SOLUTION: Two cards are drawn successively from an ordinary deck of 52 well-shuffled cards. Find the probability that (a) the first card is not a 10 of clubs or an ace, (b) the first card i

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Question 1187388: Two cards are drawn successively from an ordinary deck of 52 well-shuffled cards. Find the probability that
(a) the first card is not a 10 of clubs or an ace, (b) the first card is an ace but the second is not, (c) at least one
card is a diamond, (d) the cards are not of the same suit, (e) not more than one card is a picture card (jack,
queen, king), ( f ) the second card is not a picture card, (g) the second card is not a picture card given that the
first was a picture card, and (h) the cards are picture cards or spades or both.
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!


a) There are 5 unfavorable outcomes (10 of clubs + any of the four Aces).  This means the first card has a (52-5)/52 = 47/52 chance of being a favorable outcome. 



b) In this case, 4/52 chance of drawing an Ace on the first draw, leaving 48 non-Aces for the 2nd draw, which is drawn from the 51 remaining cards:

4/52 * 48/51 = 192/2652  (reduce or convert to decimal as needed)



c) At least one diamond -- that means the first card can be a diamond, OR the 2nd card can be a diamond, or BOTH can be a diamond.  We can solve more easily by taking (1 - inverse condition)  which is  (1 - (probability of NO diamonds in two draws)):



1 - (39/52)(38/51) = 1 - 1482/2652 = 1170/2652   (reduce, convert as needed)



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The long way is to consider each case and add the probabilities (D = diamond drawn):   P(at least one D) = P(D)*P(not D) +  P(D)P(D) + P(not D)P(D)



(13/52)(39/51) + (13/52)(12/51) + (39/52)(13/51) = (507+156+507)/2652 = 1170/2652   (same as before)



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d.  Here the first draw establishes the suit, and the 2nd draw must not match. So really it is only the 2nd draw you need to look at, or we can say the first card has "some suit" with probability 1 and include it:  (52/52) * (39/51)





... by now you should get the idea: for each problem you need to consider (and count, or compute) the number of favorable outcomes and divide that number by the total number of possible outcomes.  Remember the denominator is 52 on the first draw, 51 on the 2nd draw because you are  not replacing cards.



Good luck.