Question

Pair of dice is rolled together till a sum of either 5 or 7 is obtained. Then the probability that 5 comes before 7 is _____.

Answer 1

Let A denote the event that a sum of 5 occur,
B denote the event that a sum of 7 occur and
C denote the event that neither a sum of 5 nor of 7 occur
$P\left(A\right)=\frac{4}{36}=\frac{1}{9},P\left(B\right)=\frac{6}{36}=\frac{1}{6},P\left(C\right)=\frac{26}{36}=\frac{13}{18}$
Therefore required probability
$=P\left(A\right)+P(C\cap A)+P(C\cap C\cap A)+...=P\left(A\right)+P\left(C\right)P\left(A\right)+{\left(P\left(C\right)\right)}^{2}P\left(A\right)+...$
$=\frac{P\left(A\right)}{1-P\left(C\right)}=\frac{2}{5}$