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`        Three randomly chosen non negative integers x, y and z are found to satisfy the equation x + y + z =10. Then find the probability that z is even.`
2 years ago

Arun
24739 Points
```							Dear Ayush

Total number of solutions  = 10+3 –1 C 3 –1  = 66
favourable number of soutions = 11 C1 + 9 C1 + 7C1 + 5C1 + 3C1 + 1C1 = 36
hence probability (required) = 36/ 66 = 6/11
Regards
```
2 years ago
Yatish
12 Points
```							They have provided that x,y and z are three non negative integers(basically, whole numbers) Therefore, either x or y or z will be ==0.So, z can be equal to 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9 or 10.Among those events, 6 of them are even (,0,2,4,6,8,10)since 11 is the total possible outcomes.Probability(z is even)= 6/11
```
2 years ago
Yatish
12 Points
```							 [Edited]They have provided that x,y and z are three non negative integers(basically, whole numbers) Therefore, either x or y or z will be >=0 and =So, z can be equal to 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9 or 10.Among those events, 6 of them are even (0,2,4,6,8,10)since 11 is the total possible outcomes.Probability(z is even)= 6/11
```
2 years ago
Alexander
13 Points
```							X +Y + Z = 10Think of it as you have got 10 things and you wanna distribute  those in 3 boxes. So total solutions    10+3-1C3-1 =66Now for favourable outcomes. Make cases. CASE 1: Z=0 then X+Y = 10. Thus 10 things to arrange in 2 boxes. Hence  10+2-1C 2-1 Case 2: Z=2  then X+Y= 8. Thus 8 things to arrange in 2 boxes hence 8+2-1C2-1.and so on till z= 10   You will get favourable outcomes 36. Thus required probability  36/66. Easy.
```
one year ago
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