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 is

A. 0.024

B. 0.452

C. 0.336

D. 0.188

Answer :

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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