A fair coin tossed five times. What is the probability that it landshead - up at least twice ?

Options

A. `(3)/(16)`
B. `(1)/(4)`
C. `(7)/(16)`
D. `(13)/(16)`

Explanation

To find the probability of getting at least two heads when a fair coin is tossed five times, we can follow these steps:

### Step 1: Understand the Problem
We need to calculate the probability of getting at least 2 heads in 5 tosses of a fair coin. This means we want to find the probabilities of getting 2, 3, 4, or 5 heads.

### Step 2: Use the Complement Rule
Instead of calculating the probabilities for 2, 3, 4, and 5 heads directly, we can use the complement rule. We will calculate the probability of getting 0 heads and 1 head, and then subtract that from 1.

\[ P(\text{at least 2 heads}) = 1 - P(0 \text{ heads}) - P(1 \text{ head}) \]

### Step 3: Calculate Probability of 0 Heads
To find \(P(0 \text{ heads})\), we need to consider that all 5 tosses result in tails. The number of ways to choose 0 heads from 5 tosses is given by the combination \( \binom{5}{0} \), which is 1. The probability of getting tails in all 5 tosses is:

\[ P(0 \text{ heads}) = \binom{5}{0} \left(\frac{1}{2}\right)^5 = 1 \cdot \left(\frac{1}{2}\right)^5 = \frac{1}{32} \]

### Step 4: Calculate Probability of 1 Head
Next, we calculate \(P(1 \text{ head})\). This means we have 1 head and 4 tails. The number of ways to choose 1 head from 5 tosses is given by \( \binom{5}{1} \), which is 5. The probability of getting 1 head and 4 tails is:

\[ P(1 \text{ head}) = \binom{5}{1} \left(\frac{1}{2}\right)^5 = 5 \cdot \left(\frac{1}{2}\right)^5 = 5 \cdot \frac{1}{32} = \frac{5}{32} \]

### Step 5: Combine the Probabilities
Now we can combine the probabilities we calculated:

\[ P(0 \text{ heads}) + P(1 \text{ head}) = \frac{1}{32} + \frac{5}{32} = \frac{6}{32} \]

### Step 6: Calculate the Final Probability
Now we can find the probability of getting at least 2 heads:

\[ P(\text{at least 2 heads}) = 1 - P(0 \text{ heads}) - P(1 \text{ head}) = 1 - \frac{6}{32} = \frac{32 - 6}{32} = \frac{26}{32} \]

### Step 7: Simplify the Fraction
Finally, we simplify \(\frac{26}{32}\):

\[ \frac{26}{32} = \frac{13}{16} \]

### Final Answer
Thus, the probability of getting at least 2 heads when a fair coin is tossed five times is:

\[ \frac{13}{16} \]

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