Q. 1.18 - A First Course in Probability

Question

In a certain community, there are $3$ families consisting of a single parent and $1$ child, $3$ families consisting of a single parent and $2$ children, $5$ families consisting of $2$ parents and a single child, $7$ families consisting of $2$ parents and $2$ children, and $6$ families consisting of $2$ parents and $3$ children. If a parent and child from the same family are to be chosen, how many possible choices are there?

Step-by-Step Solution

Answer

The possible no. of choices are $83$.

Step 1. Given information.

It is given that,

No. of families consisting of a single parent and $1$ child $=3$

No. of families consisting of a single parent and $2$ children $=3$

No. of families consisting of $2$ parents and a single child $=5$

No. of families consisting of $2$ parents and $2$ children $=7$

No. of families consisting of $2$ parents and $3$ children $=6$

Step 2. Find the possible no. of choices.

No. of ways of selecting a parent and child from $3$ families consisting

of a single parent and $1$ child  $= 3\times 1\times 1=3$

No. of ways of selecting a parent and child from $3$ families consisting of a single parent and $2$ children $= 3\times 1\times 2=6$

No. of ways of selecting a parent and child from $5$ families consisting of $2$ parents and a single child $= 5\times 2\times 1=10$

No. of ways of selecting a parent and child from $7$ families consisting of $2$ parents and $2$ children $= 7\times 2\times 2=28$

No. of ways of selecting a parent and child from $6$ families consisting of $2$ parents and $3$ children $= 6\times 2\times 3=36$

Therefore, the no. of possible choices are $3+6+10+28+36 =83$.