Diane is on a game show and she has made it to the very final round where she will have an opportunity to win a prize. For the final round, there are three doors and behind each door are eight suitcases. Each suitcase either contains a money prize or is empty. Behind the first door 7 of the suitcases are empty, behind the second door 6 of the suitcases are empty and behind the third and final door 2 of the suitcases are empty. Diane must choose a door, and once she has chosen a door she can open two of the eight suitcases and keep any prize money that she finds. Rather than just choosing among the three doors completely randomly, based on how previous contestants have performed in this final round, Diane decides to choose from the first, second and third doors with probabilities 0.51, 0.15 and 0.34, respectively. Once she has chosen a door, Diana decides to randomly open two of the eight suitcases.Find the probability that Diane goes home empty-handed.Answer:Question 1
Question
Diane is on a game show and she has made it to the very final round where she will have an opportunity to win a prize. For the final round, there are three doors and behind each door are eight suitcases. Each suitcase either contains a money prize or is empty. Behind the first door 7 of the suitcases are empty, behind the second door 6 of the suitcases are empty and behind the third and final door 2 of the suitcases are empty. Diane must choose a door, and once she has chosen a door she can open two of the eight suitcases and keep any prize money that she finds. Rather than just choosing among the three doors completely randomly, based on how previous contestants have performed in this final round, Diane decides to choose from the first, second and third doors with probabilities 0.51, 0.15 and 0.34, respectively. Once she has chosen a door, Diana decides to randomly open two of the eight suitcases.Find the probability that Diane goes home empty-handed.Answer:Question 1
Solution
To solve this problem, we need to calculate the probability of Diane going home empty-handed from each door and then sum these probabilities, each weighted by the probability of Diane choosing that door.
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For the first door, there are 7 empty suitcases out of 8. The probability of Diane choosing an empty suitcase twice is (7/8)*(6/7) = 6/8 = 0.75. The probability of Diane choosing this door and going home empty-handed is 0.51 * 0.75 = 0.3825.
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For the second door, there are 6 empty suitcases out of 8. The probability of Diane choosing an empty suitcase twice is (6/8)*(5/7) = 0.5357. The probability of Diane choosing this door and going home empty-handed is 0.15 * 0.5357 = 0.08035.
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For the third door, there are 2 empty suitcases out of 8. The probability of Diane choosing an empty suitcase twice is (2/8)*(1/7) = 0.0357. The probability of Diane choosing this door and going home empty-handed is 0.34 * 0.0357 = 0.01214.
Adding these probabilities together, the total probability of Diane going home empty-handed is 0.3825 + 0.08035 + 0.01214 = 0.47499, or approximately 47.5%.
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