Answer:
P(≥ 7 males) = 0.0548
Step-by-step explanation:
This is a binomial probability distribution problem.
We are told that Before 1918;
P(male) = 40% = 0.4
P(female) = 60% = 0.6
n = 10
Thus;probability that 7 or more were male is;
P(≥ 7 males) = P(7) + P(8) + P(9) + P(10)
Now, binomial probability formula is;
P(x) = [n!/((n - x)! × x!)] × p^(x) × q^(n - x)
Now, p = 0.4 and q = 0.6.
Also, n = 10
Thus;
P(7) = [10!/((10 - 7)! × 7!)] × 0.4^(7) × 0.6^(10 - 7)
P(7) = 0.0425
P(8) = [10!/((10 - 8)! × 8!)] × 0.4^(8) × 0.6^(10 - 8)
P(8) = 0.0106
P(9) = [10!/((10 - 9)! × 9!)] × 0.4^(9) × 0.6^(10 - 9)
P(9) = 0.0016
P(10) = [10!/((10 - 10)! × 10!)] × 0.4^(10) × 0.6^(10 - 10)
P(10) = 0.0001
Thus;
P(≥ 7 males) = 0.0425 + 0.0106 + 0.0016 + 0.0001 = 0.0548
%69 of visited only one place
I don’t know what answers you have but I think it’s 23.4
V=(4/3)(pi)(r^3)
V=(4/3)(pi)(8^3)
V=(4/3)(pi)(512)
V=(2048/3)(pi), or around 2144.660585
