Question

What is the probability of getting exactly two heads when flipping four coins?

Answer 1

Just do this: 1/2. Your answer, anonymous

Answer 2

To find the probability of getting exactly two heads when flipping four coins, we can use the binomial probability formula. The formula for the probability of getting exactly \( k \) successes (in this case, heads) in \( n \) independent Bernoulli trials (coin flips) is given by:

\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]

Where:
- \( n \) is the total number of trials (coins flipped),
- \( k \) is the number of successes,
- \( p \) is the probability of success on a single trial (for a fair coin, \( p = 0.5 \)),
- \( \binom{n}{k} \) is the binomial coefficient, which gives the number of ways to choose \( k \) successes from \( n \) trials.

In this situation, we have:
- \( n = 4 \) (the number of coins),
- \( k = 2 \) (we want exactly 2 heads),
- \( p = 0.5 \) (the probability of getting heads in one flip).

Now we can calculate the probability.

1. Calculate the binomial coefficient \( \binom{4}{2} \):

\[
\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
\]

2. Calculate \( p^k \) and \( (1 - p)^{n - k} \):

\[
p^k = (0.5)^2 = 0.25
\]
\[
(1 - p)^{n - k} = (0.5)^{4 - 2} = (0.5)^2 = 0.25
\]

3. Substitute these values into the probability formula:

\[
P(X = 2) = \binom{4}{2} (0.5)^2 (0.5)^{2} = 6 \times 0.25 \times 0.25 = 6 \times 0.0625 = 0.375
\]

Therefore, the probability of getting exactly two heads when flipping four coins is

\[
\boxed{0.375}
\] or \( \frac{3}{8} \).