 My Math Forum Select two different digits. Odd sum and Conditional probability.

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 May 16th, 2015, 12:38 PM #1 Senior Member   Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7 Select two different digits. Odd sum and Conditional probability. Hi, I have the following problem: Two different digits are selected at random from the digits 1 through 5. If the sum is odd, what is the probability that 2 is one of the numbers selected? We have to calculate the conditional probability, where we can consider the event A = "2 is one of the numbers selected" that can happens only if the event B = "the sum is odd" has been happened already. Considering the event B, take a look at the following image, in the yellow highlighted cells the event B occurs: To find, in a "algoritmical way" the total number of the different ordered couples we want, we can consider the following: we can select 1 out of 5 digits for the first digit, and we can select 1 out of 4 remaining digits for the second one: $\displaystyle \binom{5}{1} \cdot \binom{4}{1} = 20$ looking at the image where the cells are with yellow background, we can count 12 couples where the sum is odd. Therefore the probability of the event B: $\displaystyle P(B) = \frac{12}{20} = \frac{3}{5}$ Now, we can calculate the conditional probability using the formula: $\displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)}$ We go to calculate $\displaystyle P(A \cap B)$: through the ordered couples where their sum is odd, we have to select only the couples that contain 2. Take a look at the above image: from a total of 20 couples, there are 6 couples, (cells that have red border), that contains 2. So, the conditional probability P(A|B): $\displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{6}{20}}{\frac{3}{5}} = \frac{3}{10} \cdot \frac{5}{3} = \frac{1}{2}$ but the answer given by the textbook is: $\displaystyle \frac{3}{10}$ So my question are: - Consider the result I have found. Is it wrong, or not? - To find the ordered couples where their sum is odd, is there any algorithmical approach, (just like we have found the two different couples using a product of two combinations)? Or I have to create a table as the one in the above image, and therfore highlight "manually", with a waste of time, the couples that I want? - Again, to find the ordered couples where their sum is odd and contains 2, is there any algorithmical approach? I have considered these reasonings, is it possible to construct an algorithm to find fastly what we want? for example: - if the first digit is 5, - the second must be different from 5, and so remains to select 1 out of 4 digits (1,2,3,4), - the condition is: if we sum every single digit with 5, only odd results are important. By definition, a number is even if its module (%) with 2 results 0. i.e. dividing the considered number by 2 we obtain a remainder as 0. If we sum 5 with an even digit from the above four digits, we obtain always a odd result: for example: $\displaystyle 5+2=7 \% 2 \ne 0$. ODD! If we sum 5 with an odd digit from the above four digits, we obtain always a even result: for example: $\displaystyle 5+1=6 \% 2 = 0$. EVEN! considering what I have said above, is it possible to find any "automation" by any algorithm on what I have said in the questions? I hope you can help me! if you have read till the end, many thanks for your great patience! many thanks! May 16th, 2015, 12:43 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra You have illustrated the ordered couples. For the unordered solution required, simply look at the entries above the diagonal. The book answer seems to be the unconditional probability that the sum is odd and contains 2. Thanks from beesee Last edited by v8archie; May 16th, 2015 at 12:45 PM. May 16th, 2015, 05:12 PM #3 Global Moderator   Joined: May 2007 Posts: 6,854 Thanks: 744 The statement of the problem and the answer (3/10) seems peculiar. For the sum of two numbers to be odd, one has to even and the other odd. If the selection probability is the same for all numbers (1/5), then the even number must be 2 or, each with a probability of 1/2. Therefore the probability that a 2 is selected, under the condition that the sum of two numbers is odd, must be 1/2. If the statement is supposed to be sum odd and two is chosen, then the result is (3/5)(1/2)=3/10, where 3/5 is the probability that the sum is odd. The two events (sum odd, two chosen) are independent. Thanks from beesee Last edited by mathman; May 16th, 2015 at 05:12 PM. Reason: style May 17th, 2015, 03:36 AM   #4
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Quote:
 Originally Posted by mathman Therefore the probability that a 2 is selected, under the condition that the sum of two numbers is odd, must be 1/2.
Therefore in this case, does my answer would be right?

Quote:
 Originally Posted by mathman If the statement is supposed to be sum odd and two is chosen, then the result is (3/5)(1/2)=3/10, where 3/5 is the probability that the sum is odd. The two events (sum odd, two chosen) are independent.
Now, I am really confused! What I don't understand is: how can be the events indipendent, if we have to choose 2 from the result of the first event?

If we consider indipendent then I say this:
P(sum odd) AND P(two chosen) = $\displaystyle \frac{12}{20} \cdot \frac{8}{20}$
and the two has been chosen from all couples (20), and not from the couples where the sum of the couple is odd (12).

Can you explain me better? May 17th, 2015, 03:52 PM   #5
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Quote:
 Originally Posted by beesee Therefore in this case, does my answer would be right? Now, I am really confused! What I don't understand is: how can be the events indipendent, if we have to choose 2 from the result of the first event? If we consider indipendent then I say this: P(sum odd) AND P(two chosen) = $\displaystyle \frac{12}{20} \cdot \frac{8}{20}$ and the two has been chosen from all couples (20), and not from the couples where the sum of the couple is odd (12). Can you explain me better?
You are right - not independent. If the sum is odd, prob of a 2 is 1/2. If the sum is even, prob of a 2 is 1/4. May 18th, 2015, 09:55 AM   #6
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Quote:
 Originally Posted by v8archie The book answer seems to be the unconditional probability that the sum is odd and contains 2.
how can it be unconditional probability, if the event "contains 2" happens only after the event "the sum is odd"? can you explain me? Because I have another similar exercise and the result I obtained is not the same of the one given by my book.

Last edited by beesee; May 18th, 2015 at 10:13 AM. May 18th, 2015, 01:46 PM   #7
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Quote:
 Originally Posted by beesee how can it be unconditional probability, if the event "contains 2" happens only after the event "the sum is odd"? can you explain me? Because I have another similar exercise and the result I obtained is not the same of the one given by my book.
It is not a question of before or after. Unconditional refers to both happening at the same time. Specifically, get two numbers. What is the probability that there is a 2 AND the sum is odd. May 20th, 2015, 09:59 AM   #8
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Quote:
 Originally Posted by beesee but the answer given by the textbook is: $\displaystyle \frac{3}{10}$
Quote:
 Originally Posted by mathman The statement of the problem and the answer (3/10) seems peculiar.
yes, some news, in fact on the errata corrige, the answer given is $\displaystyle \frac{1}{3}$, but, however, I don't obtain that result.
I believe that that error has not been corrected at all!
What do you think?

.
I have tried to follow the reasoning of unconditional probability, without having to look at a "tabular image" as above:

Quote:
 Originally Posted by v8archie You have illustrated the ordered couples. For the unordered solution required, simply look at the entries above the diagonal.
the number of ways to choose two different digits from 1 to 5 (unordered couples): $\displaystyle \frac{\binom{5}{1}\binom{4}{1}}{2!} = \frac{20}{2} = 10$

Quote:
 Originally Posted by mathman For the sum of two numbers to be odd, one has to even and the other odd.
I can select an odd number in $\displaystyle \binom{3}{1} = 3$ ways.
I can select an even number in $\displaystyle \binom{2}{1} = 2$ ways.
Given two even numbers the probability that happens 2 is : $\displaystyle \frac{1}{2}$

So the probability:
$\displaystyle P = \frac{3 \cdot 2}{10} \cdot \frac{1}{2} = \frac{3}{10}$ Tags conditional, digits, odd, probability, select, sum ,

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# find the probability that the sum of two numbers select from5,6,7 is odd

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