Find the probability of getting 5 exactly twi | vedclass

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(A) The repeated tossing of a die are Bernoulli trials. Let $X$ represent the number of times of getting $5$ in $7$ throws of the die.Probability of g

Vedclass · Accepted Answer

$21 	imes \frac{5^5}{6^7}$

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(A) The repeated tossing of a die are Bernoulli trials. Let $X$ represent the number of times of getting $5$ in $7$ throws of the die.Probability of getting $5$ in a single throw of the die,$p = \frac{1}{6}$.Therefore,$q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.Clearly,$X$ follows a binomial distribution with $n = 7$ and $p = \frac{1}{6}$.The probability mass function is given by $P(X = x) = ^{n}C_{x} \cdot q^{n-x} \cdot p^{x} = ^{7}C_{x} \cdot (\frac{5}{6})^{7-x} \cdot (\frac{1}{6})^{x}$.We need to find the probability of getting $5$ exactly twice,i.e.,$P(X = 2)$.$P(X = 2) = ^{7}C_{2} \cdot (\frac{5}{6})^{7-2} \cdot (\frac{1}{6})^{2}$.$= \frac{7 	imes 6}{2 	imes 1} \cdot (\frac{5}{6})^{5} \cdot (\frac{1}{6})^{2}$.$= 21 \cdot \frac{5^5}{6^5} \cdot \frac{1}{6^2} = 21 \cdot \frac{5^5}{6^7}$.

Find the probability of getting $5$ exactly twice in $7$ throws of a die.

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