It is observed that 30 % of the students appe | vedclass

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(A) This problem follows a binomial distribution with $n = 5$ trials and probability of success $p = 0.3$ (science student). The probability of failur

Vedclass · Accepted Answer

$0.3087$

Vedclass · Answer

(A) This problem follows a binomial distribution with $n = 5$ trials and probability of success $p = 0.3$ (science student). The probability of failure is $q = 1 - p = 0.7$. The probability of having $x = 2$ science students is given by the formula $P(X = x) = {}^nC_x \cdot p^x \cdot q^{n-x}$. Substituting the values: $P(X = 2) = {}^5C_2 \cdot (0.3)^2 \cdot (0.7)^3$. Calculating the values: ${}^5C_2 = \frac{5 	imes 4}{2 	imes 1} = 10$. $(0.3)^2 = 0.09$. $(0.7)^3 = 0.343$. Therefore,$P(X = 2) = 10 	imes 0.09 	imes 0.343 = 0.9 	imes 0.343 = 0.3087$.

It is observed that $30 \%$ of the students appearing for a certain entrance test are science students. If $5$ students are randomly selected from this group,the probability of having $2$ science students among these students is

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