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wow...keep up the good work guys..
forum is picking up momentum now..gr8...

pls help to solve this..

Find the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains atleast 1 woman ?

cannot remember the ans choices...could you pls help me with the approach..
thanx...

marcel wrote:
wow...keep up the good work guys..
forum is picking up momentum now..gr8...

pls help to solve this..

Find the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains atleast 1 woman ?

cannot remember the ans choices...could you pls help me with the approach..
thanx...

no takers for this question....

ok marcel ...here is the approach to these kind of problems.

probability is chance of something happening..

probability that a team of 4 should comprise atleast 1 woman implies :

Team of 4 can have 1 woman OR 2 women OR 3 women OR all women..
in effect, u add up individual probability of each eventuality...
only case u dont want is when a team of 4 has no women..

Also, probability of a certain event is 1.

Hence, p(atleast 1 woman) = 1 - p( no woman is selected in the team)

no of ways to select a team of 4, such that there is no woman means to select 4 people from remaining 11 people ( 6 men and 5 children )
i.e 11C4.

total poss ways to select a team of 4 is 18C4 ways .

Hence, p(no woman in the team) = 11C4/18C4.

Hence, p( atleast 1 woman) = 1 - 11C4/18C4

I am sure this would be 1 of the answer options...you may not be expected to do the calculation on GMAT..

hope this solves your query..keep solving..

am not able to solve this question...could you guys pls help..

5 girls and 3 boys are arranged randomly in a row. Find the probability that there is one boy on each end.

i dont have the answer options..can somebody pls help with this question...

thanx...

Hi Naaz,

Following is the way I am approaching the problem. Am not sure abt the answer. Would request Bhavin to validate.

Since there are 5 girls and 3 boys, I am drawing 8 lines in a row indicating the position of the girls and boys in the row

I II III IV V VI VII VIII
------- --------- --------- ---------- --------- -------- -------- ----------

Now the Ist position has to be filled in by a boy. Since there are 3 boys, the Ist place can be filled in 3C1 ways.

Left are 2 boys. The VIIIth position has to be filled by a boy among the rest of 2 boys. That can be done in 2C1 ways.

Now there are 6 vacant positions. The last boy can take any place of the 6 in 6C1 ways.

We have now accounted for the boys. Left are 5 positions for 5 girls. That cane be done in 5! ways.

So total ways 5 girls and 3 boys can stand in a row so that one boy is at each end is:
3C1*2C1*6C1*5!

Total possibe combinations among 8 people to stand in a row is 8!

So the required probability is : (3C1*2C1*6C1*5!)/ 8! = 4320/40320 = 3/28

The answer is 3/28..

Hope thats ok. Would request Bhavin to validate the answer.
Thanks

rchuck21 wrote:
Hi Naaz,

Following is the way I am approaching the problem. Am not sure abt the answer. Would request Bhavin to validate.

Since there are 5 girls and 3 boys, I am drawing 8 lines in a row indicating the position of the girls and boys in the row

I II III IV V VI VII VIII
------- --------- --------- ---------- --------- -------- -------- ----------

Now the Ist position has to be filled in by a boy. Since there are 3 boys, the Ist place can be filled in 3C1 ways.

Left are 2 boys. The VIIIth position has to be filled by a boy among the rest of 2 boys. That can be done in 2C1 ways.

Now there are 6 vacant positions. The last boy can take any place of the 6 in 6C1 ways.

We have now accounted for the boys. Left are 5 positions for 5 girls. That cane be done in 5! ways.

So total ways 5 boys and 3 girls can stand in a row so that one boy is at each end is:
3C1*2C1*6C1*5!

Total possibe combinations among 8 people to stand in a row is 8!

So the required probability is : (3C1*2C1*6C1*5!)/ 8! = 4320/40320 = 3/28

The answer is 3/28..

Hope thats ok. Would request Bhavin to validate the answer.
Thanks

Wow..lovely approach...even i thought so of starting to count the no of ways for each position starting from the most restricted position...but was somehow unable to solve till the end..
solution looks good..let us see if this gets confirmed..

hi,
could you provide me another solution to this problem without using the formula & just doing it normally.