if a fair coin is tossed four times, the probability of getting exactly three heads is 3/16 4/16 1/16 2/16

Question

Accepted Answer

# Explanation:
## Step1: Calculate the number of combinations
The formula for combinations is $C(n,k)=\frac{n!}{k!(n - k)!}$, where $n = 4$ (number of coin - tosses) and $k=3$ (number of heads).
$$
\begin{align*}
C(4,3)&=\frac{4!}{3!(4 - 3)!}\
&=\frac{4!}{3!1!}\
&=\frac{4\times3!}{3!×1}\
&=4
\end{align*}
$$
## Step2: Calculate the probability of each outcome
The probability of getting a head ($p$) or a tail ($q$) in a single fair - coin toss is $p=\frac{1}{2}$, $q = 1-p=\frac{1}{2}$.
The probability mass function of a binomial distribution is $P(X = k)=C(n,k)\times p^{k}\times q^{n - k}$.
Substitute $n = 4$, $k = 3$, $p=\frac{1}{2}$, $q=\frac{1}{2}$ into the formula:
$$
\begin{align*}
P(X = 3)&=C(4,3)\times(\frac{1}{2})^{3}\times(\frac{1}{2})^{4 - 3}\
&=4\times\frac{1}{8}\times\frac{1}{2}\
&=\frac{4}{16}
\end{align*}
$$

# Answer:
$\frac{4}{16}$

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Question

Explanation:

Step1: Calculate the number of combinations

Step2: Calculate the probability of each outcome

Answer: