Question

Four numbers are chosen at random from the numbers $1,2,3,...,20$. Let $p_e\left(p_0\right)$ denote the probability that their sum is even (odd)

• A. $p_e-p_0=3/323$
• B. $p_e=163/323$
• C. $p_0>p_e$
• D. $p_e>p_0$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $p_e-p_0=3/323$
B $p_e=163/323$
D $p_e>p_0$

Sample space$\left(N\right)=$ selection four numbers from $1,2,3.......20{=}^{20}{C}_4=17\times 19\times 15$
Let selected numbers are $a,b,c,d$
so sum of selected numbers will be even only if either all are even, all are odd or two are even and
two are odd, also we have 10 even and 10 odd numbers in sample space.
thus $n\left(e\right){=}^{10}{C}_4{+}^{10}{C}_4{+}^{10}{C}_2{\times }^{10}{C}_2=2\times 21\times 10+45\times 45$
Hence probability of sum of selected numbers to be even is $p_e=n\left(e\right)/N=\frac{163}{323}$
also selected numbers will be either even or odd
Thus probability of sum of selected numbers to be odd $=p_o=1-p_e=\frac{160}{323}$