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(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

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