To solve the given problems, we need to use the concept of binomial probability since we have a fixed number of trials (selecting 12 jurors) and each trial has two possible outcomes (being a woman or not being a woman).
Let's denote:
p = probability of selecting a woman = 0.45 (45%)
q = probability of not selecting a woman = 1 - p = 1 - 0.45 = 0.55
a) To find the probability that none of the 12 jurors is a woman:
Using the binomial probability formula, we have:
P(x = 0) = C(12, 0) * p^0 * q^(12-0) = 1 * 1 * 0.55^12 = 0.0135 (rounded to 4 decimal places)
b) To find the probability that at most 4 of the 12 jurors are women:
We need to calculate the probabilities for x = 0, 1, 2, 3, and 4 and add them together:
P(x ≤ 4) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)
= C(12, 0) * p^0 * q^12 + C(12, 1) * p^1 * q^11 + C(12, 2) * p^2 * q^10 + C(12, 3) * p^3 * q^9 + C(12, 4) * p^4 * q^8
= 0.0135 + 0.0757 + 0.1762 + 0.2590 + 0.2535 = 0.7779 (rounded to 4 decimal places)
c) Probability distribution of x:
We need to calculate the probabilities for x = 0, 1, 2, ..., 12. Each value of x corresponds to a specific combination of women and men selected for the jury.
P(x) = C(12, x) * p^x * q^(12-x) for x = 0, 1, 2, ..., 12
d) Using the probability distribution obtained in part c, let's find the requested probabilities:
i. P(x > 6) = P(x = 7) + P(x = 8) + ... + P(x = 12)
ii. P(x ≤ 3) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3)
iii. P(2 ≤ x ≤ 7) = P(x = 2) + P(x = 3) + ... + P(x = 7)
You can calculate these probabilities using the binomial probability formula as shown above.
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