Questions: You are opening boxes of cereal one at a time looking for your favorite sports player's card, which is in 40% of the boxes. You buy 5 boxes to see how many cards you might get. Complete parts (a) though (d) below.

You are opening boxes of cereal one at a time looking for your favorite sports player's card, which is in 40% of the boxes. You buy 5 boxes to see how many cards you might get. Complete parts (a) though (d) below.
Transcript text: You are opening boxes of cereal one at a time looking for your favorite sports player's card, which is in $40 \%$ of the boxes. You buy 5 boxes to see how many cards you might get. Complete parts (a) though (d) below.
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Solution

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Solution Steps

Step 1: Define the Problem

We are analyzing the probability of finding a sports player's card in boxes of cereal. Each box has a \(40\%\) chance of containing the card. We will open \(n = 5\) boxes and let \(X\) be the random variable representing the number of cards found. The probability of success (finding a card) is \(p = 0.4\) and the probability of failure (not finding a card) is \(q = 1 - p = 0.6\).

Step 2: Calculate the Probability Model

Using the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

we calculate the probabilities for \(x = 0, 1, 2, 3, 4, 5\):

  • For \(x = 0\): \[ P(X = 0) = 0.078 \]

  • For \(x = 1\): \[ P(X = 1) = 0.259 \]

  • For \(x = 2\): \[ P(X = 2) = 0.346 \]

  • For \(x = 3\): \[ P(X = 3) = 0.230 \]

  • For \(x = 4\): \[ P(X = 4) = 0.077 \]

  • For \(x = 5\): \[ P(X = 5) = 0.010 \]

Step 3: Present the Probability Model

The complete probability model for the number of cards found is as follows:

\[ \begin{align_} P(X = 0) & = 0.078 \\ P(X = 1) & = 0.259 \\ P(X = 2) & = 0.346 \\ P(X = 3) & = 0.230 \\ P(X = 4) & = 0.077 \\ P(X = 5) & = 0.010 \\ \end{align_} \]

Final Answer

The probability model for the number of sports player's cards found in 5 boxes of cereal is:

\[ \boxed{ \begin{align_} P(X = 0) & = 0.078 \\ P(X = 1) & = 0.259 \\ P(X = 2) & = 0.346 \\ P(X = 3) & = 0.230 \\ P(X = 4) & = 0.077 \\ P(X = 5) & = 0.010 \\ \end{align_} } \]

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