Find the probability of throwing at most 2 si | vedclass

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(A) The repeated tossing of the die are Bernoulli trials. Let $X$ represent the number of times of getting sixes in $6$ throws of the die.Probability

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$\frac{35}{18}\left(\frac{5}{6}\right)^{4}$

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(A) The repeated tossing of the die are Bernoulli trials. Let $X$ represent the number of times of getting sixes in $6$ throws of the die.Probability of getting a six in a single throw of a die,$p = \frac{1}{6}$.$	herefore q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.Clearly,$X$ follows a binomial distribution with $n = 6$.$	herefore P(X = x) = ^{n}C_{x} q^{n-x} p^{x} = ^{6}C_{x} \left(\frac{5}{6}ight)^{6-x} \left(\frac{1}{6}ight)^{x}$.$P(	ext{at most } 2 	ext{ sixes}) = P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$.$P(X = 0) = ^{6}C_{0} \left(\frac{5}{6}ight)^{6} = \left(\frac{5}{6}ight)^{6}$.$P(X = 1) = ^{6}C_{1} \left(\frac{5}{6}ight)^{5} \left(\frac{1}{6}ight) = 6 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}ight)^{5} = \left(\frac{5}{6}ight)^{5}$.$P(X = 2) = ^{6}C_{2} \left(\frac{5}{6}ight)^{4} \left(\frac{1}{6}ight)^{2} = 15 \cdot \frac{1}{36} \cdot \left(\frac{5}{6}ight)^{4} = \frac{15}{36} \left(\frac{5}{6}ight)^{4} = \frac{5}{12} \left(\frac{5}{6}ight)^{4}$.Summing these probabilities:$P(X \leq 2) = \left(\frac{5}{6}ight)^{4} \left[ \left(\frac{5}{6}ight)^{2} + \left(\frac{5}{6}ight) + \frac{5}{12} ight]$.$= \left(\frac{5}{6}ight)^{4} \left[ \frac{25}{36} + \frac{30}{36} + \frac{15}{36} ight] = \left(\frac{5}{6}ight)^{4} \left[ \frac{70}{36} ight]$.$= \frac{35}{18} \left(\frac{5}{6}ight)^{4}$.

Find the probability of throwing at most $2$ sixes in $6$ throws of a single die.

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