Create the probability distribution for X. Enter the values of X in ascending order from left to right. All answers should be exact.
Understand the Problem
The question describes a game where you draw a ball from a bucket numbered 1-12, and the amount you win is determined by the number on the ball you draw. You pay $8 to play and the goal is to determine the probability distribution for your net winnings (X). The values for X are given, and you need to find the probability P(X) of each net winning.
Answer
X | -8 | -7 | -6 | -5 ------- | -------- | -------- | -------- | -------- P(X) | $\frac{1}{2}$ | $\frac{1}{12}$| $\frac{1}{12}$ | $\frac{1}{3}$
Answer for screen readers
| X | -8 | -7 | -6 | -5 |
|---|---|---|---|---|
| P(X) | 1/2 | 1/12 | 1/12 | 1/3 |
Steps to Solve
- Determine possible net winnings
The possible winnings before paying to play are $0, $1, and $3. Since it costs $8 to play, the possible net winnings (X) are $$-8, $-7, $-5$. The question provides these values of $X$. It remains to determine the probability $P(X)$ for each of them.
- Calculate $P(X = -8)$
You win nothing if you draw a ball that is not less than 3 and not a multiple of 3. The numbers that are multiples of 3 from 1-12 are 3, 6, 9 and 12. The numbers that are less than 3 are 1 and 2. Thus the numbers where you win nothing are 4, 5, 7, 8, 10, and 11. There are 6 such numbers. If you win nothing, your net winning is $-8$, the cost to play.
$P(X = -8) = \frac{6}{12} = \frac{1}{2}$
- Calculate $P(X = -7)$
You win $1 if you draw a 1. In that case, your net winning is $1 - $8 = $$-7.
$P(X = -7) = \frac{1}{12}$
- Calculate $P(X = -5)$
You win $3 if you draw a number that is a multiple of 3. In that case, your net winning is $3 - $8 = $$-5$. There are 4 multiples of 3 (3, 6, 9, 12).
$P(X = -5) = \frac{4}{12} = \frac{1}{3}$
- Calculate $P(X = -6)$
You win $2 if you draw a 2. In that case, your net winning is $2 - $8 = $$-6$.
$P(X = -6) = \frac{1}{12}$
| X | -8 | -7 | -6 | -5 |
|---|---|---|---|---|
| P(X) | 1/2 | 1/12 | 1/12 | 1/3 |
More Information
The sum of all probabilities should equal 1.
$\frac{1}{2} + \frac{1}{12} + \frac{1}{12} + \frac{1}{3} = \frac{6}{12} + \frac{1}{12} + \frac{1}{12} + \frac{4}{12} = \frac{12}{12} = 1$
Tips
A common mistake is not taking into account the cost of playing the game when calculating the net winnings.
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