Grade 12th passAlgebra

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.

Ayush Singh

8 Years agoGrade 12th pass

4 Answers

Arun

8 Years ago

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)

Yatish

8 Years ago

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

Yatish

8 Years ago

[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

Alexander

7 Years ago

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+3-1C_3-1 =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+2-1C_2-1.

and so on till z= 10

You will get favourable outcomes 36.

Thus required probability 36/66.

Easy.

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View all Questions

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.

Ayush Singh

4 Answers

Arun

Yatish

Yatish

Alexander

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