
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
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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.
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