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Question

Mathematics Question on Conditional Probability

If two unbiased six-faced dice are thrown simultaneously until a sum of either 77 or 1111 occurs, then the probability that 77 comes before 1111 is

A

14\frac{1}{4}

B

34\frac{3}{4}

C

59\frac{5}{9}

D

518\frac{5}{18}

Answer

34\frac{3}{4}

Explanation

Solution

Let AA be the event of obtained sum of 77 and BB be the event of obtained sum of 1111.
n(A)=(2,5),(5,2),(3,4),(4,3),(1,1),(6,1)=6\therefore n(A)=\\{(2,5),(5,2),(3,4),(4,3),(1,1),(6,1)\\}=6
Now, P(A)=636=16P(A)=\frac{6}{36}=\frac{1}{6}
and n(B)=(5,6),(6,5)=2n(B)=\\{(5,6),(6,5)\\}=2
P(B)=236=118\therefore P(B)=\frac{2}{36}=\frac{1}{18}
C=C= Neither a sum of 1111 nor a sum of 77 shows of
P(C)=36(6+2)36\therefore P(C)=\frac{36-(6+2)}{36}
=36836=2836=79=\frac{36-8}{36}=\frac{28}{36}=\frac{7}{9}
Required probability (p)(p)
=636+636×2836+(2836)2(636)+(2836)3(636)=\frac{6}{36}+\frac{6}{36} \times \frac{28}{36}+\left(\frac{28}{36}\right)^{2}\left(\frac{6}{36}\right)+\left(\frac{28}{36}\right)^{3}\left(\frac{6}{36}\right) \ldots
=16[1179]=\frac{1}{6}\left[\frac{1}{1-\frac{7}{9}}\right]
=16×92=34=\frac{1}{6} \times \frac{9}{2}=\frac{3}{4}