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Hi I have a question about this question Use a binomial calculator to find the following probabilities for your scenario79 % of people who purchase pet insurance are woman. if 8 pet insurance owners are randomly selected. Probability of exactly 6 woman will purchase pet insurance probability of at least 4 men will purchase pet insurance

User Solana
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In this case n = 8

p = 79% = 0.79

q = 1 - p = 1 - 0.79 = 0.21

x for the first question is 6

x for the second question is <=4, so it is the sum of the probabilities when x = 1, x =2, x = 3 and x = 4

Using the formula at the bottom:


P(6)\text{ = }\frac{8!}{(8-6)!\cdot\text{ 6!}}\cdot(0.79)^6\cdot(0.21)^((8-6))=(8!)/(2!\cdot6!)\cdot(0.79)^6\cdot(0.21)^2=(8\cdot7)/(2!)(0.2430874555)\cdot(0.0441)
P(6)\text{ =28}\cdot(0.2430874555)\cdot(0.0441)\text{ = }0.3001643901\text{ }\approx\text{ 0.30}

Probability of exactly 6 woman will purchase pet insurance

P(X = 6) = 0.30 = 30%



Probability of at least 4 men will purchase pet insurance (That means that 4 or less woman out of the 8 randomly selected will purchase pet insuranse):

P(X <= 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = ​0.0659 = 6.59%

Hi I have a question about this question Use a binomial calculator to find the following-example-1
User Max Macfarlane
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