Problem Solving

In a certain board game, a stack of 48 cards, 8 of which represent shares of stock, are shuffled and then placed 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

OA is E...whats the correct method to solve this...i got D as my answer...

Answer is 4/23

probability of selecting 1 card out of 8 stock cards is 8c1 = 8

total population is 46 ( 2 cards are already drawn and is assumed not replaced)

hence probability is 8/46 which is 4/23

please let me know if u have any doubts,...

Just imagine the scenario, 48 cards, only 8 cards are good.
You place a set of cards down.
At this time the probability the first card is a good card is: 8/48

Now, the scenario question is the first txo cards are wrong ones.
So before looking at the third card you know txo things:
46 cards remain
8 cards are good (given that there was no good cards in the 1st and 2nd place)
So the probability the third one is a good card is: 8/46=4/23

@sudhir3127 #
@pepeprepa

I have a doubt guys
..when its not given in the question nethg about replacement.. do we have to assume it "without replacement"??

yeah ,, i have done lotof problem where nothing is specified abt replacement and its assumed that its " without replacement"

hope it helps u !!

here is my method:

8 good...48 possible

1st 2 card not good....leaves with prob of 8 good/46 avail
reduce you get 4/23

this is the first time, I see a question that's solve like this,


usually, if it is asking the probability of having the frist two cars not shares of stock, and thrid car is shares of stock

I would solve it like this:

40/48*39/47*8/46

but for some reason, I just no quite fitting my mind to this quesiton yet

The question is targeted at ONLY selecting the 3rd card...the first 2 cards are already selected...so you dont need to consider probability of the first 2 not being shares of stock...

Hence in your calculation, you would need to get rid of 40/48*39/47....leaving you with 8/46 -> 4/23...

Hope this helps.

Cheers.

robbie523 wrote:this is the first time, I see a question that's solve like this,


usually, if it is asking the probability of having the frist two cars not shares of stock, and thrid car is shares of stock

I would solve it like this:

40/48*39/47*8/46

but for some reason, I just no quite fitting my mind to this quesiton yet

yeah, now I got it! can you help me solve that patio and swimming pool question also? I'm posting it now! is a DS question