Aspire Faculty ID #16448 · Topic: NIMCET 2009 · Just now
NIMCET 2009

An anti-aircraft gun fires a maximum of four shots. Probabilities of hitting in the 1st, 2nd, 3rd, and 4th shot are 0.4, 0.3, 0.2 and 0.1 respectively. Find the probability that the gun hits the plane.

Solution

Hit at least once = 1 − (miss all shots) 
Miss probabilities: 
1st: $(1 - 0.4) = 0.6$ 
2nd: $(1 - 0.3) = 0.7$ 
3rd: $(1 - 0.2) = 0.8$ 
4th: $(1 - 0.1) = 0.9$ 
Miss all: $0.6 \cdot 0.7 \cdot 0.8 \cdot 0.9 = 0.3024$ 
Hit at least once: $1 - 0.3024 = 0.6976$

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