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Lera25 [3.4K]
2 years ago
12

Suppose a box of Cracker Jacks contains one of 5 toy prizes: a small rubber ball, a whistle, a Captain America decoder ring, a r

ace car, or a magnifying glass. Each prize is equally likely to be in a box. Question 1. How many boxes of Cracker Jacks would you expect to buy until you obtain a complete set of prizes
Mathematics
1 answer:
lakkis [162]2 years ago
5 0

Answer:

11.42 boxes

Step-by-step explanation:

For the first box bought, there is a 100% chance of getting a unique toy (since you still don't have any). E₁ = 1.

After that, there is a 4 in 5 chance of getting a unique toy from the next box, the expected number of boxes required is:

E_2 = (\frac{4}{5})^{-1} = 1.25

For the next unique toy, there is now a 3 in 5 chance of getting it:

E_3 = (\frac{3}{5})^{-1} = 1.67

Following that logic, there is a 2 in 5 chance of getting the 4th unique toy:

E_4 = (\frac{2}{5})^{-1} = 2.5

Finally, there is a 1 in 5 chance to get the last unique toy:

E_5 = (\frac{1}{5})^{-1} = 5

The expected number of boxes to obtain a full set is:

E=E_1+E_2+E_3+E_4+E_5\\E=1+1.25+1.67+2.5+5\\E=11.42\ boxes

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----------------------------------------------

The product of 8 and a number

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Using the slope and the y intercept, graph the line represented by the following equation.
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We have P(X=x)=\frac{1}{3} and P(Y=y)=\frac{1}{5}, because the dice are fair.

Now we use the assumption of independence to claim that

P(X=x, Y=y) = P(X=x)\cdot P(Y=y) =\dfrac{1}{3}\cdot\dfrac{1}{5} = \dfrac{1}{15}

Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:

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This implies that the probabilities of the outcomes of W=X+Y are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5

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