
How do you use the binomial probability formula to find the probability of x successes given probability p of success on a single trial for n=6, x=4, p=0.75?
Answer
541.5k+ views
Hint: In the binomial probability, the number of successes X in ‘n’ trials of a binomial experiment is called a binomial random variable. The probability distribution of the random variable X is called a binomial distribution, and is given by the formula as below:
$P\left( X \right) = C_x^np^xq^{{n - x}}$
Where n is the number of trials, x is 0, 1, 2..., n, p is the probability of success in a single trial, q is the probability of failure in a single trial and the value of q is 1-p. P(X) gives the probability of successes in n binomial trials.
The combination formula is $C_x^n = \dfrac{{n!}}{{x!\left( {n - x} \right)!}}$.
Complete step-by-step answer:
We already know the binomial probability formula
$P\left( X \right) = C_x^np^xq^{{n - x}}$
The value of q is 1-p.
$ \Rightarrow P\left( X \right) = C_x^np^xq^{{n - x}}$
Let us substitute the values n=6, x=4, and p=0.75.
$ \Rightarrow P\left( X \right) = C_4^6{\left( {0.75} \right)^4}{(1 - 0.75)^{6 - 4}}$ ...(1)
The combination formula is
$C_x^n = \dfrac{{n!}}{{x!\left( {n - x} \right)!}}$
Let us substitute the value of n and x.
$ \Rightarrow C_4^6 = \dfrac{{6!}}{{4!\left( {6 - 4} \right)!}}$
Therefore,
$ \Rightarrow C_4^6 = \dfrac{{6!}}{{4! \times 2!}}$
So,
$ \Rightarrow C_4^6 = \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1 \times 2 \times 1}}$
Let us simplify it.
$ \Rightarrow C_4^6 = 15$
Now, substitute the value of combination in equation (1).
$ \Rightarrow P\left( X \right) = 15{\left( {0.75} \right)^4}{(1 - 0.75)^{6 - 4}}$
$ \Rightarrow P\left( X \right) = 15{\left( {0.75} \right)^4}{(0.25)^2}$
So, the answer will be
$ \Rightarrow P\left( X \right) = 0.297$
Hence, the probability of success is 0.297.
Note:
A binomial probability is one that possesses the following properties:
The experiment consists of n repeated trials.
Each trial results in an outcome that may be classified as a success or a failure. Based on these two conditions the name is called binomial.
The probability of success remains constant from trial to trial and repeated trials are independent.
$P\left( X \right) = C_x^np^xq^{{n - x}}$
Where n is the number of trials, x is 0, 1, 2..., n, p is the probability of success in a single trial, q is the probability of failure in a single trial and the value of q is 1-p. P(X) gives the probability of successes in n binomial trials.
The combination formula is $C_x^n = \dfrac{{n!}}{{x!\left( {n - x} \right)!}}$.
Complete step-by-step answer:
We already know the binomial probability formula
$P\left( X \right) = C_x^np^xq^{{n - x}}$
The value of q is 1-p.
$ \Rightarrow P\left( X \right) = C_x^np^xq^{{n - x}}$
Let us substitute the values n=6, x=4, and p=0.75.
$ \Rightarrow P\left( X \right) = C_4^6{\left( {0.75} \right)^4}{(1 - 0.75)^{6 - 4}}$ ...(1)
The combination formula is
$C_x^n = \dfrac{{n!}}{{x!\left( {n - x} \right)!}}$
Let us substitute the value of n and x.
$ \Rightarrow C_4^6 = \dfrac{{6!}}{{4!\left( {6 - 4} \right)!}}$
Therefore,
$ \Rightarrow C_4^6 = \dfrac{{6!}}{{4! \times 2!}}$
So,
$ \Rightarrow C_4^6 = \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1 \times 2 \times 1}}$
Let us simplify it.
$ \Rightarrow C_4^6 = 15$
Now, substitute the value of combination in equation (1).
$ \Rightarrow P\left( X \right) = 15{\left( {0.75} \right)^4}{(1 - 0.75)^{6 - 4}}$
$ \Rightarrow P\left( X \right) = 15{\left( {0.75} \right)^4}{(0.25)^2}$
So, the answer will be
$ \Rightarrow P\left( X \right) = 0.297$
Hence, the probability of success is 0.297.
Note:
A binomial probability is one that possesses the following properties:
The experiment consists of n repeated trials.
Each trial results in an outcome that may be classified as a success or a failure. Based on these two conditions the name is called binomial.
The probability of success remains constant from trial to trial and repeated trials are independent.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

