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

Question

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 (A) (625/23329) (B) (621/25329) (C) (625/23328) (D) (620/23328)

Tardigrade Admin · Accepted Answer

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} (\frac{1}{6})^{3} (\frac{5}{6})^{3}

= \frac{625}{23328}

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

$\frac{625}{23329}$

$\frac{621}{25329}$

$\frac{625}{23328}$

$\frac{620}{23328}$

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

23329625​

25329621​

23328625​

23328620​

$\frac{625}{23329}$

$\frac{621}{25329}$

$\frac{625}{23328}$

$\frac{620}{23328}$