Medical statistics indicate that 20% of the population will have cancer in their lifetimes. A group of 12 babies is selected at random. a. What is the probability that exactly 5 in the group will get cancer in their lifetime? b. What is the probability that at least 5 will get cancer in their lifetime? Record the answer to four decimal places (x.)

a. 5.3%

b. 7.3%

Step-by-step explanation:

we have to:

n = 12

p = 0.2

q = 1 - 0.2 = 0.8

We apply the binomial formula:

p (x) = nCx * p ^ x * p ^ (n-x)

a. P (x = 5) = 12C5 * 0.2 ^ 5 * 0.8 ^ (12-5)

12C5 = 12! / (5! * (12 -5)!) = 792

P (x = 5) = 792 * 0.2 ^ 5 * 0.8 ^ (12-5)

P (x = 5) = 0.053

which means that the probability is 5.3%

b. P (x => 5) = 1 - p (x <5)

P (x => 5) = 1 - p (x <= 4)

P (x => 5) = 1 - [P (x = 0) + P (x = 1) + P (x = 2) + P (x = 3) + P (x = 4)]

P (x = 0) = 12C0 * 0.2 ^ 0 * 0.8 ^ (12-0) = 0.069

P (x = 1) = 12C1 * 0.2 ^ 1 * 0.8 ^ (12-1) = 0.206

P (x = 2) = 12C2 * 0.2 ^ 2 * 0.8 ^ (12-2) = 0.283

P (x = 3) = 12C3 * 0.2 ^ 3 * 0.8 ^ (12-3) = 0.236

P (x = 4) = 12C4 * 0.2 ^ 4 * 0.8 ^ (12-4) = 0.133

replacing:

P (x => 5) = 1 - [0.069 + 0.206 + 0.283 + 0.236 + 0.133]

P (x => 5) = 1 - 0.927

P (x => 5) = 0.073

Which means that the probability is 7.3%

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

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step-by-step explanation: