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).

2 Answers

Relevance
  • M3
    Lv 7
    1 decade ago
    Favorite Answer

    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.

Still have questions? Get your answers by asking now.