Who can crack this one ?

[ve_boss]

A buyer buys 3 different items out of the newly introduced 10 different items. If two items were to be selected at random, what is the probability that the buyer does not have both the chosen items?


A. 1/8
B. 11/17
C. 13/21
D. 7/15
E. 8/15

[Lhomme]

IMO D

first item will be selected from available 10 items. We have 2 selected items. Hence 8 non selected items. Probability of selecting non selected item = 8/10.

Now there are 9 items remaining. 2 are selected items. 7 are non - selected items. Probability of selecting non selected item = 7/9

Now we have 8 items remaining. 2 items are selected items. 6 are non -selected. probability of selecting non selected item = 6/8.

Probability of selecting 3 non selected items = 8/10 * 7/9 * 6/8 = 7/15

ANS = D. Whats the OA ?

[mickgreen58]

@Lhomme, in your first sentence, why isn't it "3 items selected and 7 non-items"?

Also, when I first saw the problem, I thought maybe the easiest way to do it was to find the probability of selecting 2 items and subtract from 1, not sure if that is the best way though.

Thanks for the breakdown of the problem.

[Lhomme]

You are choosing 3 items from 10. We need to find the probability of choosing these 3 items. We have been given the condition that 2 items have been selected at random and those items are not a part of the items that are finally selected. Hence we know for sure that 2 out of the 10 items will not be in the final selection. We can continue to keep subtracting these 2 items from the total items in order to calculate our probability. Hope this helps.

[krovvidy]

IMO D .... but I felt it can be arrived at in an easier way ... please correct me if I am wrong ...

We need to find the probability of selection 2 items which are not bought by the buyer (who bought 3 out of 10 items).

probability = no. of favorable cases/total no. of cases

total no. of cases = 10c2
total no. of favorable cases = 7c2 (7 out of 10 items are not bought)

Therefore, probability = 7c2/10c2 = (7*6)/(10*9) = 7/15

[mickgreen58]

Interesting stuff Lhomme and Krovvidy.

Let me ask you all something, can Combination (even Permutation) Theory always be applied to Probability?

[krovvidy]

Certainly mickgreen58 ... combinatorics can be applied anywhere ... in fact, some of the probability theorems are proved using principles of combinations.

Now-a-days, GMAT started testing permutations, combinations & probability concepts quite regularly ... you can be sure of 1 qn in Quant (if not 2 or 3) ...

Don't worry if you're slow to understand these ... you'll eventually do well with practice ...

[Praso]

quick way guys.. i was lost in a looooong route

[ve_boss]

OA is D

Note that there is another approach :

1 - (at east 1 stamp from John)
= 1 - [(exactly 1 stamp from John + another from the remaining 7) + (both the stamps from John)
= 1 - (3c1*7c1/10c2 + 3c2/10c2) = 1 - (21/45 + 3/45) = 21/45 = 7/15

[bose]

2 items selected at random that buyer cannot buy, so buyer can buy only from remaining 8, so 8C3/10C3=7/15