The probability of success of three students X,Y and Z in the one examination are $$\frac{{1}}{{5}}$$, $$\frac{{1}}{{4}}$$ and $$\frac{{1}}{{3}}$$ respectively. Find the probability of success of at least two.
A. $$\frac{{1}}{{6}}$$
B. $$\frac{{2}}{{5}}$$
C. $$\frac{{3}}{{4}}$$
D. $$\frac{{3}}{{5}}$$
Answer: Option A
Solution(By Examveda Team)
$$P(X) = \frac{1}{5},$$ $$P(Y) = \frac{1}{4},$$ $$P(Z) = \frac{1}{3}$$Required probability:
$$ = \left[ {P\left( A \right)P\left( B \right)\left\{ {1 - P\left( C \right)} \right\}} \right]$$ $$ + $$ $$\left[ {\left\{ {1 - P\left( A \right)} \right\}P\left( B \right)P\left( C \right)} \right] + $$ $$\left[ {P\left( A \right)P\left( C \right)\left\{ {1 - P\left( B \right)} \right\}} \right] + $$ $$\left[ {P\left( A \right)P\left( B \right)P\left( C \right)} \right]$$
$$ = \left[ {\frac{1}{4} \times \frac{1}{3} \times \frac{4}{5}} \right] + $$ $$\left[ {\frac{3}{4} \times \frac{1}{3} \times \frac{1}{5}} \right] + $$ $$\left[ {\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}} \right] + $$ $$\left[ {\frac{1}{4} \times \frac{1}{3} \times \frac{1}{5}} \right]$$
$$\eqalign{ & = \frac{4}{{60}} + \frac{3}{{60}} + \frac{2}{{60}} + \frac{1}{{60}} \cr & = \frac{{10}}{{60}} \cr & = \frac{1}{6} \cr} $$
Related Questions on Probability
A. $$\frac{{1}}{{2}}$$
B. $$\frac{{2}}{{5}}$$
C. $$\frac{{8}}{{15}}$$
D. $$\frac{{9}}{{20}}$$
A. $$\frac{{10}}{{21}}$$
B. $$\frac{{11}}{{21}}$$
C. $$\frac{{2}}{{7}}$$
D. $$\frac{{5}}{{7}}$$
A. $$\frac{{1}}{{3}}$$
B. $$\frac{{3}}{{4}}$$
C. $$\frac{{7}}{{19}}$$
D. $$\frac{{8}}{{21}}$$
E. $$\frac{{9}}{{21}}$$
What is the probability of getting a sum 9 from two throws of a dice?
A. $$\frac{{1}}{{6}}$$
B. $$\frac{{1}}{{8}}$$
C. $$\frac{{1}}{{9}}$$
D. $$\frac{{1}}{{12}}$$
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