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- Thread starter aleezay
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Can you please post a specific question here or provide the link?
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its not just 1 q...i dont understand what its all about...
khaer anyhow...tell me how to solve a q lyk ds 1:
In a game a player tosses 3 fair coins. he wins $10 if all heads occur, $x if 2 heads occur, $3 if 1 head occurs and $2 if no heads occur. express in terms of x his expected gain 4m each game.
given that he pays $4.50 to play each game,calculate the value of x (if each game is fair) and his expected loss or gain over 100 games of x=4.90
khaer anyhow...tell me how to solve a q lyk ds 1:
In a game a player tosses 3 fair coins. he wins $10 if all heads occur, $x if 2 heads occur, $3 if 1 head occurs and $2 if no heads occur. express in terms of x his expected gain 4m each game.
given that he pays $4.50 to play each game,calculate the value of x (if each game is fair) and his expected loss or gain over 100 games of x=4.90
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For aleezay: NOVEMBER 2004 PAPER Q11 (b)
(i) Relies on part (a) (ii). A total score of 3 can be obtained if one of the scores is 1 and the other is 2. So the probability = (1/9 x 2/9 x 2) = 4/81.
(ii) First determine which numbers/scores can combine to give you a total score of 6. Your possibilities can be (1,5), (5,1), (2,4), (4,2), and (3,3). Now add together the probabilities of these individual scores = (1/9 x 1/9 x 2) + (2/9 x 2/9 x 2) + (1/9 x 1/9)
= 2/81 + 8/81 + 1/81
= 11/81.
(iii) The same method as in the previous part applies here. You get a total score of 2 if both your individual scores are 1. So the probability for a total score of 2 is (1/9 x 1/9) = 1/81.Continue in the same manner for the rest of the total scores. For instance for 8, your possibilities are (2,6), (6,2) (3,5), (5,3), (4,4). Hence the probability = (2/9 x 2/9 x 2) + (1/9 x 1/9 x 2) + (2/9 x 2/9)
= 8/81 + 2/81 + 4/81
= 14/81.
And so on...
(iv) Your objective is to calculate the maximum expected profit. You already have all the probabilities for each of the total scores from (iii). If the player is lucky enough to get the same total score he chose before rolling the die, he gets the same number of dollars as his/her total score.("he is paid $1 multiplied by his total score, but nothing otherwise.")So he needs to carefully select those total scores that would give him the maximum profit.
To calculate the profit for each total score, multiply the score by its probability.
For a score of 2, the profit would be = 2 x 1/81 = $0.02. For 3 it is = 3 x 4/81 = $0.15 and so on.
You'll realise (after calculation) that 8 and 10 are the scores that generate maximum profit ( $1.38 and $ 1.11, respectively).
(v) For a game to be fair, the amount paid by player for the game must equal the amount earned by the game. If the player selects 8 and 10 as his total scores, he earns = $ (1.38 + 1.11) = $2.49.
So he must pay $2.49 if the game is to be fair. (No net gain/loss for the player is what they call a fair game.)
Correct me if I'm wrong. If your still sceptical, let me know if I can help.
(i) Relies on part (a) (ii). A total score of 3 can be obtained if one of the scores is 1 and the other is 2. So the probability = (1/9 x 2/9 x 2) = 4/81.
(ii) First determine which numbers/scores can combine to give you a total score of 6. Your possibilities can be (1,5), (5,1), (2,4), (4,2), and (3,3). Now add together the probabilities of these individual scores = (1/9 x 1/9 x 2) + (2/9 x 2/9 x 2) + (1/9 x 1/9)
= 2/81 + 8/81 + 1/81
= 11/81.
(iii) The same method as in the previous part applies here. You get a total score of 2 if both your individual scores are 1. So the probability for a total score of 2 is (1/9 x 1/9) = 1/81.Continue in the same manner for the rest of the total scores. For instance for 8, your possibilities are (2,6), (6,2) (3,5), (5,3), (4,4). Hence the probability = (2/9 x 2/9 x 2) + (1/9 x 1/9 x 2) + (2/9 x 2/9)
= 8/81 + 2/81 + 4/81
= 14/81.
And so on...
(iv) Your objective is to calculate the maximum expected profit. You already have all the probabilities for each of the total scores from (iii). If the player is lucky enough to get the same total score he chose before rolling the die, he gets the same number of dollars as his/her total score.("he is paid $1 multiplied by his total score, but nothing otherwise.")So he needs to carefully select those total scores that would give him the maximum profit.
To calculate the profit for each total score, multiply the score by its probability.
For a score of 2, the profit would be = 2 x 1/81 = $0.02. For 3 it is = 3 x 4/81 = $0.15 and so on.
You'll realise (after calculation) that 8 and 10 are the scores that generate maximum profit ( $1.38 and $ 1.11, respectively).
(v) For a game to be fair, the amount paid by player for the game must equal the amount earned by the game. If the player selects 8 and 10 as his total scores, he earns = $ (1.38 + 1.11) = $2.49.
So he must pay $2.49 if the game is to be fair. (No net gain/loss for the player is what they call a fair game.)
Correct me if I'm wrong. If your still sceptical, let me know if I can help.
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abcde your 100% right !! Verified answers via Examiner Report
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Thank you!SalmanPakRocks said: