1. a. P(3 faulty) = C(10,3) * (0.1)^3 * (0.9)^7 ≈ 0.0574
b. P(at least 1 faulty) = 1 - P(0 faulty) = 1 - (0.9)^10 ≈ 0.6513
2. a. P(none break down) = (0.8)^10 ≈ 0.1074
b. P(at least one breaks down) = 1 - P(none break down) = 1 - (0.8)^10 ≈ 0.8926
c. P(exactly three break down) = C(10,3) * (0.2)^3 * (0.8)^7 ≈ 0.2013
d. Expected repair cost = 10 * 0.2 * $125 = $250
3. a. Assuming random guessing, p = 1/5 = 0.2 for each trial.
P(X=0) = C(5,0) * (0.2)^0 * (0.8)^5 = 0.32768
P(X=1) = C(5,1) * (0.2)^1 * (0.8)^4 = 0.4096
P(X=2) = C(5,2) * (0.2)^2 * (0.8)^3 = 0.2048
P(X=3) = C(5,3) * (0.2)^3 * (0.8)^2 = 0.0512
P(X=4) = C(5,4) * (0.2)^4 * (0.8)^1 = 0.0064
P(X=5) = C(5,5) * (0.2)^5 * (0.8)^0 = 0.00032
b. Outcomes to consider psychic: X=4 or X=5, as these are highly unlikely under random guessing (combined probability ≈ 0.00672), suggesting performance beyond chance.
Parent Tip: Review the logic above to help your child master the concept of binomial probability worksheet.
