[Petr Kropotkin]

Two cards are drawn from deck
[1] what is the probability both are Aces
[2] at least one red the other black

Is one a conditional probablity problem ?
[2] (1-12/52)(1-12/51) ?

[Unknown008]

(a) I already told you how to do it in the other thread.

I'll say it again.
Drawing two cards together is the same as drawing two cards, one after the other without replacement.

P(2 Aces) = P(1st is an Ace) x P(2nd is an Ace)

P(1st is an Ace) = (no. of aces)/(number of cards)
P(2nd is an Ace) = (no. of aces left)/(number of cards left)

Can you try it?

(b) P(at least one red, the other black) is a little strange. If you get 'at least one red', that means either one red one black, or two reds. Now, saying 'the other black' just says that you have to get one red and one black. It should be either:
at least one red, or,
one red and one black.
but not both.

P(at least one red) = 1 - P(no black) =

P(one red, one black) = P(red, black) + P(black, red) =

[Petr Kropotkin]

I tried it and my answer was different that books
The book gives 32/221

P(2 aces)= 2/50*48/50=32/26*25= 32/650

For the second the book gives 26/51

It is one is red and the other black. It is rom an approach to general systems theory by Klir

[Unknown008]

There are initially 4 aces. Hence P(1st ace) = 4/52
There are then 3 aces left. Hence, P(2nd ace) = 3/51

Both give 4/52 * 3/51 = 1/221

The book might have done the number wrong. The probability of having 2 aces is 1/221, there is no other answer to that (if the pack of card is a regular pack of cards)

For the second one, you need one black and one red.

So, use the second one I gave you, giving 26/51 which is good.



Theory: There are 26 red cards and a total of 52 cards. P(1st red) = 26/52 = 1/2.
Since you removed a card, there are 51 remaining. But there are still 26 black cards. So, P(2nd black) = 26/51

Now, there is also the case that you take the first card as black, and the second as red. This is acceptable too because you only need a red and a black, irrespective of the order. The same logic applies, and when added up give the answer.

In fact, you'll find that the probability of having red first then black is the same as having a black first then red. Hence, you can also multiply the first probability by two instead of adding both, to give the answer.

[Petr Kropotkin]

There are initially 4 aces. Hence P(1st ace) = 4/52
There are then 3 aces left. Hence, P(2nd ace) = 3/51

Both give 4/52 * 3/51 = 1/221

The book might have done the number wrong. The probability of having 2 aces is 1/221, there is no other answer to that (if the pack of card is a regular pack of cards)

For the second one, you need one black and one red.

So, use the second one I gave you, giving 26/51 which is good.



Theory: There are 26 red cards and a total of 52 cards. P(1st red) = 26/52 = 1/2.
Since you removed a card, there are 51 remaining. But there are still 26 black cards. So, P(2nd black) = 26/51

Now, there is also the case that you take the first card as black, and the second as red. This is acceptable too because you only need a red and a black, irrespective of the order. The same logic applies, and when added up give the answer.

In fact, you'll find that the probability of having red first then black is the same as having a black first then red. Hence, you can also multiply the first probability by two instead of adding both, to give the answer.

I realised that. I went through all sorts of calculations to get which is I posted. I didn't get the books answer and it drove me nuts.

[Unknown008]

So, you understand how to solve those kinds of problems now?

[Petr Kropotkin]

Better understanding of it.