Samples of rejuvenated mitochondria are mutated(defective) in 1% ofcases. Suppose 15 samples are studied, and they can be consideredto be independent for mutation.

Determine the following probabilities.

a) No samples are mutated.
b) At most one sample is mutated.
c) More than half the samples are mutated.

A)0.86

B)0.99

C)6.047E-13

Step-by-step explanation:

Since p = 0.01 and q= 0.99

Number of event n = 15

Using binomial expansion

A) Probability (No samples are mutated) = 15C0 (p)^0 * (q)^15

= 1*(0.01)^0*(0.99)^15 = 0.86

B)Probability (At most one sample is mutated) = 15C0 * p^0 *q^15 + 15C1 * p^1 * q^14 = 1*(0.01)^0*(0.99)^15 + 15C1 * 0.01 * 0.99^14 = 0.86 + 15*0.01*0.8687 = 0.86 + 0.13 = 0.99

C) Probability (More than half the samples are mutated) = 15C8* 0.01^8 * 0.99^7 + 15C9* 0.01^9 * 0.99^6 +15C10* 0.01^10 *0.99^5 + 15C11 * 0.01^11 * 0.99^4 + 15C12 * 0.01^12 * 0.99^3 + 15C13 * 0.01^13 * 0.99^ 2 + 15C14 * 0.01^14 * 0.99^1 + 15C15 * 0.01^15 * 0.99^0

Probability (More than half the samples are mutated) = 6E-13 + 4.71E-15 + 2.86E-17 + 1.31E-19 + 4.41E-22 + 1.03E-24 + 1.49E-27 + 1E-30 = 6.047E-13

your answer is the frist one

hop i

are ya tryin to ask something or is this a statement