From a group of 5 men, 4 women and 3 children, if 4 persons are selected at random, what is the probability of no child being selected?
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children not selected
Source: Beat The GMAT — Problem Solving |
Good question Ruplun :!:
As with most other probability questions, there are a variety of ways to solve this, and here are a few:
1. The best and the easiest has been posted above by ColumbiaVC :!: (and Tryingtobeat :!:):
9c4/12c4 = 126/495 = [spoiler]14/55 = 0.254545[/spoiler]
where 9c4 = ways to select 4 from 9 (9 non-children, M or Wo does not really matter here)
...and 12c4 = all possible ways to select 4 from 12
2. Similarly, we can also do: (9/12)*(8/11)*(7/10)*(6/9) = 3024/11880 = [spoiler]14/55 = 0.254545[/spoiler]
This is also fairly intuitive, 9/12 ways to select the first 'non-child'
...so also 8/11 ways, 7/10 ways and 6/9 ways to select the 2nd, 3rd and 4th non-child(ren)
3. Long way - but this decomposes the problem analytically and is helpful for other variations of this and similar questions (such as pr(at least one child selected)):
Pr(No Child) = 1- pr(at least one child selected)
pr(at least one child selected)= Pr(3 Ch)+Pr(2 Ch)+ Pr(1 ch) [Ch= Child selected]
=>pr(at least one child selected)= (3*2*1*9*4c3 + 3*2*9*8*4c2 + 3*9*8*7*4c1)/(12*11*10*9)
Note that the denominator '(12*11*10*9)' has been factored out and 4cn represents the different ways in which 3, 2 or 1 child(ren) can be selected to fit the 4 spots.
=> pr(at least one child selected)= (3*2*1*9*4 + 3*2*9*8*6 + 3*9*8*7*4)/(12*11*10*9)
=> pr(at least one child selected)= [spoiler]8856/11880 = 41/55 = 0.74545[/spoiler]
=> Pr(No Child) = 1- pr(at least one child selected)= [spoiler]1- 41/55 = 14/55 = 0.254545[/spoiler]
4. Yes, even if the question were to find pr(at least one child selected), it would still be quicker to find Pr(all 4 non-children) and find pr(at least one child selected) by doing 1-Pr(all 4 non-children)
Hope this helps...
(Probability is always fun - that's where I myself make the most errors too!)
As with most other probability questions, there are a variety of ways to solve this, and here are a few:
1. The best and the easiest has been posted above by ColumbiaVC :!: (and Tryingtobeat :!:):
9c4/12c4 = 126/495 = [spoiler]14/55 = 0.254545[/spoiler]
where 9c4 = ways to select 4 from 9 (9 non-children, M or Wo does not really matter here)
...and 12c4 = all possible ways to select 4 from 12
2. Similarly, we can also do: (9/12)*(8/11)*(7/10)*(6/9) = 3024/11880 = [spoiler]14/55 = 0.254545[/spoiler]
This is also fairly intuitive, 9/12 ways to select the first 'non-child'
...so also 8/11 ways, 7/10 ways and 6/9 ways to select the 2nd, 3rd and 4th non-child(ren)
3. Long way - but this decomposes the problem analytically and is helpful for other variations of this and similar questions (such as pr(at least one child selected)):
Pr(No Child) = 1- pr(at least one child selected)
pr(at least one child selected)= Pr(3 Ch)+Pr(2 Ch)+ Pr(1 ch) [Ch= Child selected]
=>pr(at least one child selected)= (3*2*1*9*4c3 + 3*2*9*8*4c2 + 3*9*8*7*4c1)/(12*11*10*9)
Note that the denominator '(12*11*10*9)' has been factored out and 4cn represents the different ways in which 3, 2 or 1 child(ren) can be selected to fit the 4 spots.
=> pr(at least one child selected)= (3*2*1*9*4 + 3*2*9*8*6 + 3*9*8*7*4)/(12*11*10*9)
=> pr(at least one child selected)= [spoiler]8856/11880 = 41/55 = 0.74545[/spoiler]
=> Pr(No Child) = 1- pr(at least one child selected)= [spoiler]1- 41/55 = 14/55 = 0.254545[/spoiler]
4. Yes, even if the question were to find pr(at least one child selected), it would still be quicker to find Pr(all 4 non-children) and find pr(at least one child selected) by doing 1-Pr(all 4 non-children)
Hope this helps...
(Probability is always fun - that's where I myself make the most errors too!)
