Three students $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ appear for an examination. The probability of $X$…

Three students $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ appear for an examination. The probability of $X$ passing the examination is $\frac{1}{5}$, the probability of $\mathrm{Y}$ passing the examination is $\frac{1}{4}$ and the probability of $\mathrm{Z}$ failing the examination is $\frac{2}{3}$. The probability that atleast two of them pass the exam is
  1. $\frac{1}{6}$
  2. $\frac{2}{5}$
  3. $\frac{3}{4}$
  4. $\frac{3}{5}$

Solution

$P($ pass $)+P($ fail $)=1$ $\begin{aligned} & \mathrm{P}(x \text { passing })+\mathrm{P}(x \text {-fail })\}=1 \\ & \mathrm{P}(x \text { fail })=1-\frac{1}{5}=\frac{4}{5}=P\left(x^{\prime}\right) \\ & \text { Similarly } \mathrm{P}(y \text { fail })=\mathrm{P}\left(y^{\prime}\right)=1-\frac{1}{4}=\frac{3}{4} \\ & \mathrm{P}(z \text { pass })=\mathrm{P}(z)=1-\frac{2}{3}=\frac{1}{3}\end{aligned}$ Probability of atleast 2 passing $=($ Any 2 pass and 1 fail $)$ + (All 3 pass) $\begin{aligned} & =\mathrm{P}(x) \cdot \mathrm{P}(y) \cdot \mathrm{P}\left(z^{\prime}\right)+\mathrm{P}(x) \mathrm{P}\left(y^{\prime}\right) \mathrm{P}(z)+\mathrm{P}\left(x^{\prime}\right) \cdot \mathrm{P}(y) \\ & \mathrm{P}(\mathrm{z})+\mathrm{P}(x) \cdot \mathrm{P}(y) \cdot \mathrm{P}(\mathrm{z})\end{aligned}$ $\begin{aligned} & =\left(\frac{1}{5} \times \frac{1}{4} \times \frac{2}{3}\right)+\left(\frac{1}{5} \times \frac{3}{4} \times \frac{1}{3}\right)+\left(\frac{4}{5} \times \frac{1}{4} \times \frac{1}{3}\right)+\left(\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3}\right) \\ & =\frac{2}{60}+\frac{3}{60}+\frac{4}{60}+\frac{1}{60}=\frac{10}{60}=\frac{1}{6}\end{aligned}$

Asked in: AP EAMCET 2022 (05 Jul Shift 2)

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