Question:
First one number is chosen uniformly at random from {1,2,3,4,5} and is removed from the set; then a second number is chosen uniformly at random from the remaining numbers. What is the probability that a number greater than 3 is picked in the second step?
My attempt:
STEP 1: FIND THE SAMPLE SPACE
The first and second number removed from {1,2,3,4,5}
Sample space S =
{(1,2), (1,3), (1,4), (1,5),
(2,1), (2,3), (2,4), (2,5),
(3,1), (3,2), (3,4), (3,5),
(4,1), (4,2), (4,3), (4,5),
(5,1), (5,2), (5,3), (5,4)}
STEP 2: DEFINE EVENT OF INTEREST
Event of interest is a number greater than 3 is picked in the second step
E = {(1,4), (1,5), (2,4), (2,5) , (3,4), (3,5), (4,5), (5,4)}
STEP 3: DETERMINE OUTCOME PROBABILITIES:
Pr[(1,4)] = 1/5 x 1/4 = 1/20
Pr[(1,5)] = 1/5 x 1/4 = 1/20
Pr[(2,4)] = 1/5 x 1/4 = 1/20
Pr[(2,5)] = 1/5 x 1/4 = 1/20
Pr[(3,4)] = 1/5 x 1/4 = 1/20
Pr[(3,5)] = 1/5 x 1/4 = 1/20
Pr[(4,5)] = 1/5 x 1/4 = 1/20
Pr[(5,4)] = 1/5 x 1/4 = 1/20
STEP 4: COMPUTE EVENT PROBABILITIES
Pr[a number greater than 3 is picked in the second step] = Pr[(1,4)] + Pr[(1,5)] + Pr[(2,4)] + Pr[(2,5)] + Pr[(3,4)] + Pr[(3,5)] + Pr[(4,5)] + Pr[(5,4)]
= 1/20 + 2/10 + 2/10 + 1/20 + 1/20 + 1/20 + 1/20 + + 1/20
= 8/20
= 2/5 = 0.4
Best Answer
Your method is right. Another way is by using conditioning.
There are five cases at the first step. You need to condition on each of these five to get the probability of picking up a number greater than 3 in second step. \begin{align*} \mathbb P[\text{Pick Number $> 3$ in second step}] &= \mathbb P[\text{First number is $1,2,3$}]\cdot\\&\quad\qquad \mathbb P[\text{Second number $>3$}|\text{First number is $1,2,3$}]+\\&\qquad\mathbb P[\text{First number is $4,5$}]\cdot \\&\quad\qquad\mathbb P[\text{Second number $>3$}|\text{First number is $4,5$}]\\ &= \frac{3}{5}\cdot \frac{1}{2}+\frac{2}{5}\cdot\frac{1}{4}\\ &= \frac{3}{5}\cdot \frac{1}{2}+\frac{2}{5}\cdot\frac{1}{4}\\ &= \frac{2}{5} \end{align*}