The total no. of possible outcomes of rolling a fair die = 6
The possible outcomes are 1, 2, 3, 4, 5 and 6
The event to occur is getting a no. greater than 4.
The no. of possible outcomes to get a no. greater than 4 = 2
So, probability of getting a no. greater than 4 = 2/6 = 1/3
If you get a no. greater than 4 for the first time, you win $80.
So, the probability to win $80 is 1/3.
The probability to roll the die again = 1 - (1/3) = 2/3
In the second chance, if you get a no. greater than 4 you win $20, else nothing.
Since the two events are dependent, P(X and Y) = P(X) * P(Y)
Probability of getting a no. greater than 4 in the second chance = (2/3) * (1/3) =2/9
Probability of win (either in 1st or 2nd chance) = 1/3 + 2/9 =5/9
Please check attachment for probability model.
Answer:
Event Probability
Win First time 1/3
Win Second time 2/9
No win 1- 1/3 -2/9= 4/9
And the expected value for this case would be:
[tex] E(X) = \frac{1}{3} *80 + \frac{2}{9} *20 +\frac{4}{9}*0 = \frac{280}{9}[/tex]
Explanation:
For this case the sampling space when we roll a fair die is :
[tex] S= [1,2,3,4,5,6][/tex]
We define the following events
A= Obtain a number >4 , the sample space for the event A is [tex]S_A =[5,6][/tex]
And the probability for the event A would be [tex] p_A = \frac{2}{6} =\frac{1}{3}[/tex]
And the expected value if we got a number greater than 4 is $80 o we can define the amount of money for this event like [tex]X_A = 80[/tex]
If we don't win at the first roll we need to roll a second time the die and for this case we can assume that this event would be B and by the complement rule we know that the probability ofr thi event is :
[tex]p_B = 1-p_A = 1-\frac{1}{3}= \frac{2}{3}[/tex]
For this case is we roll the die a second time and we got a number >4 we win $20 and nothing if we got a number lower or equal than 4.
So then the probability of obtain a number >4 in the second chance asuming independence would be:
[tex] P_{SB} = \frac{2}{3} *\frac{1}{3}= \frac{2}{9}[/tex]
Then we can define the following model
Event Probability
Win First time 1/3
Win Second time 2/9
No win 1- 1/3 -2/9= 4/9
And the expected value for this case would be:
[tex] E(X) = \frac{1}{3} *80 + \frac{2}{9} *20 +\frac{4}{9}*0 = \frac{280}{9}[/tex]
