The probability of solving a question by thre | vedclass

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Question

(a) $(i)$ This question can also be solved by one student

$(ii)$ This question can be solved by two students simultaneously

$(iii)$ This questio

Vedclass · Accepted Answer

$\frac{{33}}{{48}}$

Vedclass · Answer

(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} 	imes \frac{1}{4} + \frac{1}{4} 	imes \frac{1}{6} + \frac{1}{6} 	imes \frac{1}{2}} ight] + \left[ {\frac{1}{2} 	imes \frac{1}{4} 	imes \frac{1}{6}} ight] = \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}} ight)\,\left( {1 - \frac{1}{4}} ight)\,\left( {1 - \frac{1}{6}} ight)$

$ = 1 - \frac{1}{2}.\frac{3}{4}.\frac{5}{6} = \frac{{33}}{{48}}$.

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

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