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goalsnr
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young_gun
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young_gun
i'm getting 13/28 but that's not a choice...

1 - (6/8)(5/7)

The answer choices are broken.
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goalsnr
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young_gun
i'm getting 13/28 but that's not a choice...

1 - (6/8)(5/7)

The answer choices are broken.


Yep Iam getting 13/28 with the same approach. But the OA given for this problem is E.The problem is from Set 25
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young_gun
i'm getting 13/28 but that's not a choice...


I am also getting 13/28

1-6c2/8c2

can you explain how you got 6c2 and 8c2? combination is one of my weak areas...
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seongbae
rpmodi
young_gun
i'm getting 13/28 but that's not a choice...


I am also getting 13/28

1-6c2/8c2

can you explain how you got 6c2 and 8c2? combination is one of my weak areas...

seongbae ,

Here total number of choice for TV set is 8 (2 b&w ; 6 color ) , by combination logic if we have to select two TV set out of 8 , it is

8c2= 8!/(2!(8-2))!

Now if we have to find the pobability that atleast 1 TV is B&W , easier method is to find the probability of no B&W TV chosen which means 2 TV sets are chosen from 6 color TV ; by the same logic explained above , number of way we can select 2 TV out of 6 will be 6c2. And so probability of chosing 2 TV set from 6 color TV set will be 6c2/8c2 .

To get aleast one B&W TV , we simply subtract 6c2/8c2 from 1 .



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