A die is thrown again and again until three s | vedclass

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Question

(A) The probability of getting a six in a single throw of a die is $p = \frac{1}{6}$,and the probability of not getting a six is $q = 1 - p = \frac{5}

Vedclass · Accepted Answer

$\frac{625}{23328}$

Vedclass · Answer

(A) The probability of getting a six in a single throw of a die is $p = \frac{1}{6}$,and the probability of not getting a six is $q = 1 - p = \frac{5}{6}$.For the third six to occur exactly on the $6^{th}$ throw,there must be exactly $2$ sixes in the first $5$ throws,and the $6^{th}$ throw must be a six.The probability of getting exactly $2$ sixes in the first $5$ throws follows the binomial distribution formula $P(X=k) = ^{n}C_{k} p^{k} q^{n-k}$,where $n=5$ and $k=2$.$P(	ext{2 sixes in 5 throws}) = ^{5}C_{2} \left(\frac{1}{6}ight)^{2} \left(\frac{5}{6}ight)^{3} = 10 	imes \frac{1}{36} 	imes \frac{125}{216} = \frac{1250}{7776}$.The probability that the $6^{th}$ throw results in a six is $\frac{1}{6}$.Thus,the required probability is $P = \left(\frac{1250}{7776}ight) 	imes \frac{1}{6} = \frac{1250}{46656} = \frac{625}{23328}$.

$A$ die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.

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