Question

In a series of 10 Bernoulli trials, if the probability of success in each trial is p = 0.3, what is the probability of exactly 4 successes?

• A. 0.2668
• B. 0.2001
• C. 0.2150
• D. 0.3823

Answer 1

Answer: (B)

0.2001

Answer 2

This problem can be solved using the binomial probability formula, which is applicable for a series of Bernoulli trials.

1. Understand the Binomial Probability Formula: The probability of exactly 'k' successes in 'n' Bernoulli trials is given by: $P(X=k) = C_k^n p^k (1-p)^{n-k}$ where:

• $n$ = total number of trials
• $k$ = number of desired successes
• $p$ = probability of success in a single trial
• $(1-p)$ = probability of failure in a single trial (often denoted as $q$)
• $C_k^n$ = binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$

2. Identify the Given Values: From the question:

• Number of trials, $n = 10$
• Number of desired successes, $k = 4$
• Probability of success in each trial, $p = 0.3$

3. Calculate the Probability of Failure (q): $q = 1 - p = 1 - 0.3 = 0.7$

4. Calculate the Binomial Coefficient ($C_k^n$): $C_4^{10} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!}$ $C_4^{10} = \frac{10 \times 9 \times 8 \times 7 \times 6!}{4 \times 3 \times 2 \times 1 \times 6!}$ $C_4^{10} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}$ $C_4^{10} = \frac{5040}{24}$ $C_4^{10} = 210$

5. Calculate the Powers of p and q: $p^k = (0.3)^4 = 0.3 \times 0.3 \times 0.3 \times 0.3 = 0.0081$ $q^{n-k} = (0.7)^{10-4} = (0.7)^6$ $(0.7)^6 = (0.7^3)^2 = (0.343)^2 = 0.117649$

6. Substitute the Values into the Formula: $P(X=4) = C_4^{10} \times (0.3)^4 \times (0.7)^6$ $P(X=4) = 210 \times 0.0081 \times 0.117649$ $P(X=4) = 1.701 \times 0.117649$ $P(X=4) = 0.200099949$

7. Round to the Nearest Option: Rounding the result to four decimal places, we get 0.2001.

The final answer is $\boxed{\text{0.2001}}$

Explanation of the solution:

The problem involves finding the probability of exactly 4 successes in 10 Bernoulli trials with a success probability of 0.3. This is a binomial probability scenario. We use the formula $P(X=k) = C_k^n p^k (1-p)^{n-k}$. Substituting $n=10$, $k=4$, $p=0.3$ (and thus $1-p=0.7$), we calculate $C_4^{10} = 210$, $(0.3)^4 = 0.0081$, and $(0.7)^6 = 0.117649$. Multiplying these values gives $210 \times 0.0081 \times 0.117649 \approx 0.2001$.