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Two boys A and B throw a ball at a target. Suppose that the probability that boy A will hit the target on any throw is 1/3 and the probability that boy B will hit the target on any throw is 1/4. Suppose also that boy A throws first and the two boys take turns throwing. Determine the probability that the target will be hit for the first time on the third throw of boy A

Respuesta :

Answer:

the probability is equal to 0.0123

Step-by-step explanation:

given,

Probability of hitting a Target by A (p)= 1/3

Probability of not hitting a Target by A (q) = 1 - 1/3

                                                                 = 2/3

given, problem follows negative binomial distribution

let X be the probability of hitting the target for first time in third throw

X follow NBin. (k = 3 , p = 1/3)

[tex]P(X) = \dfrac{\Gamma(4)}{\Gamma(2)\times \Gamma(4)}\times p^k\times q^x[/tex]

[tex]P(X) = \dfrac{\Gamma(4)}{\Gamma(2)\times \Gamma(4)}\times (\dfrac{1}{3})^k\times \dfrac{2}{3}^1[/tex]

      = [tex]\dfrac{3!}{2! \times 3!} \times (\dfrac{1}{3})^3 \times \dfrac{2}{3}^1[/tex]

      = [tex]\dfrac{1}{81}[/tex]

      = 0.0123

hence, the probability is equal to 0.0123

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