Samples of rejuvenated mitochondria are mutated (defective) with a probability 0.17. Find the probability that at most one sample is mutated in 10 samples. Report answer to 3 decimal places.
The Correct Answer and Explanation is :
To solve this problem, we can use the binomial probability formula because we are dealing with a fixed number of independent trials, where each trial has two possible outcomes (mutated or not mutated).
The binomial probability formula is given by:
[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
]
Where:
- ( n ) = number of trials = 10
- ( k ) = number of successes (mutated samples)
- ( p ) = probability of success (mutation) = 0.17
- ( \binom{n}{k} ) = (\frac{n!}{k!(n-k)!}), the binomial coefficient.
We need to find the probability that at most one sample is mutated, which means ( P(X \leq 1) ). This can be computed as:
[
P(X \leq 1) = P(X = 0) + P(X = 1)
]
Step 1: Calculate ( P(X = 0) )
[
P(X = 0) = \binom{10}{0} (0.17)^0 (1 – 0.17)^{10} = 1 \cdot 1 \cdot (0.83)^{10} = 0.1938
]
Step 2: Calculate ( P(X = 1) )
[
P(X = 1) = \binom{10}{1} (0.17)^1 (0.83)^9 = 10 \cdot 0.17 \cdot (0.83)^9 = 0.3976
]
Step 3: Add ( P(X = 0) ) and ( P(X = 1) )
[
P(X \leq 1) = 0.1938 + 0.3976 = 0.5914
]
Final Answer:
The probability that at most one sample is mutated is 0.591 (to three decimal places).
Explanation (300 words):
The problem involves calculating the probability that at most one of 10 samples is defective, given that each sample is defective with a probability of 0.17. This is a classic example of a binomial distribution, where each sample is an independent trial, and we are counting the number of successes (mutations).
The probability of success (( p )) is 0.17, and the probability of failure (( 1-p )) is 0.83. Using the binomial formula, we computed the probabilities for exactly 0 and 1 mutations, which are ( P(X = 0) ) and ( P(X = 1) ), respectively. These represent scenarios where no samples are defective or only one sample is defective.
To find the total probability of at most one mutation, we summed the probabilities for these two scenarios, ( P(X \leq 1) = P(X = 0) + P(X = 1) ). The calculations used the binomial coefficient, which determines the number of ways to select ( k ) successes from ( n ) trials.
This method ensures an exact calculation, avoiding approximations or unnecessary assumptions. The final result, 0.591, indicates that in 59.1% of cases, 10 randomly selected samples will have 1 or fewer mutations. This result is rounded to three decimal places for precision, aligning with the problem’s requirements.