You have a 3-card deck containing a king, a queen, and a jack. You draw a random card, then put it back and draw a second random card. Calculate the probability that you draw exactly 1 king.

Question

anonymous · Answer

Same as the earlier question about the jack: 2/3*1/2 = 1/3

dumbcow · Answer

P(draw both jacks) = 1/3*1/3 = 1/9
P(no jacks) = 2/3*2/3 = 4/9
P(1 jack) = 1 - 1/9 - 4/9 = 4/9

dumbcow · Answer

its different because there is replacement

anonymous · Answer

Probability of taking out one king in one attempt is 1/3

Correspondingly, prob of 'not' taking out a king in another attempt is 2/3

So the total probability of taking out 'just one' king is 2/9!

dumbcow · Answer

what if you pick the king on the 2nd attempt, not on the first

anonymous · Answer

yes! i pick it on 'any' one of the attempt! both the probs have the same sample set, so it shudnt matter

dumbcow · Answer

yes both situations have prob of 2/9
2/9 + 2/9 = 4/9

anonymous · Answer

the right answer is 4/9

anonymous · Answer

Getting king from first attempt only = 1/3 * 2/3 = 2/9
Getting king from second attempt = 2/3 * 1/3 = 2/9
Add them up , u get 4/9

anonymous · Answer

@dumbcow: I was thinking whether both are same or different, but yeah, you're right as there is actual time difference, sry!