The probability of A hitting a target is 3/7 and that of B hitting is 1/3.They both fire at the find the probability that
(i)at least one of them will hit the target .
(ii)only one of them will hit the target
Answers
2.only one of them will hit the target is zero
(i) P(atleast one of them will hit the target) = 13/21
(ii)P(only one of them will hit the target) = 10/21
- Given,
P(A hits target) = 3/7
⇒ P(A does not hit target) = 1 - (3/7) = (7-3)/7
= 4/7
P(B hits the target) = 1/3
⇒ P(B does not hit the target) = 1 - 1/3 = (3-1)/3
= 2/3
- Let us suppose that A and B are independent.
- P(none of them hits the target) = (4/7)(2/3)
= 8/21
- (i) P(at least one of them will hit the target) = 1 - P( none of them will hit the target)
= 1 - 8/21
= (21-8)/21
= 13/21
- P(A and B will hit the target) = P(A∩B)
= P(A)P((B)
= (3/7)(1/3)
= 3/21
- (ii) P(only one of them will hit the target) = P(A XOR B)
= P(A) + P(B) - 2P(A∩B)
= 3/7 + 1/3 - 2(3/21)
= (16 - 6)/21
= 10/21