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# Probability

Four different positive integers less than 10 are chosen randomly what is the probability that their sum is odd?

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1. kritika sharma

5/9

2. Abhishek Tandon

20/9?

1. Probability can never be more than 1 . Abhishek 🙂

3. 4/21

4. 10/21

5. Mayank Khanna

We know that odd+ even = odd. So it means, we have to choose some combination of odd and even. Now, as we see 2 odd + 2 even such as 2+4+5+7= 18 , 6+8+3+9= 26 gives even answer. ( As numbers are small so I just calculated the sum . However, if the numbers are big is there any method or rule for 4 numbers , please mention in reply sir) .
By observing we find , we have to keep 3 number of odd and 1 of even or 3 of even and 1 of odd. So we try and probability is
(3/5.1/5 )+(3/5+1/5)
As we consider both cases. Now we have 6/25.
Hence 6/25.
I am not sure of my approach as it works well with small number but when big numbers are there it won’t and secondly it took me more than 2 minutes to do it.

6. Mayank Khanna

I followed the rule that odd+even=odd. Till 10 we have to choose 4 numbers, so choose it in the manner as either 3 odd and 1 even or 3 even and 1 odd. Now, we have
(3/5).(i/5)+ (3/5).(1/5)= 6/25 or 0.24

1. 4 numbers. So either O,O,O,E or E,E,E,O.

7. (5C3*4C1)+(4C3*5C1)/126===> 10/21

• Nice instruction, thanks

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