Probability from a discrete distribution product and sum comparison

Key takeaway: This question mixes a probability table with an inequality test on two selected values. The first task is to determine the missing distribution constant, then list the ordered pairs that satisfy the product-sum condition.

Step 1: Find the missing probability value

The table gives

$P(X=0)=\frac{1}{12},\quad P(X=1)=\frac{1}{4},\quad P(X=2)=p,\quad P(X=3)=3p.$

Since total probability must equal 1,

$\frac{1}{12}+\frac{1}{4}+p+3p=1.$

Convert the fractions:

$\frac{1}{12}+\frac{1}{4}=\frac{1}{12}+\frac{3}{12}=\frac{4}{12}=\frac{1}{3}.$

So

$\frac{1}{3}+4p=1 \Rightarrow 4p=\frac{2}{3} \Rightarrow p=\frac{1}{6}.$

Thus the distribution is

$P(0)=\frac{1}{12},\quad P(1)=\frac{1}{4},\quad P(2)=\frac{1}{6},\quad P(3)=\frac{1}{2}.$

Step 2: Identify when product is greater than sum

We choose two values of $X$. Let them be $a$ and $b$. We need

$ab>a+b.$

Rearrange:

$ab-a-b>0$

$(ab-a-b+1)>1$

$(a-1)(b-1)>1.$

A quick way is to test the possible values $0,1,2,3$. The only pair that makes the product exceed the sum is when both values are 3:

$3\cdot 3=9,\qquad 3+3=6.$

Check other pairs:

• $2\cdot 3=6$, which equals $2+3=5$ and is greater, so this one also works.
• $2\cdot 2=4$, which is equal to $2+2=4$, so it does not work.
• Any pair involving 0 or 1 fails.

So the successful ordered pairs are $(2,3)$, $(3,2)$, and $(3,3)$.

Step 3: Compute the probability of the successful pairs

Because the values are chosen independently from the same distribution, multiply probabilities for each ordered pair:

$P(2,3)=\frac{1}{6}\cdot\frac{1}{2}=\frac{1}{12}$

$P(3,2)=\frac{1}{2}\cdot\frac{1}{6}=\frac{1}{12}$

$P(3,3)=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$

Add them:

$\frac{1}{12}+\frac{1}{12}+\frac{1}{4}=\frac{1}{6}+\frac{1}{4}=\frac{2}{12}+\frac{3}{12}=\frac{5}{12}.$

So the required probability is

$\boxed{\frac{5}{12}}.$

A structured table or ordered-pair list is the safest way to avoid missing cases.

---

Pitfalls the pros know 👇 The most common error is to assume only the pair $(3,3)$ works because it looks the largest. But $(2,3)$ and $(3,2)$ also satisfy the inequality, so leaving them out gives the wrong answer. Another frequent mistake is to forget to find $p$ first. Without the missing probability value, you cannot assign correct weights to the ordered pairs. Always normalize the distribution before doing the pair analysis.

What if the problem changes? If the condition changed to “their product is at least their sum,” then $(2,2)$ would also count because $2\cdot 2=2+2$. The successful ordered pairs would become $(2,2)$, $(2,3)$, $(3,2)$, and $(3,3)$. Using the same distribution, the probability would be $\frac{1}{6}\cdot\frac{1}{6}+\frac{1}{12}+\frac{1}{12}+\frac{1}{4}$, which is larger than the original answer. This shows how a small change in the inequality can add one or more extra cases.

Tags: discrete probability distribution, ordered pairs, inequality testing