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

the perimeter is increased by a factor of 2

3. parallelogram

step-by-step explanation:

we turn each of our sides into direction vectors

ab=b-a=(3 - 7, 1 - 0) = (-4, 1)

bc=c-b=(1,6)

cd=d-c=(4,-1)

da=a-d=(-1,-6)

we see how the opposite sides have negated direction vectors; this is characteristic of a parallelogram.   we have congruent sides ab=cd and bc=da with unequal lengths because 1^2+6^2 is different from (-4)^2+1^2.   so it's not a square or a rhombus.  

we check the dot product of direction vectors of adjacent sides.   we have

ab dot bc = -4(1) + 1(6) = 2

that's not zero so the sides are not perpendicular.   we've ruled out rectangle.

the direction vectors indicate two pairs of congruent parallel sides, so we choose parallelogram over trapezoid.