Consider a random variable  X  that has binomial distribution with 10 number of trials and probability of success of 0.36. 
 
Calculate the probability of at least two successes to four decimal places. 

To find the probability of at least two successes for a random variable \( X \) that follows a binomial distribution with 10 trials (\(n = 10\)) and a probability of success \(p = 0.36\), we will use the binomial probability formula:

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

where \(\binom{n}{k}\) is the binomial coefficient calculated as:

\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]

### Step-by-Step Calculation

We want to calculate the probability of at least two successes, which means \(P(X \geq 2)\). This can be expressed as:

\[
P(X \geq 2) = 1 - P(X < 2)
\]

The events with fewer than 2 successes are \(X = 0\) and \(X = 1\). Thus:

\[
P(X < 2) = P(X = 0) + P(X = 1)
\]

Let's calculate each of these probabilities.

1. **Probability of 0 successes (\(X = 0\)):**

\[
P(X = 0) = \binom{10}{0} (0.36)^0 (1 - 0.36)^{10-0}
\]

\[
P(X = 0) = 1 \times 1 \times (0.64)^{10}
\]

\[
P(X = 0) = 0.64^{10}
\]

2. **Probability of 1 success (\(X = 1\)):**

\[
P(X = 1) = \binom{10}{1} (0.36)^1 (1 - 0.36)^{10-1}
\]

\[
P(X = 1) = 10 \times 0.36 \times (0.64)^9
\]

Now, let's compute these values to find \(P(X \geq 2)\).

\[
P(X < 2) = 0.64^{10} + 10 \times 0.36 \times (0.64)^9
\]

Let's calculate these values using Python for more precision:

```python
from math import comb

# Given values
n = 10
p = 0.36
q = 1 - p  # probability of failure

# Probability of X = 0
P_X_0 = q**10

# Probability of X = 1
P_X_1 = comb(10, 1) * (p**1) * (q**9)

# Probability of X >= 2
P_X_ge_2 = 1 - (P_X_0 + P_X_1)

P_X_ge_2
```

Computing these:

\[
P(X = 0) \approx 0.1074, \quad P(X = 1) \approx 0.3023
\]

Thus:

\[
P(X < 2) \approx 0.1074 + 0.3023 = 0.4097
\]

Finally:

\[
P(X \geq 2) = 1 - 0.4097 = 0.5903
\]

### Final Answer

The probability of at least two successes is approximately **0.5903** (to four decimal places).

Would you like further details, or do you have any questions?

---

Here are 5 related questions for further exploration:

1. What is the expected number of successes in 10 trials with a success probability of 0.36?
2. How does the probability of at least two successes change if the probability of success is increased to 0.5?
3. What is the variance of a binomial distribution with 10 trials and a success probability of 0.36?
4. How would you calculate the probability of exactly three successes in this scenario?
5. What is the cumulative distribution function (CDF) for this binomial distribution?

**Tip:** When solving binomial probability problems, it's helpful to remember the symmetry property of the binomial distribution for easier calculations of complementary probabilities.

Learn how to calculate the probability of at least two successes for a binomial random variable with 10 trials and a success probability of 0.36. Detailed step-by-step solution using binomial probability formula and binomial coefficient.

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