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A card is drawn and replaced five times from an ordinary deck of $52$ cards and the sequence of colors is observed. What is the probability that:

a) Five red cards were drawn?

b) five black cards were drawn?

c) Three red and two black cards were drawn?

d) why is it necessary to replace the cards?

My thoughts:

a) $^5P_1\left(\dfrac{26}{52}\right)^1 (1-p)^4 + ...+ ^5P_5 \left(\dfrac{26}{52}\right)^5 (1-p)^0$

b) isn't this the same as part (a) ?

c) isn't this the same as asking exactly $5$ black or red cards were drawn ?

d) not sure about this one.

Aang
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user85956
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    d) If you don't replace the cards, your probability of picking a color gets altered, therefore you don't have a binomial distribution anymore. (But a hypergeometric one). – Raskolnikov Jul 16 '13 at 16:44

2 Answers2

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Hints: a) Each time you have $\frac 12$ chance of a given color. b) yes, this is the same as a c) no, because there are many combinations of three red and two blacks. Each specific order is the same as a or b d) it is not necessary to replace the cards, but the probabilities will change. If you are looking for five reds, after you draw the first red the chance of the second is $\frac {25}{51}$ and so on.

Ross Millikan
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  • so is a correct ? 26/52=1/2. I still cant get part c – user85956 Jul 16 '13 at 16:48
  • No, a is not correct. As you have $\frac 12$ chance each time for a red, and you need them all to be red, it is $\frac 1{2^5}$. For c, you have ${5 \choose 3}$ ways to distribute the red cards. – Ross Millikan Jul 16 '13 at 17:30
  • why is it 1/2^5, since each time i pick a card the prob of being a red is 1/2? Is it because each prob is multiplied together: 1/2x1/2x1/2x1/2x1/2? But my teacher said to use this formula nPr x (p)^r x (1-p)^r-1 – user85956 Jul 17 '13 at 02:56
  • Yes, you multiply independent probabilities for independent events. The formula you quote is not correct, even if you supply the parentheses around the $r-1$. I have never seen that one. The closest I have seen is $nCr\cdot p^r\cdot (1-p)^{n-r}$, which is what you want for c. nCr is the number of orders of r things in n spaces. – Ross Millikan Jul 17 '13 at 03:02
  • so a) is 1/32 but using the formula would it be (26C5)/(52C5) because we take 5 out of 26 red cards and divide by the whole sample space which is take 5 out of 52 – user85956 Jul 18 '13 at 01:08
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Another Hint: The probability of picking a type of card is $$ \frac{\text{# of cards of that type left in the deck}}{\text{total # of cards in the deck before drawing}} $$

c) Doesn't tell you in what order they were drawn: how many different orders are there?

d) Looking at the fraction/ratio above, what happens to the numerator and denominator if you don't replace a card before drawing?

Kendra Lynne
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