Question

Tyler rolled a six-sided number cube, recorded the results, and then rolled again. What is theprobability that the first roll was an odd number and the second roll was a number less than3?

Answer 1

NYou have an experiment where you could get the following answers, notice that I will suppose that each side represents a different number from 1 to 6

`\lbrace1,2,3,4,5,6\rbrace`

It means that when you roll the cube for the first time you could get any of this 6 numbers. You should also notice that you have three odd numbers and other three even. Then

`P(X=odd)=(|\lbrace1,3,5\rbrace|)/(|\lbrace1,2,3,4,5,6\rbrace|)=(3)/(6)=0.5`

Now, if you want to know the probability for the second event we have the following

`P(X<3)=\frac\lbrace1,2\rbrace{|{}\lbrace1,2,3,4,5,6\rbrace|}=(2)/(6)=(1)/(3)`

Now, the question for this time is what is the probability for both cases, it means

`P(X=odd,and,X<3)`

Notice that the cases are independent just because even when 1 is an odd number and at the same time it is less than 3, it will not affect the second role. Then

`P(X=odd,and,X<3)=P(X=odd)P(X<3)=(1)/(2)((1)/(3))=(1)/(6)`

Then the probability that the first roll is and odd number and the second roll a number less than 3 is 1/6.