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STD 11 - 14. probability — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 14. probability1 Mark
MCQ
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
  • ✓
    $\frac{{33}}{{48}}$
  • B
    $\frac{{35}}{{48}}$
  • C
    $\frac{{31}}{{48}}$
  • D
    $\frac{{37}}{{48}}$
✓

Answer

Correct option: A.
$\frac{{33}}{{48}}$
a
(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}}$.

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