A fair die is thrown up to 20 times. The prob | vedclass

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(A) To get the fourth six on the $10^{th}$ throw,we must have exactly $3$ sixes in the first $9$ throws and a six on the $10^{th}$ throw.The probabili

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$\frac{84 	imes 5^6}{6^{10}}$

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(A) To get the fourth six on the $10^{th}$ throw,we must have exactly $3$ sixes in the first $9$ throws and a six on the $10^{th}$ throw.The probability of getting a six in a single throw is $p = \frac{1}{6}$,and the probability of not getting a six is $q = \frac{5}{6}$.The probability of getting exactly $3$ sixes in the first $9$ throws is given by the binomial distribution formula: $P(X=3) = ^{9}C_{3} 	imes (\frac{1}{6})^3 	imes (\frac{5}{6})^6$.The probability of getting a six on the $10^{th}$ throw is $\frac{1}{6}$.Therefore,the required probability is: $P = [^{9}C_{3} 	imes (\frac{1}{6})^3 	imes (\frac{5}{6})^6] 	imes \frac{1}{6}$.Calculating the value: $P = 84 	imes \frac{1}{6^3} 	imes \frac{5^6}{6^6} 	imes \frac{1}{6} = \frac{84 	imes 5^6}{6^{10}}$.

$A$ fair die is thrown up to $20$ times. The probability that on the $10^{th}$ throw,the fourth six appears is:

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