coffeeloverfreak wrote:

OK, here's my reasoning. I hope it's right:

The theory is that this is a pick without replacement question. The first pick is a free one; you have an 100% probability that the first telephone you pick will be one of the three types. The second one is the probability that, assuming you remove one of the first type, you'll pick one of the second type.

If the first one you pick is a Telephone 1, your chances of the second one being a Telephone 1 are 64/249.

If the first one you pick is a Telephone 2, your chances of the second one being a Telephone 2 are 74/249.

If the first one you pick is a Telephone 3, your chances of the second one being a Telephone 3 are 109/249.

But of course, each of those three scenarios has a different chance of happening. The first scenario has a 65/250 probability, the second scenario has a 75/250 probability and the third scenario has a 110/250 probability.

So the total probability = (65/250x64x249) + (75/250)(74/249) + (110/250)(109/249)

= 434/1245

Exactly what I did. Anyone else?