Probability: Amy and her deck of cards

Here's my approach:

Let "k" be the number of kings (black) that may be part of the 8 BLACK cards added to Deck 1.
No. of red aces in the new deck remains unchanged i.e. 2
No. of cards in the new deck = 52+8 = 60
No. of kings in the new deck = 4+k (2 Red kings + 2 Black kings + k Black kings)

Thus, probability of picking a red ace or a king is :

2/(52+8) + (4+k)/(52+8) = (6+k)/60

Now, since it's given that the above probability is greater than 1/8
We get,
(6+k)/60 > 1/8

i.e. k > 3/2
i.e. k > 1

k can only take 3 values (0 black kings or 1 black king or 2 black kings)
Thus, since k>1, k=2.

Now, the required probability of Amy picking a black king or a red jack from the new pack can be calculated as follows:

Total no. of black kings in the new pack = 4 (2 from Deck 1 + 2 from the 8 cards added to Deck 1)
Total no. of red Jacks in the new pack = 2

Thus,

4/60 + 2/60 = 6/60 = 1/10.

I got (D) as the answer.
Did anyone else get the same answer???

Cheers,
Taz

tabsang wrote:I got (D) as the answer.
Did anyone else get the same answer???

Cheers,
Taz

D seems to be the right ans. I have a two liner solution for this:

If there are k kings in the pack, then 2+k /60 > 1/8 => k>=6 => k=6 and both the black kings have been borrowed from deck 2.

Required probabilty = (4+2)/60 = 1/10

IMO D should be the answer. What is the answer?

Btw what is source of this problem?

anirudhadhopate wrote:IMO D should be the answer. What is the answer?

Btw what is source of this problem?

This from a printed set of questions that I have.
The OA is (A) but I've verified that the correct answer is infact (D).

So Cheers, You got it spot on ,
Taz