# Probability – School lottery (advanced)

A basic version of School lottery is also available.

Scenario:

Your school launches a spring lottery in which you can win various prizes. There are 900 lottery tickets, out of which 200 guarantee a prize. These include:

• single grand prize – a weekend trip,
• nine secondary prizes – free tickets to the movies,
• forty tertiary prizes – school T-shirts,
• the rest – school mugs.

You keep your eyes on the prize – that school T-shirt you’ve been longing for. You’re the first person to try their luck, so you go ahead and buy three tickets at once.

1. How likely is it that you’ll win your beloved T-shirt?
2. What is the probability that you’ll get exactly two prizes
3. What is the probability that you’ll win three different prizes?

Useful calculators:

Question 1 hints:

Hint 1
What are the possible outcomes of each of your tickets? If a T-shirt is all you want, everything else is considered a loss.
Hint 2
If you don’t order the three tickets, does it matter which one was the winning one? And does it matter if there is more than one?
Hint 3
Be careful not to count the same event twice! But be sure to include all the possible results.
Hint 4
If, e.g., I want to see how many triples there are with two winning tickets, how many combinations of ‘two wins, one loss’ are there?

Question 2 hints:

Hint 1
How many possible pairs of prizes can you win? Do we have to count them separately or can we do them all at once?
Hint 2
Can you get two similar prizes?
Hint 3
How can we describe all the possible pairs that we can get?

Question 3 hints:

Hint 1
How many possible triples of prizes can you win? Do they have the same probability of happening?
Hint 2
Do we care about the order of the prizes we win?

Solutions:

Question 1
P ≈ 0.13
Question 2
P ≈ 0.12
Question 3
P ≈ 0.0005

Step-by-step solution:

Question 1
We need to be very careful with this question so that we don’t count the same event twice. To be sure of that, let’s consider all four possible cases:
P₁. we win only one T-shirt;
P₂. we win exactly two T-shirts;
P₃. we win three T-shirts; and
P₄. we don’t win any T-shirts.

Clearly, no two of these events can happen at the same time. What is more, exactly one of them will happen. This means that the four of them cover every possibility in our random experiment:
P₁ + P₂ + P₃ + P₄ = 1.

We want to find the probability of winning a T-shirt, which means AT LEAST one T-shirt. Therefore, we want to find the sum P = P₁ + P₂ + P₃ since those three events (and only them) support our desired outcome. Using the equation above, we can write
P = P₁ + P₂ + P₃ = 1 – P₄.

What does this mean? Well, we only need to calculate P₄ to find P. This is, in fact, the definition of the complement of P₄.

We move on to calculating P₄. In classical probability problems, such as this one, the probability of an event happening is the number of times that instances occurs divided by the number of all possible outcomes. In our scenario, this means that we need to count all times three tickets don’t give us a T-shirt. Observe that here we don’t really mind if the ticket gives us nothing or the main prize – if it doesn’t give us a T-shirt, we count it. The number of such tickets is 900 – 40 = 860. Therefore, all the possible combinations of three tickets can be found using the combination calculator:
no_Tshirt = C(860, 3) = (860 * 859 * 858) / (1 * 2 * 3) = 105,639,820.

Similarly, the number of all possible sets of three tickets, whether they support our event or not, is simply the number of all the triples from the pool of 900 tickets. The combination calculator gives this number to be
all_triples = C(900, 3) = (900 * 899 * 898) / (1 * 2 * 3) = 121,095,300.

Now we are ready to calculate P₄. This time we use the probability calculator to get
P₄ = no_Tshirt / all_triples = 105,639,820 / 121,095,300 ≈ 0.87.
Therefore,
P = 1 – P₄ ≈ 1 – 0.87 = 0.13.

NOTE: According to the description above, we could have calculated P as the sum P₁ + P₂ + P₃. This would, however, require us to calculate all three of those probabilities. It is, of course, possible, but we have to be very careful so that we don’t count the same possible outcome twice.

