $n=5$.For $n=4$,$\sum_{r=0}^{3} {}^{8}C_{r} = 1 + 8 + 28 + 56 = 93$,which is not greater than $128$.Thus,the smallest value of $n$ is $5$.

$A$ student appeared in an examination consisting of $8$ true-false type questions. The student guesses the answers with equal probability. The smallest value of $n$,so that the probability of guessing at least $n$ correct answers is less than $\frac{1}{2}$,is:

Similar Questions

If $X$ and $Y$ are two independent binomial variates,satisfying $B(10, 1/2)$ and $B(8, 1/2)$ respectively,then the probability $P(X + Y = 2)$ is

If the mean and variance of a binomial variable $X$ are $2$ and $1$ respectively,then $P(X>1)=$

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 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:

$A$ die is thrown $5$ times. Getting an odd number is considered a success. Find the variance of the distribution of successes.

Vedclass Test Series

Exam Paper Generator

Online Exam Module