Question

The number of girl child per family was recorded for 1000 families with 3 children. The data obtained was as follows: [2] Number of Girls 0 1 2 3 Number of Families 112 314 382 192 Suppose a family is chosen at random, find the probability that it has: (a). At most one girl (b). More girls than boys

Answer 1

Answer:

$a.\ Probability = 0.426$

$b.\ Probability = 0.574$

Step-by-step explanation:

Given

$Total\ Families = 1000$

Number of Girls :      0 ---->  1 -----> 2   ------->3

Number of Families: 112 -> 314 --> 382 ---> 192

Solving (a): At most one girl

From the table, the number of families with at most one girl is: 112 + 314

i.e.

Number of Girls :      0 ---->  1

Number of Families: 112 -> 314

So, we have:

$At\ Most\ One\ Girl = 112 + 314$

$At\ Most\ One\ Girl = 426$

The probability is then calculated as:

$Probability = \frac{At\ Most\ One\ Girl}{Total}$

$Probability = \frac{426}{1000}$

$Probability = 0.426$

Solving (b): More girls

From the table, the number of families with more girl is: 382 + 192

i.e.

Number of Girls :      -----> 2   ------->3

Number of Families: --> 382 ---> 192

So, we have:

$More\ Girls = 382 + 192$

$More\ Girls = 574$

The probability is then calculated as:

$Probability = \frac{More\ Girls}{Total}$

$Probability = \frac{574}{1000}$

$Probability = 0.574$