Discussion Forum

If two unbiased dice  are rolled simultaneously until a sum of the number appeared on these dice is either 7 or 11, then the probability that 7 comes before 11, is 

Answer:

Explanation:

Let A  be the events sum appeared on two unbiased dice is 7 and B  be the event sum appeared on two unbiased dice is either 7 or 11.

$\therefore$     P(A)  =$\frac{6}{36}$, P(B) = $\frac{6}{36}+\frac{2}{36}=\frac{2}{9}$

$\therefore$   Required probability,

= $P(A)+P(\overline{B}A)+P(\overline{B}\overline{B}A)+P(\overline{B}\overline{B}\overline{B}A)+.....$

   =$\frac{1}{6}+\frac{7}{9}\times\frac{1}{6}+\left(\frac{7}{9}\right)^{2}\times\frac{1}{6}+$.....

=  $\frac{1}{6}\left(\frac{1}{1-\frac{7}{9}}\right)=\frac{1}{6}\times\frac{9}{2}=\frac{3}{4}$