Q. 48

# Mark the correct alternative in each of the following:Three person, A, B and C fire a target in turn starting with A. Their probabilities of hitting the target are 0.4, 0.3 and 0.2 respectively. The probability of two hits isA. 0.024B. 0.452C. 0.336D. 0.188

A = Event of Person A hitting the target.

P(A) = 0.4

P = 1 – P(A) = 1 – 0.4

P = 0.6 (1)

B = Event of Person B hitting the target.

P(B) = 0.3

P = 1 – P(B) = 1 – 0.3

P = 0.7 (2)

C = Event of Person C hitting the target.

P(C) = 0.2

P = 1 – P(C) = 1 – 0.2

P = 0.8 (3)

Now then,

= 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

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