first 2 cards have already been selected, so we can forget about that. Now we have 46 cards, so the chance of having the third card selected represent a share of stock is the same as selecting the first card form a deck of 46 with 8 stock cards. P=8/46 or 4/23.

Answer E

dferm wrote:In a certain board game, a stack of 48 cards, 8 of which represent shares of a stock, are shuffled and then place face down. If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?

A. 1/8
B. 1/6
C. 1/5
D. 3/23
E. 4/23

Please Explain...

If there are a total of 48 cards, 8 which represent shares of a stock (lets call these cards X) then there are 40 that do not represent shares of a stock. (Called Y)

Now the first and second cards selected are both Y. This leaves 38 Y and 8 X cards or 46 cards remaining. The chance that one X card will be next is the total X cards/(total X Cards + total Y Cards). 8/46 which is 4/23 or E.

Hope that helps.

This problem is very easy, if you are organized and if you read the question carefully.See Img attached

Since first 2 cards didn't represent the shares, so left over are 46. So, probability is 8 out of 46.
Finally it comes 4/23

In a certain board game, a stack of 48 cards, 8 of which represent shares of a stock, are shuffled and then place face down. If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?

A. 1/8
B. 1/6
C. 1/5
D. 3/23
E. 4/23

Total Cards = 48, Shares of Stock = 8, NON Share of Stock = 48-8 = 40

This is a question of Conditional Probability however it's easiest to think in the way that since two of the cards have already been drawn and they are not part of Share of 8 stock cards therefore the cards remaining in the pack = 48-2 = 46
and Share of stock remaining = 8

Desired Probability = (Prob of third card being share of stock)

Desired Probability = (8/46) = 4/23

Answer: Option E