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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Three persons A, B, and C fire at a target in turn, starting with A.
Their probabilities of hitting the target are $0.4,\ 0.3,\ 0.2$ respectively.
The probability of exactly two hits is:
Solution
**Solution:** Let $A,B,C$ denote hitting events. Probability of exactly 2 hits: \[ P = P(A,B,\bar{C}) + P(A,\bar{B},C) + P(\bar{A},B,C) \] $= (0.4)(0.3)(0.8) + (0.4)(0.7)(0.2) + (0.6)(0.3)(0.2)$ $= 0.096 + 0.056 + 0.036 = 0.188$ $\boxed{\text{Answer: (B) 0.188}}$Online Test Series, Information About Examination,
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