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 Subject: Probabiility and cards Category: Science > Math Asked by: kato25-ga List Price: \$4.00 Posted: 05 Dec 2002 10:36 PST Expires: 04 Jan 2003 10:36 PST Question ID: 119815
 ```I have tried numerous times to resolve the following 2 questions, can someone please help in explaining and solving the questions. 1.In drawing 5 cards from a 52 card deck without replacement, what is the probability that the 3rd card is black? 2.Four cards are drawn in succession from a standard 52 card deck. What is the probability that 3 cards are black?```
 ```I'll answer the second question first: Whenever you draw a card, there are two options: it is either red or black. The "sample space" of an experiment is a list of all possible outcomes. Four cards with two options each will result in a sample space of 2*2*2*2 = 16 different combinations. Since this is so few, I've listed them below: RRRR RRRB RRBR RBRR BRRR RRBB RBRB RBBR BRRB BRBR BBRR BBBR BBRB BRBB RBBB BBBB You are only interested in the last five combinations--I am not sure if you want the probability of drawing exactly 3 black cards (in which case BBBB is excluded), or the probability of at least 3 black cards (in which case BBBB is included). That can easily be accounted for in the calculation. Let's look at the combination BBRB. First we must draw one black card from a deck of 52--which contains 26 black cards. So the probability of the first card being black is 26/52 or 0.500. The second card must also be black, but since we have already removed a card from the deck, the probability is no longer 0.500; instead, we now select amongst the remaining 25 black cards in the deck, which now consists of 51 cards. So the probability of the second black card is 25/51 = 0.490. Next we wish to draw a red card. Since no red cards have yet been chosen, there are 26 available reds in a deck of size 50, yielding a probability of 26/50 = 0.520. Finally, the last black card will be selected from 24 blacks in a deck of 49 cards, for a probability of 24/49 = 0.490 (I am rounding to three decimal places on all calculations). The total probability of selecting BBRB is found by multiplying the individual card probabilities (multiplication is the statistician's way of saying "and"; we want a black and a black and a red and a black, so we multiply). 0.500 * 0.490 * 0.520 * 0.490 = 0.0624. The statistician's way of saying "or" is addition. So to find the probability of BBBR or BBRB, we simply add the probabilities of each of the combinations. Below, I have the work shown (I use fractions until the end in order to preserve accuracy): BBBR: (26/52)*(25/51)*(24/50)*(26/49) = 52/833 = 0.62425 BBRB: (26/52)*(25/51)*(26/50)*(24/49) = 52/833 BRBB: (26/52)*(26/51)*(25/50)*(24/49) = 52/833 RBBB: (26/52)*(26/51)*(25/50)*(24/49) = 52/833 Total probability of selecting exactly 3 black cards = 4*(52/833) = 0.24970 If you want the probability of at least four black cards, you need to add the probability of BBBB: BBBB: (26/52)*(25/51)*(24/50)*(23/49) = 46/833 = 0.055222 Total probability of at least four black cards: 0.30492 For the first question, it does not really matter that you are drawing five cards, since the event you are concerned about takes place on the third draw. You can lay out a sample space for the first three cards like we did above. For 3 cards, there should be 2*2*2 = 8 combinations: RRR RRB RBR BRR BBR BRB RBB BBB Of these, four are of importance in solving your problem. We can approach this in a similar way as the other problem: RRB: (26/52)*(25/51)*(26/50) = 13/102 RBR: (26/52)*(26/51)*(25/50) = 13/102 RBB: (26/52)*(26/51)*(25/50) = 13/102 BBB: (26/52)*(25/51)*(24/50) = 12/102 Total probability of a black on the third draw = 3*(13/102) + (12/102) = 0.500, probably as you expected. I hope this is clear to you. Thanks for the question. xargon-ga```
 kato25-ga rated this answer: ```Thank you for your help! Your explanation was very clear and succinct. Your "Answer" will definitely help others having similar problems on probability.```