Test Index

Probability

Section: Mathematics

© examsnet.com

Question : 26 of 52

Marks: +1, -0

Three randomly chosen non-negative integers x, y and z are found to satisfy the equation

x+y+z=10

Then the probability that z is even, is :

[JEE Adv 2017 P2]

$\frac{5}{11}$
$\frac{1}{2}$
$\frac{6}{11}$
$\frac{36}{33}$

Solution:

x+y+z=10

Total number of non-negative integers satisfying this equation

={}^{10+3-1}C_{3-1}={}^{12}C_{2}

If z is even i.e,

z=2m;

m=\{0,1,2,3,4,5\}x+y

=10-2m

No. of Solution

={}^{10-2m+2-1}C_{2-1}={}^{11-2m}C_{1}

=11-2m

Therefore number of favourable cases

=\sum\limits_{m=0}^{5}(11-2m)=66-30=36

Probability

=\frac{36}{{}^{12}C_{2}}=\frac{36}{66}=\frac{6}{11}

© examsnet.com

Go to Question:

Prev Question

Next Question