The probability distribution for Michael making two free throw attempts regularly, is option D : P (X = 0, 1, 2) = 0. 3025, 0.495, 0.2025 respectively.
Probability Distribution shows probabilities of occurrence of different possible outcomes of experiment.
When there are multiple trials of an event with success failure probabilities, like 2 trials of free throw attempts with success probability = 0.45
Then Binomial probability is suitably used, where Prob = [tex]N c R . P^r .Q^(n-r)[/tex] : N = Number of trials, R = number of successes, P = success probability, Q = failure probability
- P (success = 0) = [tex]2 c 0 . (0.45)^0 . (0.55)^2[/tex] = 0.3025
- P (success = 1) = [tex]2 c 1. (0.55)^1 (0.45)^1[/tex]= 0.495
- P (success = 2) = [tex]2 c 2. (0.45)^2 (0.55)^1[/tex]= 0.2025
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