Samples of rejuvenated mitochondria are mutated (defective) in 3% of cases. Suppose 17 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
We are given a binomial probability problem:
- The probability of mutation in a single sample is p=0.03p = 0.03p=0.03
- The number of samples is n=17n = 17n=17
- Each sample is independent
Let XXX be the number of mutated samples out of 17. Then X∼Binomial(n=17,p=0.03)X \sim \text{Binomial}(n = 17, p = 0.03)X∼Binomial(n=17,p=0.03)
We now calculate the required probabilities.
(a) No samples are mutated
This means X=0X = 0X=0P(X=0)=(170)(0.03)0(0.97)17=(0.97)17P(X = 0) = \binom{17}{0} (0.03)^0 (0.97)^{17} = (0.97)^{17}P(X=0)=(017)(0.03)0(0.97)17=(0.97)17P(X=0)≈0.5966⇒59.66%P(X = 0) \approx 0.5966 \quad \Rightarrow \boxed{59.66\%}P(X=0)≈0.5966⇒59.66%
(b) At most one sample is mutated
This means X=0X = 0X=0 or X=1X = 1X=1P(X≤1)=P(X=0)+P(X=1)P(X \leq 1) = P(X = 0) + P(X = 1)P(X≤1)=P(X=0)+P(X=1)
We already have P(X=0)≈0.5966P(X = 0) \approx 0.5966P(X=0)≈0.5966P(X=1)=(171)(0.03)1(0.97)16=17⋅0.03⋅(0.97)16P(X = 1) = \binom{17}{1} (0.03)^1 (0.97)^{16} = 17 \cdot 0.03 \cdot (0.97)^{16}P(X=1)=(117)(0.03)1(0.97)16=17⋅0.03⋅(0.97)16P(X=1)≈17⋅0.03⋅0.6139≈0.3133P(X = 1) \approx 17 \cdot 0.03 \cdot 0.6139 \approx 0.3133P(X=1)≈17⋅0.03⋅0.6139≈0.3133P(X≤1)=0.5966+0.3133=0.9099⇒90.99%P(X \leq 1) = 0.5966 + 0.3133 = 0.9099 \quad \Rightarrow \boxed{90.99\%}P(X≤1)=0.5966+0.3133=0.9099⇒90.99%
(c) More than half the samples are mutated
More than half of 17 is 8.5, so we look for P(X>8)P(X > 8)P(X>8)
We will compute the complement:P(X>8)=1−P(X≤8)P(X > 8) = 1 – P(X \leq 8)P(X>8)=1−P(X≤8)
We can compute P(X≤8)P(X \leq 8)P(X≤8) using a binomial calculator or software. Given that the mutation rate is very low (3%), the chance of having 9 or more mutations in 17 samples is extremely small.
In fact, when computed:P(X>8)≈0.0000⇒0.00%P(X > 8) \approx 0.0000 \quad \Rightarrow \boxed{0.00\%}P(X>8)≈0.0000⇒0.00%
Final Answers:
- (a) The probability is 59.66%
- (b) The probability is 90.99%
- (c) The probability is 0.00%
Explanation:
This is a binomial distribution problem where we are determining the likelihood of a rare event (mutation) occurring among a fixed number of independent trials (samples). The low mutation rate (3%) means most samples are expected not to be mutated. The binomial formula uses combinations and powers of success and failure probabilities. Because the event is rare, it’s extremely unlikely to see many mutations, and the chance of more than half being mutated is nearly zero.
