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# A bag contains 5 yellow balls and 6 orange balls. When 4 bal

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Manager
Joined: 27 Oct 2009
Posts: 129
Location: Montreal
Schools: Harvard, Yale, HEC
A bag contains 5 yellow balls and 6 orange balls. When 4 bal  [#permalink]

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23 Sep 2010, 07:33
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Question Stats:

75% (00:40) correct 25% (00:03) wrong based on 6 sessions

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Hey guys, I have a problem here:

A bag contains 5 yellow balls and 6 orange balls. When 4 balls are drawn at random simultaneously from the bag, what is the probability that not all of the balls drawn are orange?

Please let me know the mistake in my reasoning here:

Probability of selecting the first orange ball is 6/11
second 5/10
third 4/9
fourth 3/8

So the probability in selecting all four orange balls is 6/11*5/10*4/9*3/8 = 1/22

and therefore the ans = 1 - 1/22 = 21/22.

But when I use different methods (combinatorics), the result is different, cant find the expln now.

Here goes:

Four balls can be drawn from a bag containing 11 balls in 11C4 ways
The number of ways in which all four balls drawn will be orange = 6C4

The probability that all four balls drawn are orange: (6C4)/(11C4)= 1/11 .......

Maybe I need to take a nap now
Math Expert
Joined: 02 Sep 2009
Posts: 49430

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23 Sep 2010, 07:37
1
ezinis wrote:
Hey guys, I have a problem here:

A bag contains 5 yellow balls and 6 orange balls. When 4 balls are drawn at random simultaneously from the bag, what is the probability that not all of the balls drawn are orange?

Please let me know the mistake in my reasoning here:

Probability of selecting the first orange ball is 6/11
second 5/10
third 4/9
fourth 3/8

So the probability in selecting all four orange balls is 6/11*5/10*4/9*3/8 = 1/22

and therefore the ans = 1 - 1/22 = 21/22.

But when I use different methods (combinatorics), the result is different, cant find the expln now.

Here goes:

Four balls can be drawn from a bag containing 11 balls in 11C4 ways
The number of ways in which all four balls drawn will be orange = 6C4

The probability that all four balls drawn are orange: (6C4)/(11C4)= 1/11 .......

Maybe I need to take a nap now

It should be $$1-\frac{C^4_6}{C^4_{11}}=\frac{21}{22}$$, $$C^4_6$$ choosing 4 orange from 6 (favorable outcome) and $$C^4_{11}$$ choosing any 4 from total of 11 (total # of outcomes).
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Manager
Joined: 27 Oct 2009
Posts: 129
Location: Montreal
Schools: Harvard, Yale, HEC

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23 Sep 2010, 08:00
I did the cal. of the combinatoric wrong now i see it at last thanks
(6C4)/(11C4)= 1/22
Manager
Joined: 20 Jul 2010
Posts: 223

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23 Sep 2010, 13:29
do we have so much time as to solve a problem in two different ways in GMAT....I would go with one sure way rather than two ways giving two different answers....

Just my 2cents....
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Intern
Joined: 10 Oct 2010
Posts: 22
Location: Texas

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11 Oct 2010, 04:47
ezinis wrote:
A bag contains 5 yellow balls and 6 orange balls.

Bag of 11 choices.

ezinis wrote:
When 4 balls are drawn at random simultaneously from the bag,

Combo box arrangement
(_)(_)(_)(_)/4!

ezinis wrote:
what is the probability that not all of the balls drawn are orange?

Probability Table. Create and work backwards.

# of Oranges: Events
0:
1:
2:
3:
4:
------------------------
Total =

****************************************************

Total = 11 pick 4 = 11C4 = 11*10*9*8/4*3*2*1 = 330

Orange = 4: 6 orange pick 4 = 6C4 = 6*5*4*3/4*3*2 = 15 <----------- at this point the question can be answered, but I'll continue filling in the table for fun.

Orange = 3: 6 orange pick 3 AND 5 yellow pick 1 = 6C3 * 5C1 = 100

Orange = 2: 6 orange pick 2 AND 5 yellow pick 2 = 6C2 * 5C2 = 150

Orange = 1: 6 orange pick 1 AND 5 yellow pick 3 = 6C1 * 5C3 = 60

Orange = 0: 5 yellow pick 4 = 5C4 = 5
OR Sum the table total: 330 - (15+100+150+60) = 5

****************************************************

# of Oranges: Events
0: 5
1: 60
2: 150
3: 100
4: 15
------------------------
Total = 330

****************************************************

Use the info in the table to answer any probability question.
P(Orange = 0) = 5/330
P(Orange = 1) = 60/330
P(Orange = 2) = 150/330
P(Orange = 3) = 100/330
P(Orange = 4) = 15/330
P(Orange <> 4) = P(Orange = 0,1,2,or3) = (5+60+150+100)/330 = 1 - 15/330 = 21/22
P(Orange = Odd) = (60+100)/330
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Re: A bag contains 5 yellow balls and 6 orange balls. When 4 bal  [#permalink]

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18 Oct 2017, 11:15
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Re: A bag contains 5 yellow balls and 6 orange balls. When 4 bal &nbs [#permalink] 18 Oct 2017, 11:15
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