For instance, to count P₁, we need to choose 1 out of 40 tickets which give a T-shirt and choose 2 out of the remaining 560 losing ones. Note that here we don’t order the tickets, and look at the tickets we bought as a three, not saying which one was first, second, or third. All in all, if we use the combination calculator, then we’ll get that there are exactly
one_Tshirt = C(40, 1) * C(860, 2) = (40 / 1) * (860 * 859) / (2 * 1) = 14,774,800.

The number of all possible triples is the same here as it was above. Therefore, with the probability calculator, we obtain
P₁ = one_Tshirt / all_triples = 14,774,800 / 121,095,300 ≈ 0.12.

Question 2
Observe that we counted something very similar in Question 1. In fact, using the notation from the above solution, the probability of winning exactly two T-shirts was P₂. This time, however, we are not so keen on the T-shirt and will be satisfied with any pair of prizes. To be precise, we want every set of three tickets that consists of any two of the 200 winning tickets, and a single one from the pool of 900 – 200 = 700 losing ones. Note that again we do not order the tickets we buy. Therefore, the combination calculator gives that the number of such is
two_prizes = C(700, 1) * C(200, 2) = (700 / 1) * (200 * 199) / (1 * 2) = 13,930,000.

The number of all possible outcomes, i.e., the number of all possible sets of three tickets, has not changed. Therefore, we use the probability calculator to obtain
P = two_prizes / all_triples = 13,930,000 / 121,095,300 ≈ 0.12.

Question 3
In this question, we want all three tickets to grant prizes, but we do not say what prizes they should be. We only want them to be different.

Let us consider a few cases:
P₁. we win the trip, a ticket to the movies, and a T-shirt,
P₂. we win the trip, a ticket to the movies, and a mug,
P₃. we win the trip, a T-shirt, and a mug,
P₄. we win a ticket to the movies, a T-shirt, and a mug.
Observe that those four are all the sets of prizes that we may obtain to meet our condition, and that no two of them can happen simultaneously. The probability we want is
P = P₁ + P₂ + P₃ + P₄.
We must calculate each of them separately and sum them together in the end.

The calculation is fairly similar for all the cases. In essence, we need to count how many sets of three tickets give P₁, P₂, P₃, and P₄, then divide the result by the number of all possible triples (which is equal to all_triples, just as it was in Questions 1 and 2). To be precise, we need to know how many tickets give us one particular prize, then multiply it by the number of tickets giving the second one, and then do the same with the third.

Note that we know all of these values but one – the number of tickets that correspond to a school mug. To find it, recall that there are 200 winning tickets altogether. Subtracting the number of other prizes (1, 9, and 40 ) from the total number gives:
mugs = winning – trip – movies – Tshirt = 200 – 1 – 9 – 40 = 150
tickets that give a school mug.

Now we are ready to count the probabilities. Let’s start with P₁, i.e., the probability of winning the trip, a movie ticket, and a T-shirt. We know that there are 1, 9, and 40 tickets granting those prizes respectively. Remember that in our calculations we do not order the tickets we buy, and so do not distinguish between winning the trip first, then a ticket to the movies, then a T-shirt, and winning first a T-shirt, then the trip and then a ticket to the movies. Therefore, the probability calculator gives:
P₁ = (trip * movies * Tshirt) / all_triples = (1 * 9 * 40) / 121,095,300 = 360/121,095,350.
Observe that we didn’t approximate the result as a decimal yet. We will do this at the end after summing together all the results, as this minimizes the error in rounding the numbers.

Similarly, we use the probability calculator to obtain:
P₂ = (trip * movies * mug) / all_triples = (1 * 9 * 150) / 121,095,300 = 1,350/121,095,300,
P₃ = (trip * Tshirt * mug) / all_triples = (1 * 40 * 150) / 121,095,300 = 6,000/121,095,300,
P₄ = (movies + Tshirt * mug) / all triples = (9 * 40 * 150) / 121,095,300 = 54,000/121,095,300.

Lastly, we sum the four probabilities and get
P = P₁ + P₂ + P₃ + P₄ = (360 + 1,350 + 6,000 + 54,000) / 121,095,300 ≈ 0.0005.

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