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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

Sagot :

MrRoyalidn

Answer:

[tex]a.\ Probability = 0.426[/tex]

[tex]b.\ Probability = 0.574[/tex]

Step-by-step explanation:

Given

[tex]Total\ Families = 1000[/tex]

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:

[tex]At\ Most\ One\ Girl = 112 + 314[/tex]

[tex]At\ Most\ One\ Girl = 426[/tex]

The probability is then calculated as:

[tex]Probability = \frac{At\ Most\ One\ Girl}{Total}[/tex]

[tex]Probability = \frac{426}{1000}[/tex]

[tex]Probability = 0.426[/tex]

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:

[tex]More\ Girls = 382 + 192[/tex]

[tex]More\ Girls = 574[/tex]

The probability is then calculated as:

[tex]Probability = \frac{More\ Girls}{Total}[/tex]

[tex]Probability = \frac{574}{1000}[/tex]

[tex]Probability = 0.574[/tex]

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