Calculating the probability of receiving all possible rewards after 15 events

Question

I encountered this question in my Data Management and Statistics textbook. I tried to calculate the probability using binomial theorem and combinations/permutations, but I could only get close to the answer. I would really appreciate any help with this question:

If Joe buys a cereal box and it has the following probabilities of giving one of the five possible rewards:

• Toy Car: 20%
• Toy Truck: 30%
• Toy Spoon: 10%
• Toy Doll: 35%
• Toy Gun: 5%

What is the probability of Joe getting all the rewards after buying 15 cereal boxes?

Thank you in advance

Answer 1

Use the Inclusion/Exclusion principle:

1. Start with $1$

2. Subtract the following:

   • The probability of not winning C is $(1-0.20)^{15}$
   • The probability of not winning T is $(1-0.30)^{15}$
   • The probability of not winning S is $(1-0.10)^{15}$
   • The probability of not winning D is $(1-0.35)^{15}$
   • The probability of not winning G is $(1-0.05)^{15}$

3. Add the following:

   • The probability of not winning C,T is $(1-0.20-0.30)^{15}$
   • The probability of not winning C,S is $(1-0.20-0.10)^{15}$
   • The probability of not winning C,D is $(1-0.20-0.35)^{15}$
   • The probability of not winning C,G is $(1-0.20-0.05)^{15}$
   • The probability of not winning T,S is $(1-0.30-0.10)^{15}$
   • The probability of not winning T,D is $(1-0.30-0.35)^{15}$
   • The probability of not winning T,G is $(1-0.30-0.05)^{15}$
   • The probability of not winning S,D is $(1-0.10-0.35)^{15}$
   • The probability of not winning S,G is $(1-0.10-0.05)^{15}$
   • The probability of not winning D,G is $(1-0.35-0.05)^{15}$

4. Subtract the following:

   • The probability of not winning C,T,S is $(1-0.20-0.30-0.10)^{15}$
   • The probability of not winning C,T,D is $(1-0.20-0.30-0.35)^{15}$
   • The probability of not winning C,T,G is $(1-0.20-0.30-0.05)^{15}$
   • The probability of not winning C,S,D is $(1-0.20-0.10-0.35)^{15}$
   • The probability of not winning C,S,G is $(1-0.20-0.10-0.05)^{15}$
   • The probability of not winning C,D,G is $(1-0.20-0.35-0.05)^{15}$
   • The probability of not winning T,S,D is $(1-0.30-0.10-0.35)^{15}$
   • The probability of not winning T,S,G is $(1-0.30-0.10-0.05)^{15}$
   • The probability of not winning T,D,G is $(1-0.30-0.35-0.05)^{15}$
   • The probability of not winning S,D,G is $(1-0.10-0.35-0.05)^{15}$

5. Add the following:

   • The probability of not winning C,T,S,D is $(1-0.20-0.30-0.10-0.35)^{15}$
   • The probability of not winning C,T,S,G is $(1-0.20-0.30-0.10-0.05)^{15}$
   • The probability of not winning C,T,D,G is $(1-0.20-0.30-0.35-0.05)^{15}$
   • The probability of not winning C,S,D,G is $(1-0.20-0.10-0.35-0.05)^{15}$
   • The probability of not winning T,S,D,G is $(1-0.30-0.10-0.35-0.05)^{15}$

Please note that the sum of the probabilities is equal to $1$.

If it was smaller, then you would also need to subtract the probability of not winning C,T,S,D,G.

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Here is a Python script for calculating that:

p = [0.20,0.30,0.10,0.35,0.05]

res = 1

for i in range(0,len(p)):
    res -= (1-p[i])**15
    for j in range(i,len(p)):
        res += (1-p[i]-p[j])**15
        for k in range(j,len(p)):
            res -= (1-p[i]-p[j]-p[k])**15
            for n in range(k,len(p)):
                res += (1-p[i]-p[j]-p[k]-p[n])**15

print res

The result is $0.54837227253$.

Answer 2

You need the inclusion-exclusion principle. Presumably you can calculate the chance that he did not receive specifically the toy car. You might then think the chance of getting them all is 1-(probability to not get car)-(probability to not get truck)-(two more terms). This would be close, but you have subtracted twice the chance he got neither the car nor the truck, so you need to add it back in once. Then the cases he doesn't get three things have been subtracted three times and added three times, so need to be subtracted once again. Finally the foursies.....

Answer 3

The probability is $$\sum_{A\subseteq \{1,2,3,4,5\}}{(-1)^{5-|A|}\left(\sum_{i\in A}{p_i}\right)^{15}}$$ To see this, note that $(p_1+p_2+p_3+p_4+p_5)^{15}=1$ gives us all possible combinations. $(p_1+p_2+p_3+p_4)^{15},(p_1+p_2+p_3+p_5)^{15}$, etc. give us all combinations where at least one doesn't occur, and we want to subtract these off. However, we have subtracted off the cases where at most three distinct events occur twice, so we have to add them back, etc. This is the principle of inclusion-exclusion.