## Quant Question Of The Day: 20

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# Probability

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

### Solve our previous ‘questions of the day’

1.Quant Question Of The Day : 19

2. Quant Question Of The Day : 18

#### Quant Lessons

1.Divisibility Rule of a Number (Divisibility Test)

2. Divisors of a Number ( Number of Divisors of a number)

5/9

5/9 is incorrect, Kritika. Take a look at the explanation of the question here : http://tathagat.mba/quant-question-of-the-day-20/

Try today’s question: http://tathagat.mba/quant-question-of-the-day-20-2/

20/9?

Probability can never be more than 1 . Abhishek 🙂

4/21

5/9 is incorrect, Abhipri. Take a look at the explanation of the question here : http://tathagat.mba/quant-question-of-the-day-20/

Try today’s question: http://tathagat.mba/quant-question-of-the-day-20-2/

10/21

Excellent, Aaradhya! You’re the only one who has given the right answer to this question. Keep it up!

Take a shot at today’s Question of the Day : http://tathagat.mba/quant-question-of-the-day-20-2/

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.

Hi Mayank!

Take a look at the explanation of the question here : http://tathagat.mba/quant-question-of-the-day-20/

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

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

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