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Five cards are numbered as the following $$5 , 5 , 5 , 4 ,4$$ Three cards are chosen randomly , whats the probability that the third card is numbered 4 ? Without replacement My turn Let the first two cards are numbered 5 and the third is4 So we get $$3C1 \times 2C1 \times 2C1$$ Let the first is 5 and the second and the third are 4 So $$3C1 \times 2C1 \times 1$$ The same thing happens if the first and the third are 4 and the second is 5 so The probability is $$\frac{ the sum }{5P3}$$ Is the solution correct ?

Hussien Mohamed
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    The probability that the $n$'th card is something is always the same as the probability that the first card is that something so it is simply $\frac{2}{5}$, no fancy calculations necessary. – JMoravitz Aug 20 '19 at 13:15
  • I have edited my turn because there was some thing missed would you give it a look ? JMoravitz – Hussien Mohamed Aug 20 '19 at 13:34
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    Your answer is correct. However, the calculations are not necessary since two of the five cards have fours on them. Therefore, the probability that the third card you pick is a $4$ is simply $\frac{2}{5}$. To see this, imagine lining up the cards in a row and picking the third card from the left. – N. F. Taussig Aug 20 '19 at 13:50
  • The OP does not seem to be debating that there are easier ways to solve the problem. The OP is asking about a particular solution. The only comments/answers that really address this are from N. F. Taussig and user247327. So, to the OP, I ask, has your question been addressed? Do you understand the more straightforward solution expressed by JMoravitz et al.? What sort of an answer are you looking for? – SlipEternal Aug 20 '19 at 14:10

2 Answers2

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(Already answered in comments - there should be an Answer explaining carefully why this is trivial...)

The answer is obviously $2/5$; the first two cards chosen are irrelevant.

To be a little more formal: Let's give names to those five cards; say they're $C_1,\dots,C_5$ (so it happens that $C_1$, $C_2$ and $C_3$ are fives, while $C_4$ and $C_5$ are fours).

Say $X$ is the third card chosen. It's clear by symmetry that $$P(X=C_j)=P(X=C_k)$$for every $j$ and $k$; there's obviously no reason why the third card is more likely to be $C_2$ than $C_1$. Hence we have $$P(X=C_j)=1/5$$for every $j$, so $$P(X\text{ is a four})=P(X=C_4)+P(X=C_5)=2/5.$$

David C. Ullrich
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I have never liked calculating probability in terms of "binomial coefficients. This is how I would do this problem:

There are four ways the first two cards can be drawn: 55, 54, 45, and 44.

The probability the first card is a 5 is 3/5. If the first card is a 5 then there are 4 cards left, 2 5s and 2 4s. The probability of "55" is (3/5)(2/4)= 3/10. In this case there are 3 cards left, one 5 and 2 4s, so the probability the third card drawn is a 4 is 2/3. The probability of 554 is (3/10)(2/3)= 1/5.

The probability the first card is a 5 is 3/5. If the first card is a 5 then there are 4 cards left, 2 5s and 2 4s. The probability of "54" is (3/5)(2/4)= 3/10. In this case there are 3 cards left, two 5s and one 4, so the probability the third card drawn is a 4 is 1/3. The probability of "544" is (3/10)(1/3)= 1/10.

The probability the first card drawn is a 4 is 2/5. If the first card is a 4 then there are 4 cards left, three 5s and one 4. The probability of "45" is (2/5)(3/4)= 3/5. I this case there are 3 cards left, two 5s and one 4, so the probability the third card is a 4 is 1/3. The probability of "454" is (3/5)(1/3)= 1/5.

Since there are only two 4s if the first two are "44" the third cannot be a 4. The probability of "444" is 0.

Since these are independent events, the probability the third card is a 4 is 1/5+ 1/10+ 1/5= 2/10+ 1/10+ 2/10= 5/10= 1/2.

That is NOT the same as your $\frac{3C2+ 2C1+ 2C1}{5P3}= \frac{3+ 2+ 2}{60}= \frac{7}{60}$.

user247327
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  • I have edited my answer just before you type your answer would you give it a look ? @user247327 – Hussien Mohamed Aug 20 '19 at 13:44
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    With or without replacement, the probability is obviously $\frac{2}{5}$, same as it would be had it been asking about the first card rather than the third. You must have made a mistake in your calculations somewhere. – JMoravitz Aug 20 '19 at 13:46
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    The probability of "454" is $$\dfrac{2}{5}\cdot \dfrac{3}{4}\cdot \dfrac{1}{3} = \dfrac{1}{10} \neq \dfrac{1}{5}$$ Now when you add them up, you have $$\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{10} = \dfrac{2}{5}$$ as expected. – SlipEternal Aug 20 '19 at 13:47
  • It seems obvious that the probability the third card is a 4 cannot be $1/2$; we start with more 5s than 4s, so the probability must be less than $1/2$. – David C. Ullrich Aug 20 '19 at 22:58