Probability

[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?

[RIJU]

7c2/10c2=21/45

[metodiev]

I think it's

7C2/10C3 = 21/120

[greycellz]

No. of ways 2 can be selected without taking the three in question. : 7C2
No. of ways 2 can be selected out of 10 = 10C2
Prob. = 7C2/10C2 = 21/45 = 7/15

[Coral]

Riju and Grecellz are probably correct. But, can you explain why the answer is not 1-3C2/10C2?

[cocoray]

Shouldn't it be:
1 - (probability of the buyer having chosen both the randomly selected items) i.e.
1 - {(8C1)/(10C3)} = 1 - 1/15 = 14/15

It should not matter how many ways 2 items can be chosen out of 10 - what should matter is for any given two items, what is the probability that the the buyer doesn't have both of them.

[greycellz]

Originally Posted by Coral

Riju and Grecellz are probably correct. But, can you explain why the answer is not 1-3C2/10C2?

Does not have both means it should not even have one so acc to your method, it should be:
1- (3C2/10C2 + 3C1.7C1/10C2) = 1 - (3/45 + 21/45) = 1 - (24/45) = 1-8/15 = 7/15
Enjoy!

[cocoray]

Greycellz,

P (buyer has both) = 8/15 - isn't that on the higher side - given that he/ she is selecting 3 out of 10, and getting both in that set of 3....?

[greycellz]

Originally Posted by cocoray

Greycellz,

P (buyer has both) = 8/15 - isn't that on the higher side - given that he/ she is selecting 3 out of 10, and getting both in that set of 3....?

Both or any one

[cocoray]

What are you taking as the complement of P (the buyer doesn't have both) ? Isn't that P (buyer has both)?