Problem 36 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

The mean and the variance of a binomial distribution are 4 and 2, respectively, then the probability of 2 successes is (a) \(28 / 256\) (b) \(42 / 256\) (c) \(56 / 256\) (d) \(72 / 256\)

Step-by-Step Solution

Answer

The probability of 2 successes is (a) \(\frac{28}{256}\).

Step 1: Establish the Parameters

We know that for a binomial distribution with parameters \(n\) (number of trials) and \(p\) (probability of success), the mean \(\mu\) is given by \(np\) and the variance \(\sigma^2\) is given by \(np(1-p)\). We are given \(\mu = 4\) and \(\sigma^2 = 2\). Thus, we have two equations: \(np = 4\) and \(np(1-p) = 2\).

Step 2: Solve for Probability of Success p

From \(np = 4\), we have \(p = \frac{4}{n}\). Substitute \(p = \frac{4}{n}\) into the variance equation \(np(1-p) = 2\):\[n \cdot \frac{4}{n} \cdot \left(1 - \frac{4}{n}\right) = 2\] Simplifying, \[4 \left(1 - \frac{4}{n}\right) = 2\] \[4 - \frac{16}{n} = 2\]\[\frac{16}{n} = 2\]Thus, \(n = 8\).

Step 3: Calculate Parameters from Step 2

Once \(n = 8\) is known, use \(np = 4\) to find \(p\):\[8p = 4\]\[p = 0.5\]

Step 4: Determine Probability of Two Successes

The probability of getting exactly 2 successes in a binomial distribution is given by the formula:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]Substituting \(n = 8\), \(p = 0.5\), and \(k = 2\):\[P(X = 2) = \binom{8}{2} (0.5)^2 (0.5)^{6}\]\[= 28 \times \frac{1}{4} \times \frac{1}{64}\]\[= \frac{28}{256}\]

Step 5: Check Against Answer Choices

The calculated probability matches the option (a) \(\frac{28}{256}\).

Key Concepts

Probability of Success

In a binomial distribution, the probability of success (denoted as \( p \)) plays a critical role. It's important to understand this concept as it sets the likelihood of a single trial being successful.

Whenever you're dealing with a binomial distribution, you'll encounter trials that can be classified as either a success or a failure. The probability of a success is kept constant across all trials. To find this probability when given the mean of the distribution, you can use the formula:

• Mean \( \mu = np \)

where \( n \) is the number of trials. In the given exercise, once the number of trials \( n = 8 \), you can find \( p \) by rearranging the formula:

\[ 8p = 4 \quad \text{thus} \quad p = 0.5 \]
The probability of success \( p \) is 0.5, meaning there is a 50% chance of a success in each trial. This balanced probability is typical in scenarios where outcomes are equally likely.

Mean of Binomial Distribution

The mean of a binomial distribution is a foundational concept that helps us understand the central tendency of the distribution.

The mean is calculated using the formula:

• Mean \( \mu = np \)

where \( n \) is the number of trials, and \( p \) is the probability of success for each trial. The mean tells us the average number of successes we might expect in \( n \) trials.

In the original exercise, it was stated that \( \mu = 4 \). This implies that out of 8 trials, on average, 4 would be successful. It gives us a mental picture of what the distribution looks like when plotted, centering around this average value.

Understanding the mean also helps predict outcomes over a large number of trials, providing valuable insight into the expected behavior of the distribution.

Variance of Binomial Distribution

The variance in a binomial distribution measures the spread or variability of the distribution. It shows how much the number of successes in different sets of trials can be expected to vary.

The formula used to find the variance is:

• Variance \( \sigma^2 = np(1-p) \)

This formula takes into account both the probability of success \( p \) and the probability of failure \( (1-p) \).

In the context of the exercise, the variance was given as 2. This tells us that there is variability in the number of successful trials, around the mean. It conveys that the actual outcomes might differ from the average (mean) number of successes, due to natural fluctuations. With the number of trials \( n = 8 \) and probability of success \( p = 0.5 \), these values indicate a relatively stable outcome in the distribution, without extreme deviations from the mean.

Understanding variance helps in assessing how tightly the successes are clustered around the mean.