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The probability of solving a question by three students are $\frac{1}{2},\,\,\frac{1}{4},\,\,\frac{1}{6}$ respectively. Probability of question is being solved will be
$\frac{{33}}{{48}}$
$\frac{{35}}{{48}}$
$\frac{{31}}{{48}}$
$\frac{{37}}{{48}}$
Solution
(a) $(i)$ 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} – \left[ {\frac{1}{2} \times \frac{1}{4} + \frac{1}{4} \times \frac{1}{6} + \frac{1}{6} \times \frac{1}{2}} \right] + \left[ {\frac{1}{2} \times \frac{1}{4} \times \frac{1}{6}} \right] = \frac{{33}}{{48}}$.
Alliter : $P(A \cup B \cup C) = 1 – P(\bar A)P(\bar B)P(\bar C)$
$ = 1 – \left( {1 – \frac{1}{2}} \right)\,\left( {1 – \frac{1}{4}} \right)\,\left( {1 – \frac{1}{6}} \right)$
$ = 1 – \frac{1}{2}.\frac{3}{4}.\frac{5}{6} = \frac{{33}}{{48}}$.
