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Question: A can hit a target 4 times in 5 shots, B can hit the same 3 times in 4 shots and C twice in 3 shots....

A can hit a target 4 times in 5 shots, B can hit the same 3 times in 4 shots and C twice in 3 shots. If they fire one each, the probability that at least two of them hit the target is

A

5/6

B

11/12

C

9/10

D

None of these

Answer

5/6

Explanation

Solution

Required probability

=

P(A∩B∩C‾)+P(A∩B‾∩C)+P(A‾∩B∩C)+P(A∩B∩C)\mathrm { P } ( \mathrm { A } \cap \mathrm { B } \cap \overline { \mathrm { C } } ) + \mathrm { P } ( \mathrm { A } \cap \overline { \mathrm { B } } \cap \mathrm { C } ) + \mathrm { P } ( \overline { \mathrm { A } } \cap \mathrm { B } \cap \mathrm { C } ) + \mathrm { P } ( \mathrm { A } \cap \mathrm { B } \cap \mathrm { C } ) = P(1) P(2)

P(C‾)+P(A)P(B‾)P(C)+P(A‾)P(B)P(C)+P(A)P(B)P(C)\mathrm { P } ( \overline { \mathrm { C } } ) + \mathrm { P } ( \mathrm { A } ) \mathrm { P } ( \overline { \mathrm { B } } ) \mathrm { P } ( \mathrm { C } ) + \mathrm { P } ( \overline { \mathrm { A } } ) \mathrm { P } ( \mathrm { B } ) \mathrm { P } ( \mathrm { C } ) + \mathrm { P } ( \mathrm { A } ) \mathrm { P } ( \mathrm { B } ) \mathrm { P } ( \mathrm { C } )

=

= 5060=56\frac { 50 } { 60 } = \frac { 5 } { 6 }