What is the most probable number of successes in 10 trials with probability of success `2//3`?

Options

A. A. 10
B. B. 7
C. C. 5
D. D. 4

Answer

To find the most probable number of successes in 10 trials with a probability of success of \( \frac{2}{3} \), we can follow these steps:

### Step 1: Understand the problem
We have a binomial distribution where:
- Number of trials \( n = 10 \)
- Probability of success \( p = \frac{2}{3} \)

### Step 2: Calculate the expected number of successes
The expected number of successes \( E(X) \) in a binomial distribution is given by:
\[ E(X) = n \cdot p \]
Substituting the values:
\[ E(X) = 10 \cdot \frac{2}{3} = \frac{20}{3} \approx 6.67 \]

### Step 3: Determine the most probable number of successes
Since the expected value is approximately \( 6.67 \), the most probable number of successes will be the integer values around this expected value. We need to check the probabilities for \( 6 \) and \( 7 \) successes.

### Step 4: Calculate the probabilities for \( k = 6 \) and \( k = 7 \)
The probability of getting exactly \( k \) successes in a binomial distribution is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

#### For \( k = 6 \):
\[ P(X = 6) = \binom{10}{6} \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^4 \]
Calculating \( \binom{10}{6} = 210 \):
\[ P(X = 6) = 210 \cdot \left(\frac{64}{729}\right) \cdot \left(\frac{1}{81}\right) = 210 \cdot \frac{64}{59049} \]

#### For \( k = 7 \):
\[ P(X = 7) = \binom{10}{7} \left(\frac{2}{3}\right)^7 \left(\frac{1}{3}\right)^3 \]
Calculating \( \binom{10}{7} = 120 \):
\[ P(X = 7) = 120 \cdot \left(\frac{128}{2187}\right) \cdot \left(\frac{1}{27}\right) = 120 \cdot \frac{128}{59049} \]

### Step 5: Compare the probabilities
We need to compare \( P(X = 6) \) and \( P(X = 7) \) to determine which is greater.

### Conclusion
After calculating both probabilities, we find that \( P(X = 7) \) is greater than \( P(X = 6) \). Therefore, the most probable number of successes in 10 trials with a probability of success of \( \frac{2}{3} \) is:

\[ \text{Most probable number of successes} = 7 \]