Question

ten slips of paper labeled from 1 to 10 are placed in a hat. the first slip of paper is not replaced before selecting the second slip of paper.
what is the probability of selecting an odd number followed by an even number?
2/9
5/18
1/3
1/4

Explanation:

Step1: Calculate the probability of selecting an odd number first

There are 5 odd numbers (1, 3, 5, 7, 9) out of 10 numbers. So the probability of selecting an odd number first is \(P(\text{odd})=\frac{5}{10}=\frac{1}{2}\).

Step2: Calculate the probability of selecting an even number second

After selecting an odd number (without replacement), there are 9 numbers left. There are 5 even numbers (2, 4, 6, 8, 10) out of the remaining 9 numbers. So the probability of selecting an even number second is \(P(\text{even}|\text{odd})=\frac{5}{9}\).

Step3: Use the multiplication rule for dependent events

The multiplication rule for dependent events is \(P(A\cap B)=P(A)\times P(B|A)\). Here \(A\) is the event of selecting an odd number and \(B\) is the event of selecting an even number. So \(P(\text{odd then even})=\frac{1}{2}\times\frac{5}{9}=\frac{5}{18}\).

Answer:

\(\frac{5}{18}\)