MTH645 Assignment 2 Solution Fall 2023/2024
Abdul Hadi E-Services
Question 1:
The probability of a certain kind of component surviving a given shock test is \( \frac{3}{4} \). We want to find the probability that exactly 2 of the next 4 components tested survive. The probability mass function (PMF) for the binomial distribution, which models the number of successes (survivals) in a fixed number of independent trials, is given by: \[ P(X = k) = \binom{n}{k} p^k q^{n-k} \] Where:
Probability and Binomial Distribution
For our case, finding the probability of exactly 2 successes (components surviving) in 4 trials:
\[ P(X = 2) = \binom{4}{2} \left(\frac{3}{4}\right)^2 \left(\frac{1}{4}\right)^2 \]
Calculating the binomial coefficient:
\[ \binom{4}{2} = 6 \]
Substituting into the formula:
\[ P(X = 2) = 6 \times \left(\frac{3}{4}\right)^2 \times \left(\frac{1}{4}\right)^2 \]
\[ P(X = 2) = \frac{27}{128} \]
Now, let's find the mean and variance of this distribution:
The mean (\( \mu \)) of a binomial distribution is given by \( \mu = np \):
\[ \mu = 4 \times \frac{3}{4} = 3 \]
The variance (\( \sigma^2 \)) of a binomial distribution is given by \( \sigma^2 = npq \):
\[ \sigma^2 = 4 \times \frac{3}{4} \times \frac{1}{4} = \frac{3}{4} \]
So, the probability that exactly 2 of the next 4 components tested survive is \( \frac{27}{128} \). The mean is 3, and the variance is \( \frac{3}{4} \).
Question No. 2
To find the probability of drawing 1 red ball and 3 green balls in 4 draws from a box containing 5 red balls and 5 green balls, we consider different combinations. The total number of ways to draw 4 balls out of 10 is given by: \[ C(10,4) = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = 210 \] Now, let's consider the combinations for drawing 1 red ball out of 5 and 3 green balls out of 5: \[ C(\text{1 red out of 5}) = C(5,1) = 5 \] \[ C(\text{3 green out of 5}) = C(5,3) = 10 \] To find the total number of ways to draw 1 red and 3 green balls, multiply these two combinations: \[ \text{Ways to draw 1 red and 3 green} = C(\text{1 red}) \times C(\text{3 green}) = 5 \times 10 = 50 \] Now, the probability is given by the ratio of the favorable outcomes to the total outcomes: \[ P(\text{1 red and 3 green}) = \frac{\text{Ways to draw 1 red and 3 green}}{\text{Total ways to draw 4 balls}} = \frac{50}{210} = \frac{5}{21} \] So, the probability of drawing 1 red ball and 3 green balls in 4 draws is \( \frac{5}{21} \).Probability of Drawing 1 Red Ball and 3 Green Balls
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