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The binomial distribution determines the probability of looking at a specific quantity of a success results in a specific quantity of trials.

We are given,

The chance of an IRS audit for a tax return with over $\$25,000$in income is about $2\%$per year with a $20$-year period.

Random variable in simple terms generally refers to variables whose values are unknown, therefore, in this case the random variable X is the number of audits in$20$year period.

Make the list of values that you want to use X may take on.

As we can see there is an upper bound for the situation at hand $20$then X is

$X=0,1,2,.......,20.$

The random variable is distributed by the data provided X will keep the track of online offerings. The probability distribution of binomial distribution has two parameters $n=numberoftrialsandp=probabilityofsuccess$.

The binomial distribution is of the form: $X~B(n,p)$

Therefore, The distribution of X is

$X~B(20,0.02)$

The expected binomial distribution is calculated as:

$渭=np$

$渭$is the number of audits expected in $20$year period,

$n=20$

$p=0.02$

Therefore, the number of audits expected in$20$ year period is:

$$\begin{gathered} 渭=np \\ 渭=20脳0.02 \\ 渭=0.4 \end{gathered}$$

The probability that a person is audited is $0.02$

Therefore, the probability that a person is not audited at all will be

$$\begin{gathered} =1-0.02 \\ =0.98 \end{gathered}$$

using Binompdf on TI calculator we can calculate the binomial distribution.

role="math" localid="1649089492311" $Binompdf(n,p,c)$where, role="math" localid="1649089497672" $n$is number of trials, $p$is probability of success, $c$is the probability of c success, for some number c.

The values are: $n=20,p=0.02,c=0$

$Binompdf(20,0.02,0)=0.6676$

The probability that a person is audited more than twice is$0.0071$