In order to play a popular “spinning wheel” game at Fred's Fun Factory Arcade, a player is required to pay a small, fixed amount of 25 cents each time he/she wants to make the wheel spin. When the wheel stops, the player is awarded tickets based on where the wheel stops -- and these tickets are then redeemable for prizes at a redemption center within the arcade.
This particular game has no skill component; each spin of the wheel is a random event, and the results from each spin of the wheel are independent of the results of previous spins.
The wheel awards tickets with the following probabilities:
1 ticket | 35% |
2 tickets | 20% |
3 tickets | 20% |
5 tickets | 10% |
10 tickets | 10% |
25 tickets | 4% |
100 tickets | 1% |
A young girl is given 2 quarters so that she can play the game two times. Let \(X\) be the number of tickets she wins based on two spins. There are 26 possible values for \(X\) that the young girl can obtain in this case, and those values are listed to the right.
Some values of \(X\) are more common than others. For example, winning only 2 tickets in two spins is a somewhat common occurrence with probability 0.1225 as it means the player earns 1 ticket on the first spin and 1 ticket on the second spin. Similarly, winning 200 tickets in two spins is a somewhat rare occurrence with probability 0.0001 as it means the player earns 100 tickets on the first spin and 100 tickets on the second spin. A full list of the possible values of \(X\) and the corresponding probabilities for almost every value of \(X\) is shown at right.
\(X = 4\) can be obtained by ordered spins of "1 then 3" or "2 then 2" or "3 then 1." Since these 3 outcomes are disjoint, the sum of the respective probabilities will yield the correct answer. Since each spin is independent, the probability of any outcome can be computed as the product of the probability of the first spin's value multiplied by the probability of the second spin's value.
Therefore, \(P(1 \text{ then } 3) = .35 \cdot .2 = .07\); \(P(2 \text{ then } 2) = .2 \cdot .2 = .04\); \(P(3 \text{ then } 1) = .2 \cdot .35 = .07\).
The sum of these 3 probabilities is 0.18.
Note: students might use other methods such as a tree diagram approach to solve these questions. Hopefully, if that approach is taken, students will be economical in developing trees such that they do not include spin values that are greater than the \(X\) value of interest.
$$ P(X = 6) = 0.11 $$Using a similar approach as above: \(P(1 \text{ then } 5) = .35 \cdot .10 = .035\); \(P(3 \text{ then } 3) = .2 \cdot .2 = .04\); \(P(5 \text{ then } 1) = .10 \cdot .35 = .035\).
The sum of these 3 probabilities is 0.11.
$$ P(X = 7) = 0.04 $$Using a similar approach as above: \(P(2 \text{ then } 5) = .20 \cdot .10 = .02\); \(P(5 \text{ then } 2) = .10 \cdot .20 = .02\).
The sum of these 2 probabilities is 0.04.
$$ P(X = 10) = 0.01 $$\(P(5 \text{ then } 5) = .10 \cdot .10 = .01\). This is the only permutation for \(X = 10\).
Note: The 4 "\(X\)" probabilities will add up to .34 since the remaining probability distribution values shown in the task add up to .66. Thus a student might compute one of the 4 probabilities using subtraction (e.g., .34 – "3 other \(X\) probabilities" = 4th \(X\) probability, and so on) based on the assumption that the 3 other calculations are correct.
Commentary
The task is intended to address standards regarding sample space, independence, probability distributions and permutations/combinations. Students might be inclined to use skills discussed in 7.SD.8. Some key components of this task that are intended to challenge some common misconceptions about probability distributions are as follows:
Students should be aware that since the entire sample space is provided, the sum of all probabilities should be 1. This may help with verifying calculations in question #1, or perhaps a student will only compute 3 of the 4 unknown values in that question using permutation/independence logic and then compute the 4th value via subtraction.
Full detail of calculations is included for reference, but this is not required in order for students to complete this task.
Note: the information in the table has appeared in a companion exercise (Fred’s Fun Factory). A picture of a wheel fitting these parameters is included in that exercise.