15 names are in a hat, 6 girls and 9 boys. What is the probability of drawing, without replacement, two girls names in a row?

2 Answers
Aug 7, 2017

#1/7#

Explanation:

There is a total of 15 names in the hat. #6/15# are girls and #9/15# are boys.

First, you pull one name. The odds of this name being a girl are #6/15#.

Next, you'll pull out a second name but the odds have changed.

The new odds of drawing a girls' name are #5/14# and the new odds of drawing a boys' name are #9/14#. You do this because there are not as many names in the hat as there were before, hence the subtraction of 1 from the denominator. There is also one fewer girls' names in the hat because you pulled one out.

The odds of you drawing a second girls' name is #5/14#.

Take both of these odds (#6/15# and #5/14#) and multiply them together. This will give you the odds of drawing two girls' names in a row without replacement.

The final answer should be #30/210# which simplifies to #1/7#.

#C_(6,2)/C_(15,2)=1/7#

Explanation:

An alternate way to work this is to calculate the number of ways 2 girls's names can be pulled and divide that by the total number of ways 2 names can be pulled. We're working with combinations (we don't care in what order the names are pulled) and the general formula is:

#C_(n,k)=(n!)/((k!)(n-k)!)# with #n="population", k="picks"#

So we'll have:

#C_(6,2)/C_(15,2)=((6!)/(2!4!))/((15!)/(2!13!))=((6!)/(2!4!))((2!13!)/(15!))=>#

#=>(6xx5xx4!)/(4!)(13!)/(15xx14xx13!)#

#=> 30/210=1/7#