The probability that a student gets distincti | vedclass

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

Question

(D) Let $n = 5$ be the number of tests and $p = \frac{2}{3}$ be the probability of getting a distinction. Then $q = 1 - p = 1 - \frac{2}{3} = \frac{1}

Vedclass · Accepted Answer

$\frac{64}{81}$

Vedclass · Answer

(D) Let $n = 5$ be the number of tests and $p = \frac{2}{3}$ be the probability of getting a distinction. Then $q = 1 - p = 1 - \frac{2}{3} = \frac{1}{3}$.Using the binomial distribution formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$,we need to find the probability of getting a distinction in at least $3$ tests,which is $P(X \ge 3) = P(X = 3) + P(X = 4) + P(X = 5)$.$P(X = 3) = \binom{5}{3} (\frac{2}{3})^3 (\frac{1}{3})^2 = 10 	imes \frac{8}{27} 	imes \frac{1}{9} = \frac{80}{243}$.$P(X = 4) = \binom{5}{4} (\frac{2}{3})^4 (\frac{1}{3})^1 = 5 	imes \frac{16}{81} 	imes \frac{1}{3} = \frac{80}{243}$.$P(X = 5) = \binom{5}{5} (\frac{2}{3})^5 (\frac{1}{3})^0 = 1 	imes \frac{32}{243} 	imes 1 = \frac{32}{243}$.Summing these probabilities: $P(X \ge 3) = \frac{80 + 80 + 32}{243} = \frac{192}{243}$.Dividing both numerator and denominator by $3$,we get $\frac{64}{81}$.

The probability that a student gets distinction in a Mathematics test is $\frac{2}{3}$. If five such tests are conducted over a certain period of time,then the probability that he gets distinction in at least $3$ tests is

Similar Questions

There are $800$ families with four children in each family. Assuming an equal chance for every child to be a boy or a girl,the number of families expected to have children of both sexes is:

If $A$ and $B$ throw two dice $100$ times each simultaneously,then the probability that both of them will get an even number as the total score at the same time in all the throws is

The probability that an event does not happen in one trial is $0.8$. The probability that the event happens at most once in three trials is:

The probability of a bomb hitting a bridge is $\frac{1}{2}$ and two direct hits are needed to destroy it. The least number of bombs required so that the probability of the bridge being destroyed is greater than $0.9$ is:

The mean and variance of a binomial variable $X$ are $2$ and $1$ respectively. The probability that $X$ takes values greater than $1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module