Question

How do you use the binomial probability formula to find the probability x successes given the probability p of success on a single trial for n = 6, x = 4, p = 0.75?
(a) 0.297
(b) 0.28
(c) 0.452
(d) 0.56

Answer 1

Hint: To start with, we are to find the probability in this problem with the given conditions. Now, we have to use the binomial probability formula $P\left( X \right)=C_{x}^{n}{{p}^{x}}{{q}^{n-x}}$ to get the proper solution. Putting the given values from the problem will give us the desired result.

Complete step by step solution:
According to the question, we are to find the probability x successes given the probability p of success on a single trial for n = 6, x = 4, p = 0.75.
We have the binomial probability formula, $P\left( X \right)=C_{x}^{n}{{p}^{x}}{{q}^{n-x}}$ where, p + q = 1.
Now, we are given, n = 6, x = 4, p = 0.75.
Then, $P\left( X \right)=C_{4}^{6}{{\left( 0.75 \right)}^{4}}{{\left( 1-0.75 \right)}^{6-4}}$….(1)
Now, we also know the fact that,
$C_{x}^{n}=\dfrac{n!}{x!\left( n-x \right)!}$
Putting the values,
We are getting, $C_{4}^{6}=\dfrac{6!}{4!\left( 6-4 \right)!}$
Therefore, $C_{4}^{6}=\dfrac{6!}{4!\times 2!}$
Simplifying,
$C_{4}^{6}=\dfrac{6\times 5\times 4\times 3\times 2}{4\times 3\times 2\times 2}$
We get, $C_{4}^{6}=15$
Now, substituting this value to equation (1),
$P\left( X \right)=15\times {{\left( 0.75 \right)}^{4}}\times {{\left( 0.25 \right)}^{2}}$
Now, by using simplification,
$\Rightarrow P\left( X \right)=15\times 0.31640625\times 0.0625$
Further simplification gives us,
$P\left( X \right)=0.297$

So, the correct answer is “Option a”.

Note: Binomial probability refers to the probability of exactly x successes on n repeated trials in an experiment which has two possible outcomes (commonly called a binomial experiment). If the probability of success on an individual trial is p , then the binomial probability is $C_{x}^{n}{{p}^{x}}{{q}^{n-x}}$. Here $C_{x}^{n}$ indicates the number of different combinations of x objects selected from a set of n objects. Some textbooks use the notation $\left( \begin{matrix}
   n \\
   x \\
\end{matrix} \right)$ instead of $C_{x}^{n}$ . If p is the probability of success of a single trial, then (1−p) is the probability of failure of a single trial.