In a particular game, a fair die is tossed. If the number of spots showing is either 4 or 5 you win $1, if the number of spots showing is 6 you win $4, and if the number of spots showing is 1, 2, or 3 you win nothing. Let X be the amount that you win.
What is the probability distribution for the situation?
I consider this to be an "expected value" question.
It usually asks for the expected profit or loss if the game is played repeatedly.
Prob(4 or 5) = 2/6 = 1/3
prob(6) = 1/6
prob(1,2, or 3) = 3/6 = 1/2
(note 1/3 + 1/6 + 1/2 = 1)
Expected value = (1/3)(1) + (1/6)(4) + (1/2)(0) = 1
This is a totally fair game, that is, if you were to pay $1 to play this game, you would expect to get back $1 each time you play , that is , you would break even.
Well, let's break it down, shall we? We have a fair die with six sides. And you want to know the probability distribution for winning different amounts.
There are three possible outcomes: winning $1, winning $4, or winning nothing. Let's calculate the probabilities for each outcome:
To win $1, you need to roll a 4 or 5. There are two ways to roll a 4 or 5 out of six possible outcomes (4, 5, 6, 1, 2, 3). So the probability of winning $1 is 2/6 or 1/3.
To win $4, you need to roll a 6. There is only one way to roll a 6 out of six outcomes. So the probability of winning $4 is 1/6.
To win nothing, you need to roll a 1, 2, or 3. There are three ways to roll a 1, 2, or 3 out of six possible outcomes. So the probability of winning nothing is 3/6 or 1/2.
So, the probability distribution for the situation is:
P(X = $1) = 1/3
P(X = $4) = 1/6
P(X = $0) = 1/2
Hope that helps! And remember, if the dice aren't rolling in your favor, at least you can always rely on your sense of humor to lift your spirits!
To find the probability distribution for this situation, we need to calculate the probability of winning each amount of money.
Let's denote the events:
A: the number of spots showing is 4 or 5
B: the number of spots showing is 6
C: the number of spots showing is 1, 2, or 3
Since the die is fair, each outcome has a 1/6 probability of occurring.
Now, let's calculate the probability of winning $1 (event A):
P(A) = P(4 or 5) = P(4) + P(5) = 1/6 + 1/6 = 2/6 = 1/3
Next, let's calculate the probability of winning $4 (event B):
P(B) = P(6) = 1/6
Finally, let's calculate the probability of winning nothing (event C):
P(C) = P(1, 2, or 3) = P(1) + P(2) + P(3) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2
Therefore, the probability distribution for this situation is:
P(X = $1) = 1/3
P(X = $4) = 1/6
P(X = $0) = 1/2
To find the probability distribution for this situation, we need to calculate the probabilities for each possible outcome.
There are six possible outcomes when tossing a fair die, which are the numbers 1, 2, 3, 4, 5, and 6.
For this game, if the number of spots showing is 4 or 5, the person wins $1. Therefore, the probability of winning $1 is:
P(win $1) = P(4 or 5) = P(4) + P(5)
Since the die is fair, each number has an equal probability of occurring. So, the probability of rolling a 4 or a 5 is:
P(4 or 5) = P(4) + P(5) = 1/6 + 1/6 = 1/3
Next, if the number of spots showing is 6, the person wins $4. The probability of winning $4 is:
P(win $4) = P(6) = 1/6
Finally, if the number of spots showing is 1, 2, or 3, the person wins nothing. Therefore, the probability of winning nothing is:
P(win nothing) = P(1 or 2 or 3) = P(1) + P(2) + P(3)
Again, since each number has an equal probability of occurring, the probability of rolling a 1, 2, or 3 is:
P(1 or 2 or 3) = P(1) + P(2) + P(3) = 1/6 + 1/6 + 1/6 = 1/2
The probability distribution can be summarized as follows:
X | $1 | $4 | $0
--------------------------------------
P(X) | 1/3 | 1/6 | 1/2
So, the probability distribution for this situation is:
P(X = $1) = 1/3
P(X = $4) = 1/6
P(X = $0) = 1/2