# Consider a binominal experiment with 2 trials and p=0.4 a. Compute

Consider a binominal experiment with 2 trials and $p=0.4$
a. Compute the probability of 1 success f(1)
b. Compute f(0)
c. Compute f(2)
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Step 1
Given binomial experiment with
The binomial probability for x number of success is calculated by the below mentioned formula
$P\left(X=x\right){=}^{n}{C}_{x}{\left(p\right)}^{x}{\left(1-p\right)}^{n-x}$
Step 2
The probability of 1 success is computed as shown below
Probability of 1 success $f\left(1\right)=0.48$
$P\left(X=x\right){=}^{n}{C}_{x}{\left(p\right)}^{x}{\left(1-p\right)}^{n-x}$
$P\left(X=1\right){=}^{2}{C}_{1}{\left(0.4\right)}^{1}{\left(1-0.4\right)}^{2-1}$
$P\left(X=1\right)=2\cdot \left(0.4\right)\cdot \left(0.6\right)=0.48$
Step 3
The probability of 0 success is computed as shown below
The probability of 0 success is $f\left(0\right)=0.36$
$P\left(X=x\right){=}^{n}{C}_{x}{\left(p\right)}^{x}{\left(1-p\right)}^{n-x}$
$P\left(X=0\right){=}^{2}{C}_{0}{\left(0.4\right)}^{0}{\left(1-0.4\right)}^{2-0}$
$P\left(X=0\right)=1\cdot \left(1\right)\cdot {\left(0.6\right)}^{2}=0.36$
Step 4
The probability of 2 success is computed as shown below
The probability of 2 success $f\left(2\right)=0.16$
$P\left(X=x\right){=}^{n}{C}_{x}{\left(p\right)}^{x}{\left(1-p\right)}^{n-x}$
$P\left(X=2\right){=}^{2}{C}_{2}{\left(0.4\right)}^{2}{\left(1-0.4\right)}^{2-2}$
$P\left(X=2\right)=1\cdot {\left(0.4\right)}^{2}\cdot {\left(0.6\right)}^{0}=0.16$