Question

Household size In government data, a household consists of all occupants of a dwelling unit. Choose an American household at random and count the number of people it contains. Here is the assignment of probabilities for the outcome. The probability of finding $3$people in a household is the same as the probability of finding $4$people.

a. What probability should replace “?” in the table? Why?

b. Find the probability that the chosen household contains more than $2$people.

Short answer

Part(a) $0.16$ is the probability that should replace “?” in the table.

Part(b) $0.43$ is the probability that the chosen household contains more than $2$ people.

Step-by-step solution

Part(a) Step 1 : Given information

We are given a table. We need to find what probability should replace “?” in the table.

Number of people	$1$	$2$	$3$	$4$	$5$	$6$	$7+$
probability	$0.25$	$0.32$	?	?	$0.07$	$0.03$	$0.01$

Part(a) Step 2 : Simplify

As we know, Sum of all probabilities should be $1$and probability should lie between $0$and $1$.

Let the missing probabilities be X. and both are equal.

Now,

Sum of probabilities is $$\begin{gathered} 0.25+0.32+x+x+0.07+0.03+0.01=1 \\ 2x+0.68=1 \\ x=0.16 \end{gathered}$$

Hence, $0.16$should replace “?” in the table.

Part(b) Step 1 : Given information

We are given a table. We need to find probability that the chosen household contains more than $2$people.

Number of people	$1$	$2$	$3$	$4$	$5$	$6$	$7+$
Probability	$0.25$	$0.32$	?	?	$0.07$	$0.03$	$0.01$

Part(b) Step 2 : Simplify

Using addition rule for disjoint events,

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

Now,

Probability of $1$person $\mathrm{P}\left(1\right)=0.25$

Probability of $2$person $\mathrm{P}\left(2\right)=0.32$

Now,

P( At most $2$people) $=\mathrm{P}\left(1\right)+\mathrm{P}\left(2\right)$

$=0.25+0.32=0.57$

Now,

Using complement rule $\mathrm{P}\left({\mathrm{A}}^{\mathrm{c}}\right)=\mathrm{P}\left(\overline{\mathrm{A}}\right)=1-\mathrm{P}\left(\mathrm{A}\right)$

P(more than $2$people ) $=1-0.57=0.43$

Hence, $0.43$is the probability that the chosen household contains more than $2$people.