Question

use the following formula for calculating binomial probabilities to answer the question.
$_{n}c_{k}(p)^{k}(1 - p)^{n - k}$
what is the probability of getting exactly 5 \heads\ in 10 coin flips?
1/32
63/256
1/2
193/256
done

Explanation:

Step1: Identify values for formula

For coin - flips, $n = 10$ (number of trials), $k = 5$ (number of successes), and $p=\frac{1}{2}$ (probability of getting heads in a single flip). The binomial coefficient $_{n}C_{k}=\frac{n!}{k!(n - k)!}$. So, $_{10}C_{5}=\frac{10!}{5!(10 - 5)!}=\frac{10!}{5!5!}=\frac{10\times9\times8\times7\times6}{5\times4\times3\times2\times1}=252$.

Step2: Calculate probability

Using the binomial - probability formula $_{n}C_{k}p^{k}(1 - p)^{n - k}$, substitute the values: $_{10}C_{5}(\frac{1}{2})^{5}(1-\frac{1}{2})^{10 - 5}=252\times(\frac{1}{2})^{5}\times(\frac{1}{2})^{5}=252\times\frac{1}{32}\times\frac{1}{32}=\frac{252}{1024}=\frac{63}{256}$.

Answer:

63/256