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7 Cards in this Set
- Front
- Back
On April 15, 1912, the Titanic struck an iceberg and rapidly sank with only 710 of her 2,204 passengers and crew survivng. Some believe that the rescue procedures favored the wealthier first class passengers. Data on survival of passengers are summarized in the table. Are the events "passenger survived" and "passenger was in first class" independent events? Support your answer using appropriate calculations.
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P(passenger survived|passenger was in first class)=201/324 =0.620 P(passenger survived)= 500/1317= 0.380 0.620=/ 0.380, so the two events are not independent. (note that P(passenger was in first class|passenger survived) could also be compared to P(passenger was in first class))
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On April 15, 1912, the Titanic struck an iceberg and rapidly sank with only 710 of her 2,204 passengers and crew survivng. Some believe that the rescue procedures favored the wealthier first class passengers. Data on survival of passengers are summarized in the table. Are the events "passenger survived" and "passenger was in third class" independent events? Support your answer using appropriate calculations.
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P(Passenger survived|passenger was in third class) =181/709 =0.255 P(passenger survived) = 500/1317=0.380 0.255=/0.380, so the two events are not independent.
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On April 15, 1912, the Titanic struck an iceberg and rapidly sank with only 710 of her 2,204 passengers and crew survivng. Some believe that the rescue procedures favored the wealthier first class passengers. Data on survival of passengers are summarized in the table.Did all passengers aboard the Titanic have the same probability of surviving? Support your answer using appropriate calculations.
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Based on the comparisons made in the previous two qiestions, we can say that not all passengers aboard the Titanic had the same chance of surviving. The first class passengers had the greatest chance of survival, and the third class passengers had the smallest chance of being rescued.
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At Mom's diner, everyone drinks coffee. Let C= the event that a randomly-selected customer puts cream in their coffee. Let S= the event that a randomly-selected customer puts sugar in their coffee. Suppose that after years of collecting data, Mom has estimated the following probabilities: P(C)=0.6 P(S)=0.5 P(C or S)=0.7. Estimate P(C and S) and interpret this value in the context of the problem.
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P(C or S) = P(C)+P(S)-P(C and S) 0.7=0.6+0.5-P(C and S) P(C and S)=0.6+0.5-0.7 = 0.4 The probability that a randomly-selected customer at Mom's has both cream and sugar in their coffee is 0.4
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Dena has a box with 7 blue marbles and 3 pink marbles. Two marbles are drawn without replacement from the box. What is the probability that both of the marbles are blue?
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Let A= the event that the first marble is blue; and let B= the event that the second marble is blue. In the beginning, there are 10 marbles in the box, 8 of which are blue, therefore, P(A)=7/10. After the first selection, there are 9 marbles in the box, 6 of which are blue, therefore P(B|A)=6/9. P(A ∩ B) = P(A)*P(B|A) P(A ∩ B)= (7/10)(6/9)= 42/90 = 7/15
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John is going to draw two cards from a standard deck. What is the probability that the first card is a queen and the second card is a jack?
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P(queen) * P(jack)= (4/52)(4/51)=16/2652 = 4/663
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Alex, Mel, and Chelsea play a game that has 6 rounds. In each round there is a single winner, and the outcomes of the rounds are independent. For each round the probability that Alex wins is 1/2, and Mel is twice as likely to win as Chelsea. What is the probability that Alex wins three rounds, Mel wins two rounds, and Chelsea wins one round?
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We are given that Mel is twice as likely to win a round as Chelsea. Their combined probability of winning a round is 1/2 since this number combined with Alex's probability of winning a round must add up to 1. 2x+x=1/2 x=1/6, so the probability that Mel wins a round is 2/6=1/3 and the probability that Chelsea wins a round is 1/6. Suppose we let AAAMMC denote the situation where Alex wins the first three rounds, Mel wins the fourth and fifth rounds, and Chelsea wins the 6th round. P(AAAMMC)= (1/2)³ * (1/3)² * (1/6). Since there are 6 rounds, the number of ways Alex can win 3 of them is 20. Once the 3 games which alex won have been chose, Mel must win two of the other three, so the number of ways this can happen is 3. The remaining game, won by chelsea, can only happen one way. So, (1/2)³ * (1/3)² * (1/6) * 20 * 3= 5/36
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