Probability

[greycellz]

complement of 'the buyer has exactly 1 and the buyer has exactly 2'

[Sniper]

7C2/10C3...

Cheers,

[Vakul]

Originally Posted by Coral

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?

I guess here the probability asked is on the the number of items by buyer and not on the number of 2 selected items. thus total number of ways should not be 10C2.

Most of us agree on the favorable ways... that it is 7C2

Total # of ways = 10C3
# of favorable ways = 7C2

[upchucks]

can someone break this down for me. i am totally lost.

[lego2401]

Originally Posted by upchucks

can someone break this down for me. i am totally lost.

Ok here is how I saw this problem.

First we bought 3 items (or picked 3 items). So now we have two sets, one of 7 items, and one of 3.

Next we need to pick two items. So as now to have both the chosen items, we would prefer that one comes from the set of 7 left, and one from the set of 3 we bought (rather than both from the set we bought).

So the numerator is 7C1 * 3C1 = 21
As for the denominator,

From 10 items in total, the total number of way to chose 2 items is
10C2 = 10!/(8! * 2!) = 10*9/2 = 45

So probability of 2 items where they are not both from the set we bought is
21/45, or 7/15.

Had they said the wanted both to be from the ones we bought, then we'd do
3C2 / 10C2 = 3/45 = 1/15

and if they said both not from the set we bought

7C2 / 10C2 = 21/45 = 7/15

And as you can see, we covered all cases, the total sum of probabilities is 7/15+7/15+1/15.

Now my only concern is that they said that not all from the ones we bought. That doesnt mean that one from each only in my opinion, that could actually cover one from each or both from the ones we didnt buy. In this case,
Our answer would be 14/15.

Hope that covers it for you ......