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HI

 

I was going through the thread in http://www.www.urch.com/forums/gmat-math/24356-complete-guide-permutation-combination-probability-all-you-need-gmat-3.html

 

 

Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that NONE of them are blue?

 

 

The problem above is similar to the problem in question 18 as given below in the word doc.

 

 

Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that they are not all blue?

 

 

for the first one Twinnsplitter gave the solution (4/7) x (3/6) x (2/5) = 4/35

 

 

but in her/his answer in the blog her answer is 34/35 ( D) for the 18 question . There is no option of 4/35

 

 

[TABLE=width: 400]

[TR]

[TD]Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that they are not all blue?

[/TD]

[/TR]

[TR]

[TD]

[/TD]

[/TR]

[TR]

[TD] A. 5/7

[/TD]

[/TR]

[TR]

[TD] B. 23/24

[/TD]

[/TR]

[TR]

[TD] C. 6/7

[/TD]

[/TR]

[TR]

[TD] D. 34/35

[/TD]

[/TR]

[TR]

[TD] E. 8/13

 

 

so should we work like this : probability of all blue 3/7*2/6*1/6=1/35

 

 

none blue : 1-(1/35)= 34/35 one of the given options

 

 

but then is 4/7*3/6*2/5= 4/35 wrong !! as stated in the blog

 

 

if 4/35 is wrong then even question 3( below) could be wrong

 

 

 

[TABLE=width: 400]

[TR]

[TD]A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

[/TD]

[/TR]

[TR]

[TD]

[/TD]

[/TR]

[TR]

[TD] A. 15/28

[/TD]

[/TR]

[TR]

[TD] B. 1/4

[/TD]

[/TR]

[TR]

[TD] C. 9/16

[/TD]

[/TR]

[TR]

[TD] D. 1/32

[/TD]

[/TR]

[TR]

[TD] E. 1/16

[/TD]

[/TR]

[/TABLE]

 

 

 

 

[/TD]

[/TR]

[/TABLE]

so we cannot solve this as 6/8*5/7= 15/28 as this method did not work for question 18 there we got 4/35 which is not among the options .

and we should solve this as all blue prob. 2/8*1/7= 1/28 and none blue 1-(1/28)=27/28

 

but this is not among the options , but if we solve this as 6/8*5/7=15/28 , this is among the options .So why are we solving two similar sums in different ways . What is the difference between question 3 and question 18..

 

 

 

 

appreciate your help .

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