Question

use the following formula for calculating binomial probabilities to answer this 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 binomial formula values

The binomial probability formula is $_{n}C_{k}p^{k}(1 - p)^{n - k}$. Here, $n = 10$ (number of trials, coin - flips), $k = 5$ (number of successes, heads), and for a fair coin $p=\frac{1}{2}$. First, calculate the combination $_{n}C_{k}=\frac{n!}{k!(n - k)!}$. 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 the probability

Substitute $n = 10$, $k = 5$, and $p=\frac{1}{2}$ into the binomial - probability formula: $P(X = 5)=_{10}C_{5}(\frac{1}{2})^{5}(1-\frac{1}{2})^{10 - 5}$. Since $1-\frac{1}{2}=\frac{1}{2}$, we have $P(X = 5)=252\times(\frac{1}{2})^{5}\times(\frac{1}{2})^{5}=252\times\frac{1}{32}\times\frac{1}{32}=\frac{252}{1024}=\frac{63}{256}$.

Answer:

63/256 (corresponding to the second option)