Answer:
the correct answer is B. 10%
Step-by-step explanation:
<em>From the given question, we recall the following,</em>
<em>The probability of Wayne making a three pointer = 21%</em>
<em>Now, what is the the probability he makes his first three pointer on the fourth shot.</em>
<em>The next step is to find a solution to the question</em>
<em>The probability of not shooting a 3 pointer is 1- 0.79 = 1-p=q</em>
<em>All the throws of wayne are independent of each other</em>
<em>If two or ore event are independent we have,</em>
<em>P (AnB) = P(A) * P(B)</em>
<em>P (AnBnC) = P(A) *P (B) *P (C)</em>
<em>P (AnBnCnD) = P(A) * P(B)* P(C) *P(D)</em>
<em>Now, let make first 3 pointer on the 4th shot, we have:</em>
<em>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</em>
<em>which is =10%</em>