Question

If the probability of a basketball player scoring on any shot is .75, what is the probability that it will take her more than four shots to first scores on a shot?

a..0156 c) .0039 e) none of these.
b..9961 d) .0118

Answer 1

Option a is correct that is .0156

Explanation:

We have been given the probability of a basketball player scoring is p which is 0.75

So, q is 1-p

`q=1-0.75=0.25`

Here, we need to find the probability that it will take her more than four shots

We will first find the shots for less than 4 in which we take cases for 1,2 and 3 shots.

We know that total probability is 1.

So, probability of more than four shots + probability of less than 4 shots =1

So, probability of less than 4 shots we will use binomial distribution which is

`P(r)=^nC_r \cdotp^r\cdot q^(n-r)`

Here,n=3 and r will be 1,2 and 3 we will consider all three cases

We will add all three cases when r=1,2 and 3

`^3C_1(.75)(.25)^2+^3C_2(.75)^2(.25)^1+^3C_3(.75)^3(.25)^0`

On simplification we will get:0.984375

Probability of three cases that is less than four shots is:=0.984375

So, the required probability is: 1-0.984375=0.015625

Therefore, Option a is correct.

Answer 2

Probability of a basketball player scoring on any shot = .75

Probability of a basketball player does not hit on the right point = 1-0.75=0.25

Probability that it will take her more than four shots to first scores on a shot = F F F F S +F F F F S S+ F F F F S S S+ F F F F S S S S F F F F S SS SS+...........

= F F F F S[1+S+S S+S S S+ S S S S+.......upto infinity]

= F F F F S [
`(1)/(1-S)`],→→→[ sum to infinity of a geometric progression a+a r+ a r²+ a r³,... .....= [
`(a)/(1-r)`], ]

= (0.25) × (0.25) ×(0.25)× (0.25)×(0.75)×[
`(1)/(0.25)`],

= 0.0625 × 0.25×0.75

=0.01171875

= 0.0118 (Approx)→→→→Option (d) is correct .