Jui's chance of winning a single game aginst Vrushali is `(3)/(4) ` probability that , in a series of 5 games ,jui wins at least 3 games is ….

Options

A. `(135)/(512)`
B. `(459)/(512)`
C. `(324)/(512)`
D. none of these

Answer

To solve the problem of finding the probability that Jui wins at least 3 out of 5 games against Vrushali, we can use the concept of binomial distribution. Here’s a step-by-step solution:

### Step 1: Define the Parameters
- Let \( n = 5 \) (the total number of games).
- Let \( p = \frac{3}{4} \) (the probability of Jui winning a single game).
- Let \( q = 1 - p = \frac{1}{4} \) (the probability of Jui losing a single game).

### Step 2: Identify the Required Probability
We need to find the probability that Jui wins at least 3 games, which can be expressed as:
\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \]

### Step 3: 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 combination of \( n \) items taken \( r \) at a time.

### Step 4: Calculate Each Probability
1. **For \( P(X = 3) \)**:
\[ P(X = 3) = \binom{5}{3} \left(\frac{3}{4}\right)^3 \left(\frac{1}{4}\right)^{5-3} \]
\[ = 10 \cdot \left(\frac{27}{64}\right) \cdot \left(\frac{1}{16}\right) = 10 \cdot \frac{27}{1024} = \frac{270}{1024} \]

2. **For \( P(X = 4) \)**:
\[ P(X = 4) = \binom{5}{4} \left(\frac{3}{4}\right)^4 \left(\frac{1}{4}\right)^{5-4} \]
\[ = 5 \cdot \left(\frac{81}{256}\right) \cdot \left(\frac{1}{4}\right) = 5 \cdot \frac{81}{1024} = \frac{405}{1024} \]

3. **For \( P(X = 5) \)**:
\[ P(X = 5) = \binom{5}{5} \left(\frac{3}{4}\right)^5 \left(\frac{1}{4}\right)^{5-5} \]
\[ = 1 \cdot \left(\frac{243}{1024}\right) \cdot 1 = \frac{243}{1024} \]

### Step 5: Combine the Probabilities
Now, we add the probabilities:
\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \]
\[ = \frac{270}{1024} + \frac{405}{1024} + \frac{243}{1024} \]
\[ = \frac{270 + 405 + 243}{1024} = \frac{918}{1024} \]

### Step 6: Simplify the Result
We can simplify \( \frac{918}{1024} \):
\[ \frac{918 \div 2}{1024 \div 2} = \frac{459}{512} \]

### Final Answer
Thus, the probability that Jui wins at least 3 games is:
\[ \frac{459}{512} \]