A) \[\frac{9}{16}\]
B) \[\frac{13}{16}\]
C) \[\frac{11}{16}\]
D) \[\frac{15}{16}\]
Correct Answer: C
Solution :
[c] Second A comes before third B, so this process will be finished either in two draws or in three draws or in four draws. If process is finished in two draws, \[P\,\,(AA)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}\] If process is finished in three draws P (ABA or BAA) \[=2\left( \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2} \right)=\frac{1}{4}\] If process is finished in four draws, P(ABBA or BABA or BBAA) \[=3\left( \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2} \right)=\frac{3}{16}\] Total probability \[=\frac{1}{4}+\frac{1}{4}+\frac{3}{16}=\frac{11}{16}\]
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