Wayne is at a basketball practice preparing to shoot three pointers. the probability of wine making a three pointer is 21%. what is the probability he makes his first three pointer on the fourth shot.
a. 8%
b. 10%
c. 15%
d. 17%

the correct answer is B. 10%

Step-by-step explanation:

From the given question, we recall the following,

The probability of Wayne making a three pointer = 21%

Now, what is the the probability he makes his first three pointer on the fourth shot.

The next step is to find a solution to the question

The probability of not shooting a 3 pointer is  1- 0.79 = 1-p=q

All the throws of wayne are independent of each other

If two or  ore event are independent we have,

P (AnB) = P(A) * P(B)

P (AnBnC) = P(A) *P (B) *P (C)

P (AnBnCnD) = P(A) * P(B)* P(C) *P(D)

Now, let make first 3 pointer on the 4th shot, we have:

q³= p (0.79)³ (0.21) =10% or (1-0.21) x (1-0.21)x (1-0.21) x 0.21=(0.79) x (0.79) x (0.79) x(0.21)=0.1

which is =10%

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1

i use my brain for prombles like this maybe you could try it

step-by-step explanation: