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
Question 1
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.
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} )
Calculate Expected Value of Winnings:
E(X)=(201×9000)+(101×7000)+(41×3000)+(2012×0)E(X)=450+700+375+0=1525