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

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  • A random variable $X$ following binomial distribution with parameter n and p its probability of 'r' succeses $\rightarrow$ $p(X=r)=C^{n}_{r} p^{r} q^{n-r}$, where p is probability of success and $q=1-p$ and $r=0,1,\dots,n$
We need to find the P (obtaining the third six in 6th throw of the die) = P (2 sixes in first 5 throws) + P (1 six in sixth throw)
Since P (getting a six) = $\large\frac{1}{6} \rightarrow$$ P (\text{obtaining the third six in 6th throw of the die}) = P (\text{2 sixes in first 5 throws}) \times \large\frac{1}{6}$
P (getting 2 sixes in first 5 throws) $= C^{5}_{2} \large\frac{1}{6}^{2} \large\frac{5}{6}^{5-2} = \frac{5 \times 4}{1 \times 2} \frac{5^3}{6^5} = \frac{625}{3888}$
$P (\text{obtaining the third six in 6th throw of the die}) = \large \frac{625}{3888} \times \frac{1}{6} = \frac{625}{23328}$
answered Sep 23, 2013 by sreemathi.v
 

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