alleryProbability of bomb hitting target = 0.8. Out of 10 bombs dropped, probability that **exactly 2 miss** equals (with $p = 0.2$ miss):
A${}^{10}C_2 (0.2)^2 (0.8)^8$
B${}^{10}C_2 (0.8)^2 (0.2)^8$
C$0.2 \times 10$
D$0.8 \times 10$
Answer & Solution
Correct answer: A. ${}^{10}C_2 (0.2)^2 (0.8)^8$
**Treat 'miss' as success:** $p = 0.2$.
$P(2 \text{ misses out of 10}) = {}^{10}C_2 (0.2)^2 (0.8)^8 = 45 \times 0.04 \times 0.168 \approx 0.302$.
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