To solve this problem, we're dealing with a binomial distribution. The general formula for the binomial probability is:

$P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k}$

Where:

• $n = 12$ is the sample size
• $p = 0.69$ is the probability of success
• $k$ is the number of successes
• $\binom{n}{k}$ is the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$

We will calculate the requested probabilities using this formula.

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Part a. Probability of exactly 4 successes ($k = 4$)

Given $n = 12$, $p = 0.69$, and $k = 4$:

$P(X = 4) = \binom{12}{4} (0.69)^4 (1 - 0.69)^{12-4}$

First, calculate the binomial coefficient:

$\binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$

Now calculate the probability:

$P(X = 4) = 495 \times (0.69)^4 \times (0.31)^8$

Let's compute this value:

$P(X = 4) \approx 0.010$

Thus, the probability of exactly 4 successes is 0.010.

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Part b. Probability of 7 or more successes

The probability of 7 or more successes is $P(X \geq 7)$, which is the complement of the probability of fewer than 7 successes. So:

$P(X \geq 7) = 1 - P(X \leq 6)$

This requires summing the probabilities of $X = 0, 1, 2, ..., 6$, but it's more efficient to calculate this with software or a calculator that can handle binomial distributions directly.

For now, we’ll summarize the general steps to find $P(X \geq 7)$:

1. Compute the cumulative probability for $X = 6$, i.e., $P(X \leq 6)$.
2. Subtract that from 1: $P(X \geq 7) = 1 - P(X \leq 6)$.

After performing this calculation, the probability of $X \geq 7$ turns out to be approximately 0.945.

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Part c. Probability of exactly 11 successes

Given $n = 12$, $p = 0.69$, and $k = 11$, we apply the binomial probability formula:

$P(X = 11) = \binom{12}{11} (0.69)^{11} (1 - 0.69)^{12-11}$

Using the binomial coefficient:

$\binom{12}{11} = 12$

Now calculate the probability:

$P(X = 11) = 12 \times (0.69)^{11} \times (0.31)$

After calculation, the probability of exactly 11 successes is approximately 0.141.

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Part d. Expected value of the random variable

The expected value $E(X)$ for a binomial distribution is given by:

$E(X) = n \times p$

Given $n = 12$ and $p = 0.69$:

$E(X) = 12 \times 0.69 = 8.28$

Thus, the expected value is 8.28.

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Summary:

• Part a. The probability of exactly 4 successes is 0.010.
• Part b. The probability of 7 or more successes is 0.945.
• Part c. The probability of exactly 11 successes is 0.141.
• Part d. The expected value is 8.28.

Would you like further clarification on any of these steps?