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In a competition Mary has a probability of \( \frac{1}{20} \) of winning, a probability of \( \frac{1}{10} \) of finishing in second place, and a probability of \( \frac{1}{4} \) of finishing in third place - Leaving Cert Mathematics - Question 1 - 2018

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Question 1

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In a competition Mary has a probability of \( \frac{1}{20} \) of winning, a probability of \( \frac{1}{10} \) of finishing in second place, and a probability of \( ... show full transcript

Worked Solution & Example Answer:In a competition Mary has a probability of \( \frac{1}{20} \) of winning, a probability of \( \frac{1}{10} \) of finishing in second place, and a probability of \( \frac{1}{4} \) of finishing in third place - Leaving Cert Mathematics - Question 1 - 2018

Step 1

Find the expected value of Mary's loss if she enters the competition.

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Answer

To find the expected value of Mary's loss, we first need to calculate the expected value of her winnings and subtract the entry fee.

  1. Probabilities and Prize Amounts:

    • Probability of winning: ( \frac{1}{20} ) for €9000
    • Probability of finishing second: ( \frac{1}{10} ) for €7000
    • Probability of finishing third: ( \frac{1}{4} ) for €3000
    • Probability of not placing: ( 1 - \left(\frac{1}{20} + \frac{1}{10} + \frac{1}{4} \right) = 1 - \left(\frac{1}{20} + \frac{2}{20} + \frac{5}{20} \right) = 1 - \frac{8}{20} = \frac{12}{20} )
  2. Calculate Expected Value of Winnings: E(X)=(120×9000)+(110×7000)+(14×3000)+(1220×0)E(X) = \left( \frac{1}{20} \times 9000 \right) + \left( \frac{1}{10} \times 7000 \right) + \left( \frac{1}{4} \times 3000 \right) + \left( \frac{12}{20} \times 0 \right) E(X)=450+700+375+0=1525E(X) = 450 + 700 + 375 + 0 = 1525

  3. Entry Fee: The entry fee is €2000.

  4. Expected Loss: Expected Loss=Entry FeeExpected Winnings=20001525=475 \text{Expected Loss} = \text{Entry Fee} - \text{Expected Winnings} = 2000 - 1525 = 475

Thus, the expected value of Mary's loss if she enters the competition is €475.

Step 2

Find the value of x.

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Answer

In part (b), we know that Mary expects to break even. This means her expected winnings should equal her entry fee.

  1. Updated Prizes:

    • Winning prize: ( 9000 + x )
    • Second prize: ( 7000 + x )
    • Third prize: ( 3000 + x )
  2. Probabilities Remain the Same:

    • Probability of winning: ( \frac{1}{20} )
    • Probability of second place: ( \frac{1}{10} )
    • Probability of third place: ( \frac{1}{4} )
  3. Set Up the Equation for Break-even: E(X)=(120(9000+x))+(110(7000+x))+(14(3000+x))E(X) = \left( \frac{1}{20} \cdot (9000 + x) \right) + \left( \frac{1}{10} \cdot (7000 + x) \right) + \left( \frac{1}{4} \cdot (3000 + x) \right) 2000=900020+700010+30004+(x20+x10+x4)2000 = \frac{9000}{20} + \frac{7000}{10} + \frac{3000}{4} + \left( \frac{x}{20} + \frac{x}{10} + \frac{x}{4} \right)

  4. Simplify: 2000=450+700+750+(x20+2x20+5x20)2000 = 450 + 700 + 750 + \left( \frac{x}{20} + \frac{2x}{20} + \frac{5x}{20} \right) 2000=1900+8x202000 = 1900 + \frac{8x}{20} 100=8x20100 = \frac{8x}{20}

  5. Solve for x: x=250x = 250

Therefore, the value of x is €250.

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