To find the expected value of tickets per throw, we need to calculate the expected value of the number of tickets for one throw of the bean bag. The expected value \(E\) can be calculated using the formula:
\[ E = \sum (p(x) \cdot x) \]
where \(p(x)\) is the probability of each outcome and \(x\) is the value associated with that outcome.
First, let's determine the probability of the bean bag landing in the center ring. Since the probabilities for the outer and middle rings are given:
- Probability of Outer Ring: \(P(O) = 0.63\)
- Probability of Middle Ring: \(P(M) = 0.15\)
The probability of landing in the Center Ring \(P(C)\) can be calculated as follows:
\[ P(C) = 1 - P(O) - P(M) = 1 - 0.63 - 0.15 = 0.22 \]
Now, we can set up the expected value calculation based on the number of tickets awarded:
- Inner Ring (Outer): Probability \(= 0.63\), Tickets \(= 1\)
- Inner Ring (Middle): Probability \(= 0.15\), Tickets \(= 2\)
- Inner Ring (Center): Probability \(= 0.22\), Tickets \(= 3\)
Now, we can compute the expected value:
\[ E = (0.63 \cdot 1) + (0.15 \cdot 2) + (0.22 \cdot 3) \]
Calculating each term:
- Outer Ring: \(0.63 \cdot 1 = 0.63\)
- Middle Ring: \(0.15 \cdot 2 = 0.30\)
- Center Ring: \(0.22 \cdot 3 = 0.66\)
Now we sum these values:
\[ E = 0.63 + 0.30 + 0.66 = 1.59 \]
Thus, the expected value of the tickets per throw is \(1.59\).
Since you plan to play the game 10 times, the total expected value for 10 throws would be:
\[ E_{total} = E \cdot 10 = 1.59 \cdot 10 = 15.9 \]
Therefore, the expected number of tickets you will get after playing the game 10 times is 15.9 tickets.