Samples of rejuvenated mitochondria are mutated (defective) with probability 0.16. Find the probability you need to examine at least 6 samples to find 2 samples containing mutations. Report answer to 3 decimal places
The Correct Answer and Explanation is:
We are given:
- Probability of mutation (success) = p = 0.16
- We are looking for the probability that at least 6 samples are needed to find 2 mutated samples.
This is a Negative Binomial Distribution problem, where:
- We are looking for the probability that the 2nd success occurs on or after the 6th trial.
Step 1: Understanding the Setup
Let XXX be the trial on which the 2nd mutation (success) occurs. Then XXX follows a Negative Binomial distribution:P(X=k)=(k−1r−1)pr(1−p)k−rP(X = k) = \binom{k-1}{r-1} p^r (1-p)^{k – r}P(X=k)=(r−1k−1)pr(1−p)k−r
Where:
- r=2r = 2r=2 (we are looking for the 2nd mutated sample)
- p=0.16p = 0.16p=0.16
- k≥2k \geq 2k≥2 is the trial number where the 2nd success occurs
We want:P(X≥6)=1−P(X≤5)P(X \geq 6) = 1 – P(X \leq 5)P(X≥6)=1−P(X≤5)
So we will calculate:P(X≤5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)P(X \leq 5) = P(X=2) + P(X=3) + P(X=4) + P(X=5)P(X≤5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)
Step 2: Calculate Individual Probabilities
Use the negative binomial formula for each value from 2 to 5:
P(X=2)P(X = 2)P(X=2):
(11)(0.16)2(0.84)0=1⋅0.0256⋅1=0.0256\binom{1}{1} (0.16)^2 (0.84)^0 = 1 \cdot 0.0256 \cdot 1 = 0.0256(11)(0.16)2(0.84)0=1⋅0.0256⋅1=0.0256
P(X=3)P(X = 3)P(X=3):
(21)(0.16)2(0.84)1=2⋅0.0256⋅0.84=0.0430\binom{2}{1} (0.16)^2 (0.84)^1 = 2 \cdot 0.0256 \cdot 0.84 = 0.0430(12)(0.16)2(0.84)1=2⋅0.0256⋅0.84=0.0430
P(X=4)P(X = 4)P(X=4):
(31)(0.16)2(0.84)2=3⋅0.0256⋅0.7056=0.0542\binom{3}{1} (0.16)^2 (0.84)^2 = 3 \cdot 0.0256 \cdot 0.7056 = 0.0542(13)(0.16)2(0.84)2=3⋅0.0256⋅0.7056=0.0542
P(X=5)P(X = 5)P(X=5):
(41)(0.16)2(0.84)3=4⋅0.0256⋅0.5927=0.0607\binom{4}{1} (0.16)^2 (0.84)^3 = 4 \cdot 0.0256 \cdot 0.5927 = 0.0607(14)(0.16)2(0.84)3=4⋅0.0256⋅0.5927=0.0607
Step 3: Add Probabilities
P(X≤5)=0.0256+0.0430+0.0542+0.0607=0.1835P(X \leq 5) = 0.0256 + 0.0430 + 0.0542 + 0.0607 = 0.1835P(X≤5)=0.0256+0.0430+0.0542+0.0607=0.1835
Step 4: Subtract from 1
P(X≥6)=1−0.1835=0.816P(X \geq 6) = 1 – 0.1835 = \boxed{0.816}P(X≥6)=1−0.1835=0.816
Final Answer: 0.816
Explanation
This problem is modeled using the Negative Binomial Distribution, which describes the number of trials required to achieve a fixed number of successes (in this case, 2 mutations). The key here is understanding that the samples are being examined one by one until the second mutated sample is found. The probability of mutation (success) is 0.16, meaning the probability that a sample is not mutated (failure) is 0.84.
We are asked to find the probability that at least 6 samples must be examined to observe the second mutation. This is equivalent to saying that the second success does not occur in the first 5 samples. To compute this, we first calculate the probability that the second success occurs on trials 2 through 5 and subtract this from 1.
The Negative Binomial formula gives us the probability of the r-th success occurring on the k-th trial. We compute these probabilities for k = 2 through 5, and sum them to get the total probability that the second mutation occurs within the first 5 trials. Subtracting this sum from 1 gives the probability that the second mutation happens on the 6th trial or later.
After computing each component carefully and summing them, we find that the probability of needing at least 6 samples is 0.816, or 81.6%. This high value makes sense because with a relatively low mutation rate (16%), it usually takes a number of samples to encounter multiple mutations.
