Six coins are tossed simultaneously. The probability of getting at least 4 heads, is

Options

A. `11//64`
B. `11//32`
C. `15//44`
D. `21//32`

Answer

To find the probability of getting at least 4 heads when tossing 6 coins, we can follow these steps:

### Step 1: Understand the Problem
We need to calculate the probability of getting at least 4 heads when tossing 6 coins. This means we need to find the probability of getting 4, 5, or 6 heads.

### Step 2: Define the Variables
- Let \( n = 6 \) (the number of coins tossed).
- The probability of getting heads in a single toss, \( p = \frac{1}{2} \).
- The probability of getting tails, \( q = 1 - p = \frac{1}{2} \).

### Step 3: Use the Binomial Probability Formula
The probability of getting exactly \( r \) heads in \( n \) tosses is given by the binomial probability formula:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
Where:
- \( \binom{n}{r} \) is the binomial coefficient, representing the number of ways to choose \( r \) successes (heads) from \( n \) trials (tosses).

### Step 4: Calculate the Probabilities for 4, 5, and 6 Heads
1. **For 4 heads**:
\[ P(X = 4) = \binom{6}{4} \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^{6-4} = \binom{6}{4} \left(\frac{1}{2}\right)^6 \]
\[ = 15 \cdot \frac{1}{64} = \frac{15}{64} \]

2. **For 5 heads**:
\[ P(X = 5) = \binom{6}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^{6-5} = \binom{6}{5} \left(\frac{1}{2}\right)^6 \]
\[ = 6 \cdot \frac{1}{64} = \frac{6}{64} \]

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

### Step 5: Sum the Probabilities
Now, we sum the probabilities of getting 4, 5, or 6 heads:
\[ P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6) \]
\[ = \frac{15}{64} + \frac{6}{64} + \frac{1}{64} = \frac{22}{64} \]

### Step 6: Simplify the Result
Now we simplify \( \frac{22}{64} \):
\[ \frac{22}{64} = \frac{11}{32} \]

### Final Answer
Thus, the probability of getting at least 4 heads when tossing 6 coins is:
\[ \boxed{\frac{11}{32}} \]