If in a trial the probability of a success is twice the probability of failure , the probability of atleast four sucesses in six trials is

Options

A. `496/729`
B. `400/729`
C. `500/729`
D. `600/729`

Answer

To solve the problem step by step, we will follow the logical sequence outlined in the video transcript.

### Step 1: Define the probabilities
Let \( p \) be the probability of success and \( q \) be the probability of failure. According to the problem, the probability of success is twice the probability of failure:
\[ p = 2q \]
We also know that the sum of the probabilities must equal 1:
\[ p + q = 1 \]

### Step 2: Substitute and solve for \( p \) and \( q \)
Substituting \( p = 2q \) into the equation \( p + q = 1 \):
\[ 2q + q = 1 \implies 3q = 1 \implies q = \frac{1}{3} \]
Now, substituting back to find \( p \):
\[ p = 2q = 2 \times \frac{1}{3} = \frac{2}{3} \]

### Step 3: Identify the number of trials
The number of trials \( n \) is given as 6.

### Step 4: Calculate the probability of at least 4 successes
We need to find the probability of getting at least 4 successes in 6 trials, which means we need to calculate:
\[ P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6) \]

### Step 5: Use the binomial probability formula
The binomial probability formula is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where \( \binom{n}{r} \) is the binomial coefficient.

### Step 6: Calculate \( P(X = 4) \)
For \( r = 4 \):
\[ P(X = 4) = \binom{6}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^{6-4} \]
Calculating \( \binom{6}{4} \):
\[ \binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15 \]
Now substituting:
\[ P(X = 4) = 15 \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^2 = 15 \cdot \frac{16}{81} \cdot \frac{1}{9} = 15 \cdot \frac{16}{729} = \frac{240}{729} \]

### Step 7: Calculate \( P(X = 5) \)
For \( r = 5 \):
\[ P(X = 5) = \binom{6}{5} \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^{6-5} \]
Calculating \( \binom{6}{5} \):
\[ \binom{6}{5} = 6 \]
Now substituting:
\[ P(X = 5) = 6 \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^1 = 6 \cdot \frac{32}{243} \cdot \frac{1}{3} = 6 \cdot \frac{32}{729} = \frac{192}{729} \]

### Step 8: Calculate \( P(X = 6) \)
For \( r = 6 \):
\[ P(X = 6) = \binom{6}{6} \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^{6-6} \]
Calculating \( \binom{6}{6} \):
\[ \binom{6}{6} = 1 \]
Now substituting:
\[ P(X = 6) = 1 \cdot \left(\frac{2}{3}\right)^6 = \frac{64}{729} \]

### Step 9: Combine the probabilities
Now we can combine all the probabilities:
\[ P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6) = \frac{240}{729} + \frac{192}{729} + \frac{64}{729} \]
Combining:
\[ P(X \geq 4) = \frac{240 + 192 + 64}{729} = \frac{496}{729} \]

### Final Answer
The probability of at least 4 successes in 6 trials is:
\[ \frac{496}{729} \]