To find the probability of getting exactly 5 "heads" in 10 coin flips, we will use the binomial probability formula.
The formula for calculating binomial probabilities is given by:
[tex]\[
\binom{n}{k} p^k (1-p)^{n-k}
\][/tex]
where:
- [tex]\( n \)[/tex] is the number of trials (flips).
- [tex]\( k \)[/tex] is the number of successful trials (heads).
- [tex]\( p \)[/tex] is the probability of success on a single trial (getting heads in one flip).
- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient which can be calculated by:
[tex]\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\][/tex]
Now, let's plug in the values:
1. Number of trials, [tex]\( n = 10 \)[/tex].
2. Number of successful trials, [tex]\( k = 5 \)[/tex].
3. Probability of success on an individual trial, [tex]\( p = 0.5 \)[/tex] (since it's a fair coin).
First, we calculate the binomial coefficient [tex]\(\binom{10}{5}\)[/tex]:
[tex]\[
\binom{10}{5} = \frac{10!}{5! (10-5)!} = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252
\][/tex]
Next, we calculate the probability [tex]\( p^k (1-p)^{n-k} \)[/tex]:
[tex]\[
p^k (1-p)^{n-k} = (0.5)^5 (0.5)^{10-5} = (0.5)^5 (0.5)^5 = (0.5)^{10} = \frac{1}{1024}
\][/tex]
Now, multiply the binomial coefficient [tex]\(\binom{10}{5}\)[/tex] by the probability:
[tex]\[
252 \times \frac{1}{1024} = \frac{252}{1024} = \frac{63}{256}
\][/tex]
Therefore, the probability of getting exactly 5 "heads" in 10 coin flips is:
[tex]\(\boxed{\frac{63}{256}}\)[/tex]
