Seven unbiased 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

Answer

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

### Step 1: Understand the Problem
We are tossing 7 unbiased coins, and we want to find the probability of getting at least 3 heads in a single throw.

### Step 2: Define the Random Variable
Let \( X \) be the random variable representing the number of heads obtained when tossing 7 coins. \( X \) follows a binomial distribution with parameters \( n = 7 \) (the number of trials) and \( p = \frac{1}{2} \) (the probability of getting heads).

### Step 3: Calculate the Probability of Getting at Least 3 Heads
To find the probability of getting at least 3 heads, we can use the complement rule:
\[ P(X \geq 3) = 1 - P(X < 3) \]
This means we need to calculate \( P(X < 3) \), which includes the probabilities of getting 0, 1, or 2 heads.

### Step 4: Calculate \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \)
Using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where \( \binom{n}{k} \) is the binomial coefficient.

- **For \( k = 0 \)**:
\[ P(X = 0) = \binom{7}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^7 = 1 \cdot 1 \cdot \frac{1}{128} = \frac{1}{128} \]

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

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

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

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

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

### Final Answer
Thus, the number of throws in which we would find at least three heads is **99**.

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