12 coins are tossed 4096 times. The number of times that one can get atleast 2 heads is

Options

A. 4080
B. 4081
C. 4082
D. 4083

Answer

To solve the problem of finding the number of times at least 2 heads appear when 12 coins are tossed 4096 times, we can follow these steps:

### Step 1: Understand the Problem
We need to find the probability of getting at least 2 heads when 12 coins are tossed. The total number of tosses is 4096.

### Step 2: Define the Probability of Heads and Tails
When tossing a single coin, the probability of getting heads (P) is \( \frac{1}{2} \) and the probability of getting tails (Q) is also \( \frac{1}{2} \).

### Step 3: Calculate the Probability of Getting At Least 2 Heads
To find the probability of getting at least 2 heads, we can use the complement rule:
\[ P(X \geq 2) = 1 - P(X = 0) - P(X = 1) \]

### Step 4: Calculate \( P(X = 0) \) and \( P(X = 1) \)
Using the binomial probability formula:
\[ P(X = r) = \binom{n}{r} P^r Q^{n-r} \]
where \( n = 12 \), \( P = \frac{1}{2} \), and \( Q = \frac{1}{2} \).

1. **Calculate \( P(X = 0) \)**:
\[ P(X = 0) = \binom{12}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^{12} = 1 \cdot 1 \cdot \left(\frac{1}{2}\right)^{12} = \frac{1}{4096} \]

2. **Calculate \( P(X = 1) \)**:
\[ P(X = 1) = \binom{12}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{11} = 12 \cdot \frac{1}{2} \cdot \left(\frac{1}{2}\right)^{11} = 12 \cdot \frac{1}{4096} = \frac{12}{4096} \]

### Step 5: Combine the Probabilities
Now, substitute back into the equation for \( P(X \geq 2) \):
\[ P(X \geq 2) = 1 - P(X = 0) - P(X = 1) = 1 - \frac{1}{4096} - \frac{12}{4096} \]
\[ P(X \geq 2) = 1 - \frac{13}{4096} = \frac{4096 - 13}{4096} = \frac{4083}{4096} \]

### Step 6: Calculate the Expected Number of Times
To find the expected number of times at least 2 heads appear in 4096 tosses:
\[ \text{Expected number} = 4096 \times P(X \geq 2) = 4096 \times \frac{4083}{4096} = 4083 \]

### Final Answer
The number of times that one can get at least 2 heads when tossing 12 coins 4096 times is **4083**.