Mathematics - Probability Question with Solution | TestHub

Let $p_i$ be the probability of getting number $i$ on the die in a single throw ($1\le i\le 6$). We have:

$$P=p_1^2+p_2^2+\dots +p_6^2$$

$$Q=(p_1+p_3+p_5)(p_2+p_4+p_6)$$

As $ab\le \frac{1}{4}(a+b{)}^2$, we get:

$$Q\le \frac{1}{4}(p_1+p_2+\dots +p_6{)}^2=\frac{1}{4}$$

We have $P+2Q=p_1^2+p_2^2+\dots +p_6^2+2(p_2+p_4+p_6)(p_1+p_3+p_5)$.

This simplifies to:

$$P+2Q\le (p_1+p_2+\dots +p_6{)}^2=1$$

Also,

$$3P+2Q-1=3(p_1^2+p_2^2+\dots +p_6^2)+2(p_1+p_3+p_5)(p_2+p_4+p_6)-(p_1+p_2+\dots +p_6{)}^2$$

$$=2(p_1^2+p_2^2+\dots +p_6^2)-2p_1(p_3+p_5)-2p_2(p_4+p_6)-2p_3{p}_5-2p_4{p}_6$$

$$=(p_1-p_3{)}^2+(p_1-p_5{)}^2+(p_2-p_4{)}^2+(p_2-p_6{)}^2+(p_3-p_5{)}^2+(p_4-p_6{)}^2\ge 0$$

This implies $3P+2Q\ge 1$, which further implies

$P\ge \frac{1}{3}-\frac{2}{3}Q$.