Probability - four random integers between 0-9, that not more than two are the same

Four integers are chosen at random between 0 and 9, inclusive. Find the probability that (a) not more than 2 are the same.

What I tried: all unique numbers: 63/125,
two same numbers: 72/1000

And then add them both. But the answer in the book is 963/1000. I'm not getting such a high probability. Where have I made the mistake?

Are you sure about the answer in the book? I got $frac9361,000$.
– José Carlos Santos
Sep 4 '18 at 9:32

But 63/125 + 72/100 is more than 1...
– TonyK
Sep 4 '18 at 9:37

@JoséCarlosSantos yes re-checked it. Could you please explain how did you get 936/1000 as I'm not even getting close to such a high probability.
– user585380
Sep 4 '18 at 9:38

In how many cases we get $4$ distinct integers? Answer: $10times9times8times7$. In how many ways are the first $2$ integers equal but after that you get $2$ new (distinct) integers? Answer: $10times 9times 8$. But this is also the answer to the question: in how many ways are the first and the third ingeres equal and the other $2$ are distinct from each other and the $2$ others? Actually, there are $6$ questions of this sort. Therefore, the answer should be$$frac10times9times8times7+6times10times9times810,000=frac9361,000.$$
– José Carlos Santos
Sep 4 '18 at 9:44

@user585380 Sure: $binom42=6$.
– José Carlos Santos
Sep 4 '18 at 9:48

3 Answers
3

You're missing a factor of $6$ in the second case, since you can choose two slots for the two equal numbers in $binom42$ ways. Then you get $frac9361000$, as José wrote. Since the book says $frac9631000$, apparently they're also counting the case of two pairs, and "more than $2$ are the same" means that there's at least a triple. That adds another $fracbinom42binom10210^4=frac271000$, so the total is then $frac9631000$.

You can solve it by subtraction: the complementary of the event in your problem consists in having a quartet of equal numbers or a triplet and a different number.

The probability of getting four equal numbers is $frac1010000,$ because there are ten numbers available to be all equal.

The probability of having a triplet is $frac4times10times 910000$ because there are four positions where to place the "single one", 10 numbers that the triplet can assume and nine that the single can have.

Summing these two probabilities, you obtain $frac37010000$ which is exactly $1-frac9631000.$

Thanks for the edit :)
– Riccardo Ceccon
Sep 4 '18 at 9:51

You may first calculate the complementary probability.

All together, the probability of getting 3 or 4 equal digits is
$$P(mbox3 or 4 equal digits) =frac10 + 4 choose 3 cdot 10 cdot 910^4 = frac3710^3 Rightarrow $$
$$P(mboxat most 2 equal) = 1-frac3710^3 = frac9631000 $$

Thanks! But could you point out my mistake in doing the normal way without complementary?
– user585380
Sep 4 '18 at 9:56

@user585380: Your first number is right. The second is wrong. But you did not show your reasoning how you got your second number. You may split the second case (2 equal digits) into two subcases: (1) - exactly one pair; (2) - exactly 2 different (!) pairs.
– trancelocation
Sep 4 '18 at 10:35

We can't have 2 different pairs, right? I got the second number by: 9/10*1/10*8/10 (to get two similar numbers).
– user585380
Sep 4 '18 at 14:28

@user585380: Your number $frac910cdot frac110cdot frac810$ corresponds to $frac110$ for the double digit and $frac910$ and $frac810$ for the two remaining digits. But your calculation does not distinguish, for example, between $1123$, $1213$, $1231$, .... Besides this your calculation excludes cases with two pairs like $1122$. I hope this helps you understand what you overlooked in your calculation.
– trancelocation
Sep 5 '18 at 3:41

Thanks for contributing an answer to Mathematics Stack Exchange!

But avoid …

Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

Some of your past answers have not been well-received, and you're in danger of being blocked from answering.

Please pay close attention to the following guidance:

But avoid …

To learn more, see our tips on writing great answers.

Required, but never shown

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

搜尋此網誌

Dfyjkt