Question

4 multiple choice 2 points a fair coin is tossed ten times and a head appears each time. what is the probability that on the next three tosses of the coin, exactly two heads will appear? $\frac{3}{8}$ $\frac{5}{10}$ $\frac{2}{8}$ $\frac{1}{10}$

Explanation:

Step1: Recall binomial probability formula

The binomial - probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successful trials, $p$ is the probability of success in a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Identify values of $n$, $k$, and $p$

For coin - tossing, $n = 3$ (number of coin tosses), $k = 2$ (number of heads), and $p=\frac{1}{2}$ (probability of getting a head in a single fair - coin toss).

Step3: Calculate the combination $C(n,k)$

$C(3,2)=\frac{3!}{2!(3 - 2)!}=\frac{3!}{2!1!}=\frac{3\times2!}{2!×1}=3$.

Step4: Calculate the probability

$P(X = 2)=C(3,2)\times(\frac{1}{2})^{2}\times(1-\frac{1}{2})^{3 - 2}=3\times\frac{1}{4}\times\frac{1}{2}=\frac{3}{8}$.

Answer:

$\frac{3}{8}$