Answer:
The most likely outcome is exactly 4 free throws
Step-by-step explanation:
Given
[tex]n = 5[/tex] --- attempts
[tex]p = 82.6\%[/tex] ---- probability of a successful free throw
[tex]p = 0.826[/tex]
Required
A histogram to show the most likely outcome
From the question, we understand that the distribution is binomial.
This is represented as:
[tex]P(X = x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]
For x = 0 to 5, where x represents the number of free throws; we have:
[tex]P(X = x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]
[tex]P(X = 0) = ^5C_0 * 0.826^0 * (1 - 0.826)^{5-0}[/tex]
[tex]P(X = 0) = ^5C_0 * 0.826^0 * (0.174)^{5}[/tex]
[tex]P(X = 0) = 1 * 1 * 0.000159 \approx 0.0002[/tex]
[tex]P(X = 1) = ^5C_1 * 0.826^1 * (1 - 0.826)^{5-1}[/tex]
[tex]P(X = 1) = ^5C_1 * 0.826^1 * (0.174)^4[/tex]
[tex]P(X = 1) = 5 * 0.826 * 0.000917 \approx 0.0038[/tex]
[tex]P(X = 2) = ^5C_2 * 0.826^2 * (1 - 0.826)^{5-2}[/tex]
[tex]P(X = 2) = ^5C_2 * 0.826^2 * (0.174)^{3}[/tex]
[tex]P(X = 2) = 10 * 0.682 * 0.005268 \approx 0.0359[/tex]
[tex]P(X = 3) = ^5C_3 * 0.826^3 * (1 - 0.826)^{5-3}[/tex]
[tex]P(X = 3) = ^5C_3 * 0.826^3 * (0.174)^2[/tex]
[tex]P(X = 3) = 10 * 0.5636 * 0.030276 \approx 0.1706[/tex]
[tex]P(X = 4) = ^5C_4 * 0.826^4 * (1 - 0.826)^{5-4}[/tex]
[tex]P(X = 4) = 5 * 0.826^4 * (0.174)^1[/tex]
[tex]P(X = 4) = 5 * 0.4655 * 0.174 \approx 0.4050[/tex]
[tex]P(X = 5) = ^5C_5 * 0.826^5 * (1 - 0.826)^{5-5}\\[/tex]
[tex]P(X = 5) = ^5C_5 * 0.826^5 * (0.174)^0[/tex]
[tex]P(X = 5) = 1 * 0.3845 * 1 \approx 0.3845[/tex]
From the above computations, we have:
[tex]P(X = 0) \approx 0.0002[/tex]
[tex]P(X = 1) \approx 0.0038[/tex]
[tex]P(X = 2) \approx 0.0359[/tex]
[tex]P(X = 3) \approx 0.1706[/tex]
[tex]P(X = 4) \approx 0.4050[/tex]
[tex]P(X = 5) \approx 0.3845[/tex]
See attachment for histogram
From the histogram, we can see that the most likely outcome is at: x = 4
Because it has the longest vertical bar (0.4050 or 40.5%)
