Before 1918, approximately 40% of the wolves in a region were male, and 60% were female. However, cattle ranchers in this area have made a determined effort to exterminate wolves. From 1918 to the present, approximately 60% of wolves in the region are male, and 40% are female. Biologists suspect that male wolves are more likely than females to return to an area where the population has been greatly reduced. (Round your answers to three decimal places.) (a) Before 1918, in a random sample of 10 wolves spotted in the region, what is the probability that 7 or more were male

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

i think the answer would be c