Probability Question 192
Question: One dice is thrown three times and the sum of the thrown numbers is 15. The probability for which number 4 appears in first throw
[MP PET 2004]
Options:
A) $ \frac{1}{18} $
B) $ \frac{1}{36} $
C) $ \frac{1}{9} $
D) $ \frac{1}{3} $
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Answer:
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( \frac{A}{B} ) $ .
But $ P( \frac{A}{B} )=\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(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(A\cap B)=2 $ , $ n\text{ }(B)=36 $ ;
$ \therefore $ $ P( \frac{A}{B} )=\frac{2}{36}=\frac{1}{18} $ .
BETA
