# Conditional probability question?

Suppose the two joker cards are left in a standard deck of cards. one of the jokers is red, and the other is black. A single card is drawn from the deck of 54 cards but not returned to the deck, and the second card is drawn. determine the probability of the red joker or a red ace being drawn on either draw.

The answer is 1 - (51/54, 50/53).

I understand that the second part of the equation will find me the probability of me |not| drawing a red joker or ace, but why would I need to do it like this?

Update:

Whoops I mean

The answer is 1 - (51/54 * 50/53).

Relevance
• M3
Lv 7

P[red joker or red ace] = 1 - P[not red joker nor red ace]

= 1 - [ 51/54 * 50/53] = 312 /2862 = 312 / D

why like this ? because it's the simplest way, else something like

P[red ace only] = 2/54* 51/53 *2 (sequences) = 204/D

P[2 red aces] = 2/54*1/53 = 2/D

P[red ace & red joker] = 2/54 * 1/53 *2 = 4/D

P[red joker only] 1/54 * 51/53 *2 = 102/D

and most of the time you might forget to multiply by 2 for sequences !

• Anonymous
5 years ago

(a) P(B) = P(B and A1) + P(B and A2) = P(B|A1)P(A1) + P(B|A2)P(A2) = (2/5)(2/6) + (5/7)(4/6) = 2/15 + 10/21 = 14/105 + 50/105 = 64/105 (b) We wish to compute P(A1|B): P(A1|B) = P(A1 and B)/P(B) = P(B|A1)P(A1)/P(B) = (2/5)(2/6)/(64/105) = (2/15)/(64/105) = (2/15)(105/64) = 7/32 (c) P(A1|B) = 7/32 wherease P(A1) = 2/6 = 1/3 Therefore, P(A1|B) < P(A1). This should agree with intuition given that if a white chip has been chosen, the chances that jar A1 was chosen initially are less likely because jar A1 has far smaller percentage of white chips than jar A2.