Question

Samples of rejuvenated mitochondria are mutated (defective) in 2% of cases. Suppose 18 samples are studied, and they can be considered to 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. Round your answers to two decimal places (e.g. 98.76). (a) The probability is (b) The probability is (c) The probability is

Answer 1

Kindly check explanation

Explanation:

Given the following :

Probability of mutation = 2% = 0.02

Number of samples(n) = 18

nCk × (p^k) × (1 - p)^(n-k)

P = probability of success

n = number of trials

k = number of success desired

A) P(n, k) = p(18, 0) = 18C0 × 0.02^0 × 0.98^18

P(18,0) = 1 × 1 ×0.6951353 = 0.6951

0.6951 × 100% = 69.50%

B.) At most one Sample is mutated

P(x =0) + p(x = 1)

P(18,0) = 69.5%

P(18, 1) = 18C1 × 0.02^1 × 0.98^17

P(18,0) = 18 × 0.02 × 0.70932176618 = 0.25535

0.2553 × 100% = 25.5%

69.5% + 25.5% = 95.00%

C) More than half the samples are mutated

Half of samples (18 / 2) = 9

P(x> 9) : p(x = 9) + p(x = 10)... +p(x = 18)

Using the binomial distribution calculator to save time ;

P(x > 9) = 0.00