**MTH645 Assignment 2 Solution Fall 2023/2024**

**Abdul Hadi E-Services**

__Question 1:__** **

# Probability and Binomial Distribution

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:

- \( n \) is the number of trials,
- \( k \) is the number of successes,
- \( p \) is the probability of success on a single trial,
- \( q \) is the probability of failure on a single trial, and
- \( \binom{n}{k} \) is the binomial coefficient.

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__

# Probability of Drawing 1 Red Ball and 3 Green Balls

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 \( \frac{5}{21} \).

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