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

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

Vittal,

the same happens to me. I cannot properly see the attachments on posts.

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

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...

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

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.

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

avenus wrote:

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

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...

Thanks avenus. As u mentioned, I right clicked on the Ad and chose "block images". This blocked images from page2ad.google.com.

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

you can also right click and select "do not show images from this site" to block advertisements

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

I am confused here....appreciate some help clearing out the cloud for me. Thanks in advance.

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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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

Safari, Chrome and Firefox also. Think about Mac OS, which does not have IE.

Work around: to view properly right click on image and view it in another tab.

I feel its more with formatting than with browsers.