Probability Question 37
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} $
Show Answer
Answer:
Correct Answer: A
Solution:
(i) This question can also be solved by one student
(ii) This question can be solved by two students simultaneously
(iii) 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)P(B)+P(B)P(C)+P(C)P(A)]+[P(A)P(B)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{11}{16} $
BETA
