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Q.
A die is thrown again and again until three sixes are obtained. The probability of obtaining third six in the sixth throw of the die, is
Probability - Part 2
Solution:
Here we do not have a strictly binomial distribution. However, each trial is independent and the probability of obtaining a six in a throw.
$p =\frac{1}{6}$ and $q =\frac{5}{6}$
Now, as the third six is obtained in the six throw, the required probability
= Prob(exactly two sixes in 5 throws). Prob ( a six in sixth throw).
$={ }^{5} C _{2} p ^{2} q ^{3} \cdot p ={ }^{5} C _{2} p ^{3} q ^{3}=\frac{5.4}{1.2}\left(\frac{1}{6}\right)^{3}\left(\frac{5}{6}\right)^{3}$
$=\frac{625}{23328}$