Finding the probability of selecting cards that are not pairs.
Probability of selecting first card = 12/12 = 1 ( It can be any card).
Probability of selecting second card = 10/11 ( 11 cards are left for selection. We shouldn't select the matching card. )
Probability of selecting the third card = 8/10 ( 10 cards left we shouldn't select the two cards that match the already selected 2 cards)
Probability of selecting the 4th card = 6/9
Probability of selecting the cards with no pairs = 1* 10/11*8/10*6/9 = 16/33
Probability of selecting at least one pair = 1-16/33 = 17/33.
Thanks
Raama
cards
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vittalgmat
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On an slightly related topic. (sorry to post here. I will mail eric and mozilla guys about this issue)
Does everyone see jpg attachments properly in firefox for windows. The community sponsors banner drops down over the picture and I cant read completely.
thanks
-V
Does everyone see jpg attachments properly in firefox for windows. The community sponsors banner drops down over the picture and I cant read completely.
thanks
-V
A quick solution: place the mouse pointer over the image, click on the right button and select show image or the equivalent in the language/browser you're using. The image on its own will be shown in a separate tab...On an slightly related topic. Does everyone see jpg attachments properly in firefox for windows. The community sponsors banner drops down over the picture and I cant read completely. thanks -V
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kanha81
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I really like the way you solve the problems, makes it look simpler and understandable. I always fight between 2 methods. This is the method that I chose; however, it took more than 2 mins, but less than 4 mins.krisraam wrote:Finding the probability of selecting cards that are not pairs.
Probability of selecting first card = 12/12 = 1 ( It can be any card).
Probability of selecting second card = 10/11 ( 11 cards are left for selection. We shouldn't select the matching card. )
Probability of selecting the third card = 8/10 ( 10 cards left we shouldn't select the two cards that match the already selected 2 cards)
Probability of selecting the 4th card = 6/9
Probability of selecting the cards with no pairs = 1* 10/11*8/10*6/9 = 16/33
Probability of selecting at least one pair = 1-16/33 = 17/33.
Thanks
Raama
P(E): Finding at least 1 pair that have same value.
4 cards chosen from 12: 12C4 ways
We find P(E`): Finding none of the 4 cards chosen have same value
=6C4 * 4C1 * 4 (ie choose 4 cards from a pair of 6 cards, such that, you choose 1 card from each pair and this can be done in 4 ways) / (12C4)
= 16/33
p(E) = 1-P(E`) = 1 - 16/33 = 17/33
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vittalgmat
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Thanks avenus. As u mentioned, I right clicked on the Ad and chose "block images". This blocked images from page2ad.google.com.avenus wrote:A quick solution: place the mouse pointer over the image, click on the right button and select show image or the equivalent in the language/browser you're using. The image on its own will be shown in a separate tab...On an slightly related topic. Does everyone see jpg attachments properly in firefox for windows. The community sponsors banner drops down over the picture and I cant read completely. thanks -V
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RBurroughs
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you can also right click and select "do not show images from this site" to block advertisementsvittalgmat wrote:On an slightly related topic. (sorry to post here. I will mail eric and mozilla guys about this issue)
Does everyone see jpg attachments properly in firefox for windows. The community sponsors banner drops down over the picture and I cant read completely.
thanks
-V
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Vemuri
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I am confused here....appreciate some help clearing out the cloud for me. Thanks in advance.krisraam wrote:Finding the probability of selecting cards that are not pairs.
Probability of selecting first card = 12/12 = 1 ( It can be any card).
Probability of selecting second card = 10/11 ( 11 cards are left for selection. We shouldn't select the matching card. )
Probability of selecting the third card = 8/10 ( 10 cards left we shouldn't select the two cards that match the already selected 2 cards)
Probability of selecting the 4th card = 6/9
Probability of selecting the cards with no pairs = 1* 10/11*8/10*6/9 = 16/33
Probability of selecting at least one pair = 1-16/33 = 17/33.
Thanks
Raama
When I look at the way the question is framed, I am considering that 4 cards have been laid out & we have 8 cards remaining to find the probability of not finding a match. Can someone please help with explaining the approach.
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rs2010
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Safari, Chrome and Firefox also. Think about Mac OS, which does not have IE.vittalgmat wrote:On an slightly related topic. (sorry to post here. I will mail eric and mozilla guys about this issue)
Does everyone see jpg attachments properly in firefox for windows. The community sponsors banner drops down over the picture and I cant read completely.
thanks
-V
Work around: to view properly right click on image and view it in another tab.
I feel its more with formatting than with browsers.
