Question:
For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate solve any problem is \(\frac{4}{5}\), then the probability that he is unable to solve less than two problems is :
For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate solve any problem is \(\frac{4}{5}\), then the probability that he is unable to solve less than two problems is :
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Unable to solve fewer than two problems means 0 or 1 failures out of 50; add the two binomial probabilities.
Updated On: Nov 11, 2026
- \(\frac{164}{25}\left(\frac{1}{5}\right)^{48}\)
- \(\frac{316}{25}\left(\frac{4}{5}\right)^{48}\)
- \(\frac{201}{5}\left(\frac{1}{5}\right)^{49}\)
- \(\frac{54}{5}\left(\frac{4}{5}\right)^{49}\)
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The Correct Option is D
Solution and Explanation
Let X be the number of problems solved: \(X\sim\) Binomial\((50, \frac{4}{5})\). Unable to solve less than two means solving 49 or 50.
\(P = {}^{50}C_{50}\left(\frac{4}{5}\right)^{50} + {}^{50}C_{49}\left(\frac{4}{5}\right)^{49}\left(\frac{1}{5}\right)\).
\(= \left(\frac{4}{5}\right)^{49}\left[\frac{4}{5} + 50\cdot\frac{1}{5}\right] = \left(\frac{4}{5}\right)^{49}\left[\frac{4}{5}+10\right] = \frac{54}{5}\left(\frac{4}{5}\right)^{49}\).
Answer: (4) \(\frac{54}{5}\left(\frac{4}{5}\right)^{49}\)
\(P = {}^{50}C_{50}\left(\frac{4}{5}\right)^{50} + {}^{50}C_{49}\left(\frac{4}{5}\right)^{49}\left(\frac{1}{5}\right)\).
\(= \left(\frac{4}{5}\right)^{49}\left[\frac{4}{5} + 50\cdot\frac{1}{5}\right] = \left(\frac{4}{5}\right)^{49}\left[\frac{4}{5}+10\right] = \frac{54}{5}\left(\frac{4}{5}\right)^{49}\).
Answer: (4) \(\frac{54}{5}\left(\frac{4}{5}\right)^{49}\)
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