1. ## Card Probability

I have some problems similar to this, but in the back of the book it just says answers will vary...

Could someone please break this down to me?

Given events A and B within the sample space S, the following sequence of steps establishes formulas that can be used to compute conditional probabilities. Justify each statement.

a) $\displaystyle P(A and B) = P(A) * P(B|A)$

b)Therefore, $\displaystyle P(B|A) = P(A and B)/P(A)$

c) Therefore, P(B|A) = n(A and B)/n(S) over n(A)/n(S)

d)Therefore, $\displaystyle P(B|A) = n(A and B)/ n(A)$

Use the results from above to find each probability in a standard deck of 52 cards.

Like this one for example:

$\displaystyle P (red | diamond)$

2. Anyone, Blam da Blam...

I was wondering if it all meant the same thing...

I would think it is saying red intersects diamond/ red, but this just doesn't seem right to me.

3. Originally Posted by WhoaBlackBetty
I have some problems similar to this, but in the back of the book it just says answers will vary...

Could someone please break this down to me?

Given events A and B within the sample space S, the following sequence of steps establishes formulas that can be used to compute conditional probabilities. Justify each statement.

a) $\displaystyle P(A and B) = P(A) * P(B|A)$

b)Therefore, $\displaystyle P(B|A) = P(A and B)/P(A)$

c) Therefore, P(B|A) = n(A and B)/n(S) over n(A)/n(S)

d)Therefore, $\displaystyle P(B|A) = n(A and B)/ n(A)$

Use the results from above to find each probability in a standard deck of 52 cards.

Like this one for example:

$\displaystyle P (red | diamond)$ Mr F says: This probability is obviously equal to 1.
..

4. Ok...thank you. Its kind of what I thought. I will just go through the other ones. I thought maybe this was some kind of trick question.