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Farah Puri
Asked: 2022-11-02T04:12:42+05:30 In:

Person A can hit a target 4 times in 5 attempts. Person B – 3 times in four attempts. Person C – 2 times in 3 attempts. They fire a volley. The probability that the target is hit atleast two times is


1. 3/4
2. 1/2
3. 5/6
4. 1

Person A can hit a target 4 times in 5 attempts. Person B – 3 times in four attempts. Person C – 2 times in 3 attempts. They fire a volley. The probability that the target is hit atleast two times is


1. 3/4
2. 1/2
3. 5/6
4. 1
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  1. Correct Answer – Option 3 : 5/6

    Concept:

    P(\(\bar A\) : Not hitting the target)  = 1 – P(A: hitting the target )

     

    Calculations:

    Person A can hit a target 4 times in 5 attempts. Person B – 3 times in four attempts. Person C – 2 times in 3 attempts.

    P(A : hitting the target ) = \(\rm \dfrac 45\)

    ⇒P(\(\bar A\) :Not hitting the target) = \(\rm \dfrac 15\)

    P(B: hitting the target) = \(\rm \dfrac 34\)

    ⇒P(\(\bar B\) :Not hitting the target) = \(\rm \dfrac 14\)

    P(C: hitting the target) = \(\rm \dfrac 23\)

    ⇒P(\(\bar C\) : Not hitting the target) = \(\rm \dfrac 13\)

    The probability that the target is hit atleast two times = \(P(A)P(B)P(\bar C)+P(A)P(\bar B)P(C) + P(\bar A)P(B)P(C) + P(A)P(B)P(C)\)

    ⇒ The probability that the target is hit atleast two times = \(\dfrac 15+ \dfrac 1 {10}+ \dfrac {2}{15}+ \dfrac{2}{5}\)

    ⇒ The probability that the target is hit atleast two times = \(\dfrac 5 6\)

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