Statistics And Probability Question 565
The probability of solving a question by three students are $ \frac{1}{2},\frac{1}{4},\frac{1}{6} $ respectively. Probability of the question being solved will be
[UPSEAT 1999]
Options:
A) $ \frac{33}{48} $
B) $ \frac{35}{48} $
C) $ \frac{31}{48} $
D) $ \frac{37}{48} $
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Answer:
Correct Answer: A
Solution:
This question can also be solved by one student This question can be solved by two students simultaneously This question can be solved by three students all together. $ P(A)=\frac{1}{2},P(B)=\frac{1}{4},,P(C)=\frac{1}{6} $ $ P(A\cup B\cup C)= $ $ P(A),+P(B)+P(C) $ $ -[P(A)\cdot P(B)+P(B)\cdot P(C)+P(C)\cdot P(A)]+[P(A)\cdot P(B)\cdot P(C)] $ $ =\frac{1}{2}+\frac{1}{4}+\frac{1}{6}-[ \frac{1}{2}\times \frac{1}{4}+\frac{1}{4}\times \frac{1}{6}+\frac{1}{6}\times \frac{1}{2} ]+[ \frac{1}{2}\times \frac{1}{4}\times \frac{1}{6} ]=\frac{33}{48} $ .
BETA
