To pass a certain marksmanship test, an individual is required to shoot at a target until he hits it six times. He is judged on the number of trials that are necessary to achieve this. If the probability of his hitting a target on any trial is 0.25, what is the probability that he requires 18 shots?

The probability that he requires 18 shots is 0.04785

Step-by-step explanation:

To answer this, we shall be using the negative binomial distribution

From the question;

P = 0.25 , r = 6

q will be 1-p = 1-0.25 = 0.75 Which is the probability of missing a target on any trial

P(X = 18) = (18-1)C(6-1) (0.25)^6 (0.75)^(18-6)

P(X = 28) = 17C5 (0.25)^6 (0.75)^12) = 0.04785

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.

step-by-step explanation:

for exempl this   system of equations   y = x - 3

                                                                y = 2x - 5

look   the line : y = x -3     ( red)

                            the line : y = 2x -5 ( bleus)