Question

What is the probability of getting exactly 3 heads in 5 coin tosses?

Answer 1

To calculate the probability of getting exactly 3 heads in 5 coin tosses, we can use the binomial probability formula. The formula is given by:

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

where:
- \( n \) is the total number of trials (coin tosses),
- \( k \) is the number of successful trials (number of heads),
- \( p \) is the probability of success on an individual trial (probability of getting heads), and
- \( \binom{n}{k} \) is the binomial coefficient.

For a fair coin, the probability of heads (success) is \( p = 0.5 \), and the probability of tails (failure) is also \( 1 - p = 0.5 \).

In this case:
- \( n = 5 \) (the number of coin tosses),
- \( k = 3 \) (the number of heads we want).

First, we calculate the binomial coefficient \( \binom{5}{3} \):

\[
\binom{5}{3} = \frac{5!}{3!(5 - 3)!} = \frac{5!}{3! \cdot 2!} = \frac{5 \times 4}{2 \times 1} = 10
\]

Now, we can plug the values into the binomial probability formula:

\[
P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^{5 - 3} = 10 \cdot (0.5)^3 \cdot (0.5)^2 = 10 \cdot (0.5)^5
\]

Calculating \( (0.5)^5 \):

\[
(0.5)^5 = \frac{1}{32}
\]

Thus,

\[
P(X = 3) = 10 \cdot \frac{1}{32} = \frac{10}{32} = \frac{5}{16}
\]

Therefore, the probability of getting exactly 3 heads in 5 coin tosses is:

\[
\frac{5}{16}
\]