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

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)

      =

      =

      = 0.0123

hence, the probability is equal to 0.0123

c: y > - 1/6x + 1

step-by-step explanation:

dashed line - meaning this < or > also not equal

solid line - meaning this < or > also equal

your answer is c because y > means up & it's shaded up, but y < means down

hope this ! : )

c

step-by-step explanation: