A) \[\frac{1}{18}\]
B) \[\frac{1}{36}\]
C) \[\frac{1}{9}\]
D) \[\frac{1}{3}\]
Correct Answer: A
Solution :
We have to find the bounded probability to get sum 15 when 4 appears first. Let the event of getting sum 15 of three thrown number is A and the event of apearing 4 is B. So we have to find \[P\text{ }\left( \frac{A}{B} \right)\]. But \[P\,\left( \frac{A}{B} \right)=\frac{n\,(A\cap B)}{n(B)}\] When \[n\,(A\cap B)\] and \[n(B)\] respectively denote the number of digits in \[A\cap B\] and B. Now \[n\text{ }(B)=36\], because first throw is of 4. So another two throws stop by \[6\times 6=36\] types. Three dices have only two throws, which starts from 4 and give sum 15 i.e., (4, 5, 6) and (4, 6, 5). So, \[n\text{ }(A\cap B)=2\], \[n\text{ }(B)=36\]; \[\therefore \]\[P\,\left( \frac{A}{B} \right)=\frac{2}{36}=\frac{1}{18}\].
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