Question

Three randomly chosen natural numbers $x,y$ and $z$ satisfy $x+y+z=10$, then

• A. Probability that $z$ is even is $\frac{1}{2}$
• B. Probability that $z$ is even is $\frac{4}{9}$
• C. Probability that $z$ is odd is $\frac{5}{9}$
• D. Probability that $x$ is even is $\frac{5}{12}$

Answer 1

Answer: (B), (C)

The correct options are
B Probability that $z$ is even is $\frac{4}{9}$
C Probability that $z$ is odd is $\frac{5}{9}$

$x+y+z=10$, $x,y,z\in N$

Put $x=a+1$, $y=b+1$, $z=c+1$, $a,b,c\in W$

$a+b+c=7$

Total solutions $={}^{7+3-1}{C}_{3-1}=36$

For $z$ to be even, $c$ must be an odd number.
If $c=1\Rightarrow a+b=6.$
Number of solutions $={}^{6+2-1}{C}_{2-1}=7$
If $c=3\Rightarrow a+b=4.$
Number of solutions $={}^{4+2-1}{C}_{2-1}=5$
If $c=5\Rightarrow a+b=6.$
Number of solutions $={}^{2+2-1}{C}_{2-1}=3$
If $c=7\Rightarrow a+b=6.$
Number of solutions $={}^{0+2-1}{C}_{2-1}=1$

Number of solution when $c$ is odd = $16$

Probability that $z$ is even $=\frac{16}{36}=\frac{4}{9}$

Probability that $z$ is odd $=\frac{20}{36}=\frac{5}{9}$