Question

The mean and variance of a binomial distribution are 30 and 5 respectively. The probability of two successes is (1) \(\frac{875}{6^{34}}\) (2) \(\frac{875}{2.6^{34}}\) (3) \(\frac{875}{6^{36}}\) (4) \({ }^{36} \mathrm{C}_{2}\left(\frac{5}{6}\right)^{34}\left(\frac{1}{6}\right)^{2}\)

Short answer

The probability of two successes is option 4: ^36 C_2 (5/6)^34 (1/6)^2.

Step-by-step solution

Identify the Binomial Distribution Parameters

Let the number of trials be denoted by (n) and the probability of success by (p). The mean () and variance () of a binomial distribution are given by the formulas: = np and = np(1-p). We know that = 30 and = 5.

Solve for the Number of Trials

We have the equations: = np and = np(1-p). Substituting the known values, we get: = np = 30 and = np(1-p) = 5. Dividing the two equations we get, np/np(1-p) = 30/5 which simplifies to 1/(1-p) = 6. Solving for p, we get p=5/6.

Determine the Number of Trials

We already have p=5/6 and we know from step 1 that np=30. Solving for n, we get n = 30 / (5/6) = 36.

Calculate the Probability

Using the binomial probability formula: P(X=k) = C_k p^k (1-p)^(n-k). We want the probability of two successes (k=2): C_2 (5/6)^2 (1/6)^34. Simplifying this: 36! / ((36-2)! 2!) (5/6)^2 (1/6)^34 = ^36C_2(5/6)^34(1/6)^2.

Key concepts

mean and variance

In a binomial distribution, the mean \(\bar{x}\) and variance \(\sigma^2\) are key parameters that help describe the distribution. The mean is simply the expected average number of successes over many trials, while the variance measures how much those successes vary from the mean. For a binomial distribution, the formulas for mean \(\bar{x}\) and variance \(\sigma^2\) are as follows:

Mean: \(\bar{x} = np\)
Variance: \(\sigma^2 = np(1 - p)\)

Here, \('n'\) represents the number of trials, and \('p'\) is the probability of success in each trial. A higher mean indicates more expected successes, while a higher variance suggests greater fluctuation in the number of successes. For example, in our problem, we're given \(\bar{x} = 30\) and \(\sigma^2 = 5\). Substituting these values, we get the following two equations:

\(30 = np\) and \(5 = np(1 - p)\)

These equations help us determine the values of \('n'\) and \('p'\).

probability of success

The probability of success, denoted as \(p\), is a crucial part of a binomial distribution. It represents the likelihood of achieving success in a single trial. Given the mean \(\bar{x} = np\) and variance \(\sigma^2 = np(1 - p)\), we can solve for \(p\).

First, by dividing the two given formulas, we get:
\(\frac{30}{5} = \frac{np}{np(1 - p)} = \frac{1}{1 - p}\)
Solving this, we find that \(6 = \frac{1}{1 - p}\), thus \(1 - p = \frac{1}{6}\).
So, \(p = \frac{5}{6}\)

This means the probability of success in each trial is \(\frac{5}{6}\), or roughly 83.33%. This value is important as it influences the overall distribution and helps us compute further probabilities.

binomial probability formula

The binomial probability formula calculates the likelihood of a given number of successes in a fixed number of trials. The formula is:

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

Here, \(X\) is the random variable (number of successes), \(n\) is the total number of trials, \(k\) is the number of successes you are interested in, \(p\) is the probability of success, and \(\binom{n}{k}\) is the binomial coefficient, representing combinations.
In our problem, we're calculating the probability of exactly 2 successes with \(n = 36\) trials, \(p = \frac{5}{6}\), and \(k = 2\). Using the formula:

\[^36C_2 \left ( \frac{5}{6} \right )^2 \left( \frac{1}{6} \right)^{36 - 2}\]

First, compute the binomial coefficient \(\binom{36}{2} = \frac{36!}{2!(36 - 2)!}\), then:

\[^36C_2 \left ( \frac{5}{6} \right )^2 \left ( \frac{1}{6} \right )^{34}\]

Substituting the numbers, you find the probability of getting exactly 2 successes: The answer confirms that it matches option (4), verifying the calculation.

This formula is powerful for determining specific event probabilities in binomial settings.