VeritasPrepKarishma wrote:
Another thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).
Hi Karishma, can you please give the answer to the above query?
I have 2 other questions.
First question: I solved this question using conditional probability. Is my solution correct?
Probability (at least 2 non-share cards) = Probability (2 non-share AND 1 share cards) + Probability (3 non-share cards)
\(= \frac{40*39*8}{48*47*46} + \frac{40*39*38}{48*47*46}\)
\(=\frac{40*39*46}{48*47*46}\)
Desired probability = Probability(2 non-share AND 1 share cards)/Probability(at least 2 non-share cards)
\(=\frac{40*39*8}{40*47*46}\)
\(=\frac{8}{46}=\frac{4}{23}\)
Second question: I cannot tell the difference between the problem under discussion and this problem:
A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?In the latter problem, Bunuel and other experts confirmed that the probability of getting a certain ball (black ball) will not change for any successive drawing. This means the probability of getting a black ball remains 5/8 no matter what.
I thought the same line of thinking would apply here. On the third drawing, the probability of getting a share remains 8/48. I know it is dead wrong, but I cannot understand the difference between the two questions. Could you please clarify?
Thank you very much!