Question

Income tax returns Here is the distribution of the adjusted gross income (in thousands of dollars) reported on individual federal income tax returns in a recent year:

a. What is the probability that a randomly chosen return shows an adjusted gross income of $\backslash (50,000$ or more?

b. Given that a return shows an income of at least $\backslash )50,000$, what is the conditional

probability that the income is at least $\$100,000$?

Short answer

a. Probability shows an adjusted gross income of $\$50,000$or more is $0.321$

b. Required conditional probability is$0.3302$

Step-by-step solution

Given Information

It is given that

Determining Probability for the randomly chosen return shows an adjusted gross income of $ 50,000 or more. 

As per addition rule

$P\left(\mathrm{AorB}\right)=P(\mathrm{A})+P(\mathrm{B})$

From table, probability of adjusted gross income of $50-99(\text{in thousands of}\$)$

$P(50-99)=0.215$

Similarly $P(100-499)=0.100$

$P(\ge 500)=0.006$

Now, $P(\text{income}>\$50,000)=P(>50)$

$=P(50-99)+P(100-499)+P(\ge 500)$

$=0.215+0.100+0.006$

$=0.321$

Hence, Probability shows an adjusted gross income of$\$50,000$or more is$0.321$

Conditional Probability for income is at least $100,000

We know that $P(\mathrm{A}\mid \mathrm{B})=\frac{P(\mathrm{A}\text{and}\mathrm{B})}{P(\mathrm{B})}$

From above part: $P(\text{income}>\$50,000)=0.321$

As return shows income of atleast $\$50,000$

$\Rightarrow P(\ge \$50,000\text{and}\ge \$100,000)=P(\ge \$100,000)$

$=P(100-499)+P(\ge 500)$

$=0.100+0.006$

$=0.106$

Using Conditional Probability

$P(\ge \$100,000\mid \ge \$50,000)=\frac{P(\ge \$50,000\text{and}\ge \$1,00,000)}{P(\ge \$50,000)}$

$=\frac{0.106}{0.321}$

$\approx 0.3302$