Question

I need help on math questions 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? 2/91551814

Answer 1

2/9

Step-by-step explanation

The probability of an event is the number of favorable outcomes divided by the total number of outcomes possible

Step 1

Let

`\begin{gathered} \text{slip 1}\rightarrow\text{label}\rightarrow1\Rightarrow odd \\ \text{slip 2}\rightarrow\text{label}\rightarrow2\Rightarrow\text{even} \\ \text{slip 3}\rightarrow\text{label}\rightarrow3\Rightarrow odd \\ \text{slip 4}\rightarrow\text{label}\rightarrow4\Rightarrow\text{even} \\ \text{slip 5}\rightarrow\text{label}\rightarrow5\Rightarrow odd \\ \text{slip 6}\rightarrow\text{label}\rightarrow6\Rightarrow\text{even} \\ \text{slip 7}\rightarrow\text{label}\rightarrow7\Rightarrow odd \\ \text{slip 8}\rightarrow\text{label}\rightarrow8\Rightarrow\text{even} \\ \text{slip 9}\rightarrow\text{label}\rightarrow9\Rightarrow odd \\ \text{slip 10}\rightarrow\text{label}\rightarrow10\Rightarrow\text{even} \end{gathered}`

hence:

total of even numbers: 5

total odd numbers: 5

total outcomes= 10

Step 2

a)

now, the probability of selecting an odd number

`\begin{gathered} P=\frac{favorable\text{ outcomes}}{\text{total outcomes}} \\ \text{replace} \\ P=(5)/(10) \\ P=0.5 \end{gathered}`

b) now, if the slip is not replaced we would have

total of even numbers: 5

total odd numbers: 4

total outcomes= 9

`\begin{gathered} P=\frac{favorable\text{ outcomes}}{\text{total outcomes}} \\ \text{replace} \\ P=(4)/(9) \\ P=(4)/(9) \end{gathered}`

finally, the probality of both events happen is the product of the individual probabilities

so

`P(A\cap B)=(1)/(2)\cdot(4)/(9)==(4)/(18)=(2)/(9)`

therefore, the answer is 2/9

I hope this helps you