Probability
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Q.

The probability of success of three students $X, Y$ and $Z$ in the one examination are 1/5, 1/4 and 1/3 respectively.

Find the probability of success of at least two.

 A.

1/6

 B.

2/5

 C.

3/4

 D.

3/5

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Solution:
Option(A) is correct

$P(X)= 1/5, P(Y)= 1/4, P(Z)= 1/3$

Required probability:

$\begin{align*}
=& [P(A)P(B)\{1-P(C)\}] + [\{1-P(A)\}P(B)P(C)]\\
& + [P(A)P(C)\{1-P(B)\}] + P(A)P(B)P(C)
\end{align*}$

$\begin{align*}
=&\left[\dfrac{1}{4} × \dfrac{1}{3} × \dfrac{4}{5}\right] + \left[\dfrac{3}{4} × \dfrac{1}{3} × \dfrac{1}{5}\right]\\
&+ \left[\dfrac{2}{3} × \dfrac{1}{4} × \dfrac{1}{5}\right] + \left[\dfrac{1}{4} × \dfrac{1}{3} × \dfrac{1}{5}\right]
\end{align*}$

$= \dfrac{4}{60} + \dfrac{3}{60} + \dfrac{2}{60} + \dfrac{1}{60}$

$= 10/60$

$= \textbf{1/6}$


(1) Comment(s)


Sam
 ()

Can't I do it like this,

Probability of failure of $A = 4/5$

Probability of failure of $B = 3/4$

Probability of failure of $C = 2/3$

No one succeeds $= 4/5 * 3/4 * 2/3 = 2/5$

Probability that at at least 1 succeeds $= 1 - 2/5 = 3/5$