An experiment succeeds twice as often as it fails. The probability of at least 5 successes in the six trials of this experiment is

Options

A. `(496)/(729)`
B. `(192)/(729)`
C. `(240)/(729)`
D. `(256)/(729)`

Answer

To solve the problem step by step, we will first define the probabilities and then calculate the required probability of at least 5 successes in 6 trials.

### Step 1: Define the probabilities of success and failure
Let the probability of success be \( p \) and the probability of failure be \( q \). According to the problem, the experiment succeeds twice as often as it fails. This can be expressed as:
\[ p = 2q \]
Since the total probability must equal 1, we have:
\[ p + q = 1 \]
Substituting \( p = 2q \) into the equation gives:
\[ 2q + q = 1 \implies 3q = 1 \implies q = \frac{1}{3} \]
Now substituting back to find \( p \):
\[ p = 2q = 2 \times \frac{1}{3} = \frac{2}{3} \]

### Step 2: Calculate the probability of at least 5 successes
We need to find the probability of getting at least 5 successes in 6 trials. This can be calculated as:
\[ P(X \geq 5) = P(X = 5) + P(X = 6) \]
Where \( X \) is the number of successes in 6 trials.

### Step 3: Calculate \( P(X = 5) \) and \( P(X = 6) \)
Using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
For \( n = 6 \), \( p = \frac{2}{3} \), and \( q = \frac{1}{3} \):

1. **Calculate \( P(X = 5) \)**:
\[ P(X = 5) = \binom{6}{5} \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^{6-5} \]
\[ = 6 \cdot \left(\frac{2}{3}\right)^5 \cdot \left(\frac{1}{3}\right)^1 \]
\[ = 6 \cdot \frac{32}{243} \cdot \frac{1}{3} = 6 \cdot \frac{32}{729} = \frac{192}{729} \]

2. **Calculate \( P(X = 6) \)**:
\[ P(X = 6) = \binom{6}{6} \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^{0} \]
\[ = 1 \cdot \left(\frac{2}{3}\right)^6 = \frac{64}{729} \]

### Step 4: Combine the probabilities
Now, we can combine the probabilities to find \( P(X \geq 5) \):
\[ P(X \geq 5) = P(X = 5) + P(X = 6) = \frac{192}{729} + \frac{64}{729} = \frac{256}{729} \]

### Final Answer
Thus, the probability of at least 5 successes in 6 trials is:
\[ \boxed{\frac{256}{729}} \]