A basketball player has a 60% accuracy rate for making free throws. During practice, this player fails to make a free throw three times in a row, but is finally successful on the fourth attempt. P (X equals k )equals (1 minus p )to the power of k minus 1 end exponent p Using the geometric distribution formula, what is the probability of this player successfully making a free throw on the fourth attempt

This question is incomplete, the complete question is;

A basketball player has a 60% accuracy rate for making free throws. During practice, this player fails to make a free throw three times in a row, but is finally successful on the fourth attempt.

p(x=k) = p^k (1-p)^k-1

Using the geometric distribution formula, what is the probability of this player successfully making a free throw on the fourth attempt? Each attempt is independent of each other and answer choices are rounded to the hundredths place.

a.) 0.04

b.) 0.60

c.) 0.40

d.) 0.10

a.) 0.04

Step-by-step explanation:

Given that;

The basketball player has 60% accuracy of  free throws

Now the probability of the player successfully making a free throw on the fourth attempt.

probability of success p = 0.60

number of failure k = 3

the given p.d.f of geometric distribution p(x=k) = p^k (1-p)^k-1

{ 0 <p<= 1 , k=0,1,2,3... }

we substitute our given data

p(x=3) = 0.60^3 (1-0.6)^3-1

= (0.60)^3 (0.40)^2

= 0.216 * 0.16

= 0.035 = 0.04 ( OPTION A)

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.

answer: idk wbu

step-by-step explanation: