If getting a '5' or a '6' in the throw of an unbiased die is a 'success' and the random variable 'X' denotes the number of succeses in six throws of the die, find `P_(X)[X = 4, 5 , 6]`.

Answer

To solve the problem, we need to find the probabilities \( P(X = 4) \), \( P(X = 5) \), and \( P(X = 6) \) where \( X \) is the number of successes (getting a '5' or '6') in six throws of an unbiased die.

### Step 1: Define the Success Probability
In a single throw of an unbiased die, the outcomes are {1, 2, 3, 4, 5, 6}. The successful outcomes are '5' and '6'. Therefore, the probability of success (getting a '5' or '6') is:

\[ P(\text{Success}) = \frac{\text{Number of successful outcomes}}{\text{Total outcomes}} = \frac{2}{6} = \frac{1}{3} \]

The probability of failure (not getting '5' or '6') is:

\[ P(\text{Failure}) = 1 - P(\text{Success}) = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 2: Use the Binomial Probability Formula
Since we are throwing the die 6 times, we can model this situation using the binomial distribution. The probability of getting exactly \( k \) successes in \( n \) trials is given by:

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

Where:
- \( n = 6 \) (number of trials)
- \( k \) is the number of successes
- \( p = \frac{1}{3} \) (probability of success)
- \( 1 - p = \frac{2}{3} \) (probability of failure)

### Step 3: Calculate \( P(X = 4) \)
Using the formula for \( k = 4 \):

\[ P(X = 4) = \binom{6}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^{2} \]

Calculating \( \binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15 \):

\[ P(X = 4) = 15 \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^{2} = 15 \cdot \frac{1}{81} \cdot \frac{4}{9} = 15 \cdot \frac{4}{729} = \frac{60}{729} = \frac{20}{243} \]

### Step 4: Calculate \( P(X = 5) \)
Using the formula for \( k = 5 \):

\[ P(X = 5) = \binom{6}{5} \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^{1} \]

Calculating \( \binom{6}{5} = 6 \):

\[ P(X = 5) = 6 \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^{1} = 6 \cdot \frac{1}{243} \cdot \frac{2}{3} = 6 \cdot \frac{2}{729} = \frac{12}{729} \]

### Step 5: Calculate \( P(X = 6) \)
Using the formula for \( k = 6 \):

\[ P(X = 6) = \binom{6}{6} \left(\frac{1}{3}\right)^6 \left(\frac{2}{3}\right)^{0} \]

Calculating \( \binom{6}{6} = 1 \):

\[ P(X = 6) = 1 \cdot \left(\frac{1}{3}\right)^6 \cdot 1 = \frac{1}{729} \]

### Final Results
Now we have:

- \( P(X = 4) = \frac{20}{243} \)
- \( P(X = 5) = \frac{12}{729} \)
- \( P(X = 6) = \frac{1}{729} \)

### Summary
Thus, the probabilities are:
\[ P(X = 4) = \frac{20}{243}, \quad P(X = 5) = \frac{12}{729}, \quad P(X = 6) = \frac{1}{729} \]