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Medical statistics indicate that 20% of the population will have cancer in their lifetimes.
A group of 12 babies is selected at random.
What is the probability that at least 5 will get cancer in their lifetime?
Record the answer to four decimal places (x.xxxx)
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This is a binomial distribution problem.
You can use a general formula for a binomial distribution
P( >= 5) = P(5) + P(6) + P(7) + P(8) + P9) + P10) + P(11) + P(12).
Or a more effective formula
P( >= 5) = via the complement = 1 - (P(0) + P(1) + P(2) + P(3) + P(4)) =
= 1 - (C{12,0)*0.2^0*(1-0.2)^12 + C{12,1)*0.2^1*(1-0.2)^(12-1) + C{12,2)*0.2^2*(1-0.2)^(12-2) + C{12,3)*0.2^3*(1-0.2)^(12-3) + C{12,4)*0.2^4*(1-0.2)^(12-4)) =
= 1 - (1*0.2^0*0.8^12 + 12*0.2^1*0.8^11 + 66*0.2^2*0.8^10 + 220*0.2^3*0.8^9 + 495*0.2^4*0.8^8) = 0.072555 (approximately).
ANSWER. This probability is 0.0726 to 4 decimal places.
Solved.
