Answer:
0.8973
Step-by-step explanation:
Relevant data provided in the question as per the question below:
Free throw shooting percentage = 0.906
Free throws = 6
At least = 5
Based on the above information, the probability is
Let us assume the X signifies the number of free throws
So, Then X ≈ Bin (n = 6, p = 0.906)
[tex]P = (X = x) = $\sum\limits_{x}^6 (0.906)^x (1 - 0.906)^{6-x}, x = 0,1,2,3,.., 6[/tex]
Now
The Required probability = P(X ≥ 5) = P(X = 5) + P(X = 6)
[tex]= $\sum\limits_{5}^6 (0.906)^5 (1 - 0.906)^{6-5} + $\sum\limits_{6}^6 (0.906)^6 (1 - 0.906)^{6-6}[/tex]
= 0.8973
