Question:
Three students \(X,Y\) and \(Z\) appear for an examination. The probability of \(X\) passing the examination is \(\frac{1}{5}\), the probability of \(Y\) passing the examination is \(\frac{1}{4}\), and the probability of \(Z\) failing the examination is \(\frac{2}{3}\). The probability that at least two of them pass the exam is
Three students \(X,Y\) and \(Z\) appear for an examination. The probability of \(X\) passing the examination is \(\frac{1}{5}\), the probability of \(Y\) passing the examination is \(\frac{1}{4}\), and the probability of \(Z\) failing the examination is \(\frac{2}{3}\). The probability that at least two of them pass the exam is
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For “at least two” out of three events, include all cases where exactly two events occur and where all three events occur.
Updated On: Jun 24, 2026
- \(\frac{1}{6}\)
- \(\frac{2}{5}\)
- \(\frac{3}{4}\)
- \(\frac{3}{5}\)
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The Correct Option is A
Solution and Explanation
Step 1: Write the given probabilities.
Probability that \(X\) passes: \[ P(X)=\frac{1}{5} \] Probability that \(Y\) passes: \[ P(Y)=\frac{1}{4} \] Probability that \(Z\) fails: \[ P(Z')=\frac{2}{3} \] Therefore, probability that \(Z\) passes is \[ P(Z)=1-\frac{2}{3} \] \[ P(Z)=\frac{1}{3} \]
Step 2: Find probabilities of failing for \(X\) and \(Y\).
\[ P(X')=1-\frac{1}{5}=\frac{4}{5} \] \[ P(Y')=1-\frac{1}{4}=\frac{3}{4} \]
Step 3: Cases for at least two students passing.
At least two passing means either exactly two pass or all three pass.
So, required probability is \[ P(XYZ')+P(XY'Z)+P(X'YZ)+P(XYZ) \]
Step 4: Calculate each case.
\[ P(XYZ')=\frac{1}{5}\cdot \frac{1}{4}\cdot \frac{2}{3} \] \[ =\frac{1}{30} \] \[ P(XY'Z)=\frac{1}{5}\cdot \frac{3}{4}\cdot \frac{1}{3} \] \[ =\frac{1}{20} \] \[ P(X'YZ)=\frac{4}{5}\cdot \frac{1}{4}\cdot \frac{1}{3} \] \[ =\frac{1}{15} \] \[ P(XYZ)=\frac{1}{5}\cdot \frac{1}{4}\cdot \frac{1}{3} \] \[ =\frac{1}{60} \]
Step 5: Add all probabilities.
\[ \frac{1}{30}+\frac{1}{20}+\frac{1}{15}+\frac{1}{60} \] Taking LCM \(60\), \[ \frac{2}{60}+\frac{3}{60}+\frac{4}{60}+\frac{1}{60} \] \[ =\frac{10}{60} \] \[ =\frac{1}{6} \]
Step 6: Final conclusion.
Therefore, \[ \boxed{\frac{1}{6}} \]
Probability that \(X\) passes: \[ P(X)=\frac{1}{5} \] Probability that \(Y\) passes: \[ P(Y)=\frac{1}{4} \] Probability that \(Z\) fails: \[ P(Z')=\frac{2}{3} \] Therefore, probability that \(Z\) passes is \[ P(Z)=1-\frac{2}{3} \] \[ P(Z)=\frac{1}{3} \]
Step 2: Find probabilities of failing for \(X\) and \(Y\).
\[ P(X')=1-\frac{1}{5}=\frac{4}{5} \] \[ P(Y')=1-\frac{1}{4}=\frac{3}{4} \]
Step 3: Cases for at least two students passing.
At least two passing means either exactly two pass or all three pass.
So, required probability is \[ P(XYZ')+P(XY'Z)+P(X'YZ)+P(XYZ) \]
Step 4: Calculate each case.
\[ P(XYZ')=\frac{1}{5}\cdot \frac{1}{4}\cdot \frac{2}{3} \] \[ =\frac{1}{30} \] \[ P(XY'Z)=\frac{1}{5}\cdot \frac{3}{4}\cdot \frac{1}{3} \] \[ =\frac{1}{20} \] \[ P(X'YZ)=\frac{4}{5}\cdot \frac{1}{4}\cdot \frac{1}{3} \] \[ =\frac{1}{15} \] \[ P(XYZ)=\frac{1}{5}\cdot \frac{1}{4}\cdot \frac{1}{3} \] \[ =\frac{1}{60} \]
Step 5: Add all probabilities.
\[ \frac{1}{30}+\frac{1}{20}+\frac{1}{15}+\frac{1}{60} \] Taking LCM \(60\), \[ \frac{2}{60}+\frac{3}{60}+\frac{4}{60}+\frac{1}{60} \] \[ =\frac{10}{60} \] \[ =\frac{1}{6} \]
Step 6: Final conclusion.
Therefore, \[ \boxed{\frac{1}{6}} \]
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