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Grade: 12th pass
        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.
one year ago

Answers : (4)

Arun
22818 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
Arun (askIITians forum expert)
one year 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
one year 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
one year ago
Alexander
13 Points
							
X +Y + Z = 10
Think of it as you have got 10 things and you wanna distribute  those in 3 boxes. So total solutions 
   10+3-1C3-1 =66
Now 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.
 
 
                   
6 months ago
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