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diazemerson
09.10.2019 •
Mathematics
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
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Ответ:
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)
=![\dfrac{3!}{2! \times 3!} \times (\dfrac{1}{3})^3 \times \dfrac{2}{3}^1](/tpl/images/0302/0318/0fcc4.png)
=![\dfrac{1}{81}](/tpl/images/0302/0318/9f60b.png)
= 0.0123
hence, the probability is equal to 0.0123
Ответ: