Seven unbaised coins are tossed 128 times, In how many throws would you find at least three heads?

Options

A. 99
B. 102
C. 103
D. 104

Explanation

To solve the problem of finding how many times at least three heads appear when tossing seven unbiased coins 128 times, we can follow these steps:

### Step 1: Understand the Problem
We need to find the probability of getting at least 3 heads when tossing 7 unbiased coins. This can be modeled using the binomial distribution.

**Hint:** Remember that the binomial distribution is used for experiments with two possible outcomes (like heads or tails).

### Step 2: Define the Variables
Let \( X \) be the random variable representing the number of heads obtained when tossing 7 coins. Here, \( n = 7 \) (the number of trials) and \( p = \frac{1}{2} \) (the probability of getting heads).

**Hint:** The probability of tails is \( q = 1 - p \).

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

**Hint:** Use the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]

### Step 4: Calculate \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \)
1. **Calculate \( P(X = 0) \)**:
\[ P(X = 0) = \binom{7}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^7 = 1 \cdot 1 \cdot \frac{1}{128} = \frac{1}{128} \]

2. **Calculate \( P(X = 1) \)**:
\[ P(X = 1) = \binom{7}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^6 = 7 \cdot \frac{1}{2} \cdot \frac{1}{64} = \frac{7}{128} \]

3. **Calculate \( P(X = 2) \)**:
\[ P(X = 2) = \binom{7}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^5 = 21 \cdot \frac{1}{4} \cdot \frac{1}{32} = \frac{21}{128} \]

**Hint:** Use the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \) to calculate the combinations.

### Step 5: Sum the Probabilities
Now, sum these probabilities:
\[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = \frac{1}{128} + \frac{7}{128} + \frac{21}{128} = \frac{29}{128} \]

**Hint:** Ensure you have a common denominator when adding fractions.

### Step 6: Calculate \( P(X \geq 3) \)
Now, substitute back to find \( P(X \geq 3) \):
\[ P(X \geq 3) = 1 - P(X < 3) = 1 - \frac{29}{128} = \frac{99}{128} \]

**Hint:** Subtracting from 1 gives you the probability of the complementary event.

### Step 7: Calculate the Expected Number of Throws
Since the coins are tossed 128 times, the expected number of throws with at least 3 heads is:
\[ \text{Expected number of throws} = P(X \geq 3) \times \text{Total throws} = \frac{99}{128} \times 128 = 99 \]

**Hint:** The total number of trials (128) cancels out the denominator.

### Conclusion
Thus, the number of throws in which at least 3 heads are expected is **99**.

**Final Answer:** 99