beeblebrox wrote:
This is how I solved it:
for positive difference to be 0 we will draw the cards like this:
(5,5)
for positive difference to be 1 we will draw the cards like this:
(6,5),(5,6),(5,4),(4,5)
for positive difference to be 2 we will draw the cards like this:
(2,4),(4,2),(6,4),(4,6)
for positive difference to be 3 we will draw the cards like this:
(5,2),(2,5)
==11 ways
Total ways of selecting any two cards:(5,5),(5,2),(2,5), (2,4),(4,2),(6,4),(4,6), (6,5),(5,6),(5,4),(4,5), (6,2), (2,6) ==13 ways
So the ans== 11/13
What did I do wrong? I literally thought that this is the only way we can pick cards without replacement. Also, since we are not replacing the cards, I thought we need to ensure that order matters.
beeblebroxI'll try to explain your mistake by first asking a question. If the deck instead included only 2, 4, 5, 5, 6 (missing one of the 5s), you would have ended up with the same result by deploying your method, right? Does having three 5s instead of two 5s matter? Of course it does! So your method can't be the right one. Why? When you listed (6,5), for example, you really needed to list (6,5a), (6,5b),(6,5c). The same is true for each time you listed a 5.
Okay, so how do we solve it?
How many total ways are there? We have 6 options for which card we draw first. We have 5 options for which card we draw second. So there are 6*5 = 30 total possible ordered pairs. Of those, the only ways to "lose" are (6,2) and (2,6). So 28/30 "win." That's 14/15.
Answer choice E.