There are two boxes - one holds 11 cards numbered 1 through 11, the other 5 cards numbered 1 through 5. Two cards are randomly drawn. What is the probability of drawing 2 even cards in a row?

Ok... let's walk through this one slowly...

The first thing we have is 2 boxes, which I'll call Box A (which holds 11 cards) and Box B (which holds 5 cards).

#P("Box A")=P("Box B")=1/2#

Draw 1, Box A, chance of drawing an even card

Let's say we draw from Box A. We have cards numbered 1-11, and of the 11 cards, 5 are even. And so we can say that if we draw from Box A, the probability of drawing an even card is #5/11#

Draw 1, Box B, chance of drawing an even card

But what if instead we draw from Box B. We have cards numbered 1-5, and of the 5 cards, 2 are even. And so we can say that if we draw from Box B, the probability of drawing an even card is #2/5#

Draw 1, probability of drawing an even card

We can now put the above together to calculate the probability of drawing an even card:

#P("drawing an even card")=1/2xx5/11+1/2xx2/5=5/22+1/5=>#

#=> 5/22(5/5)+1/5(22/22)=25/110+22/110=47/110#

We can now move on to the second draw.

Draw 2, Box A (where draw 1 was also Box A)

In this possibility, we drew from Box A in draw 1 and are doing so again in draw 2. There are now 10 cards and 4 of them are even, and so we get #4/10=2/5#

Draw 2, Box A (where draw 1 was Box B)

In this possibility, since we drew from Box B before, Box A still has 11 cards and 5 evens, and so that's #5/11#

Draw 2, Box B (where draw 1 was also Box B)

In this possibility, we drew from Box B in draw 1 and are doing so again in draw 2. There are now 4 cards and 1 of them is even, and so we get #1/4#

Draw 2, Box B (where draw 1 was Box A)

In this possibility, since we drew from Box A before, Box B still has 5 cards and 2 evens, and so that's #2/5#

Draw 2, probability of drawing an even card

All four of the above possibilities is equally likely, and so we'll multiply each of the above probabilities by #1/4# and then sum them up:

#P("drawing an even card on second draw")=1/4xx2/5+1/4xx5/11+1/4xx2/5+1/4xx1/4=1/10+5/44+1/10+1/16=>#

#1/10(88/88)+5/44(20/20)+1/10(88/88)+1/16(55/55)#

#88/880+100/880+88/880+55/880=331/880#

And now we can finalize this calculation by multiplying in the probability of getting an even card on the first draw:

#P("drawing two even cards")=47/110xx331/880=(15,557)/(96,800)~=16.07%#