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Question

A, B, C are aiming to shoot a ballon, A will succeed 4 times out of 5 attempts. The chance of B to shoot the ballon is 3 out of 4 and that of C is 2 out of 3. If the three aim the ballon simultaneously, then find the probability that atleast two of them hit the balloon.

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Solution

Let the event of A hitting the ballon be A and the event of A missing the balloon be ¯A .
Similarly, we define B,¯B and C,¯C for B and C, respectively.

The event of atleast 2 hitting the balloon can be denoted as
(A,B,¯C)(A,¯B,C)(¯A,B,C)(A,B,C)

Let the events below be denoted as
(A,B,¯C)X
(A,¯B,C)Y
(¯A,B,C)Z
(A,B,C)T.

The events X,Y,Z,T are mutually exclusive. Therefore, the probability

P(XYZT)=P(X)+P(Y)+P(Z)+P(T)
=P(A,B,¯C)+P(A,¯B,C)+P(¯A,B,C)+P(A,B,C)

Thus, we have:
P(atleast 2 hitting the balloon) = P(A,B,¯C)+P(A,¯B,C)+P(¯A,B,C)+P(A,B,C)

Given that:
P(A)=45
P(B)=34
P(C)=23

P(¯A)=145=15
P(¯B)=134=14
P(¯C)=123=13

The events A, B, C are independent. Hence,

P(A,B,C)=P(A)×P(B)×P(C)
=45×34×23=25

Similarly,

P(A,B,¯C)=P(A)×P(B)×P(¯C)
=45×34×13=15
P(A,¯B,C)=P(A)×P(¯B)×P(C)
=45×14×23=215
P(¯A,B,C)=P(¯A)×P(B)×P(C)
=15×34×23=110

The probability of atleast 2 of them hitting the balloon:

P(A,B,¯C)+P(A,¯B,C)+P(¯A,B,C)+P(A,B,C)
=25+15+215+110

=56

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