b 085 9 0232 The probability that the 11 am flight takes off on time is 02316

B 085 9 0232 the probability that the 11 am flight

This preview shows page 9 - 11 out of 11 pages.

b) (0.85) 9 = 0.232 The probability that the 11 a.m. flight takes off on time is 0.2316 (23.16%). c) 1 – (0.85) 14 = 0.8972 The probability that the 1:30 p.m. flight is delayed is 0.8972 (89.72%). 7. a) 30 C 2 x (0.05) 2 x (0.95) 28 = 0.258636738 b) 1- P(2) – P1(1) – P(0) 1 – 30 C 2 x (0.05) 2 x (0.95) 28 - 30 C 1 x (0.05) 1 x (0.95) 29 - (0.95) 30 = 0.187821187 8. a) I have created this table to help me solve this question. Shot Number Attempt # Probability 1 Person A, 1 st Attempt 0.166666667 2 Person B, 1 st Attempt 0.138888889 3 Person A, 2 nd Attempt 0.115740741 4 Person B, 2 nd Attempt 0.096450617 5 Person A, 3 rd Attempt 0.080375514 6 Person B, 3 rd Attempt 0.066979595
Image of page 9
7 Person A, 4 th Attempt 0.055816329 8 Person B, 4 th Attempt 0.046513608 9 Person A, 5 th Attempt 0.03876134 10 Person B, 5 th Attempt 0.032301117 11 Person A, 6 th Attempt 0.026917597 12 Person B, 6 th Attempt 0.022431331 13 Person A, 7 th Attempt 0.018692776 14 Person B, 7 th Attempt 0.015577313 15 Person A, 8 th Attempt 0.012981094 16 Person B, 8 th Attempt 0.010817579 17 Person A, 9 th Attempt 0.009014649 18 Person B, 9 th Attempt 0.007512207 19 Person A, 10 th Attempt 0.006260173 20 Person B, 10 th Attempt 0.005216811 In the table above, if Person A goes first, he would take shots 1,3,5,7,9,11,13, 15,17, and 19. If Person B goes second, he would take shots 2,4,6,8,10,12,14, 16, 18, and 20. Since the cartridge is spun before every shot attempt, the probability of "success" p is the same for each outcome. Since we are looking at the number of trials required to obtain the first hit, we can use the geometric distribution formula to help us determine whether or not it would be advantage to go second. By looking at the graph below, we can deduce a general trend, which indicates that the probability of Person A getting hit on their collective “attempts” is slightly higher than Person B. This is because the probability of getting hit by the first shot on the first attempt is the greatest, and each
Image of page 10
succeeding attempt has a slightly lower probability than the previous. For example, the probability of the first hit being on the third shot (Person A) would be much higher than the probability of the first hit being on the fourth shot (Person B). Therefore, the probability of Person A getting hit first is higher than the probability that Person B gets hit first. This pattern is characteristic of any geometric distribution. To prove this, we can add up the probabilities of the attempts and determine whether Person A or Person B has a higher chance of being shot before their tenth attempt. If we add up the probabilities of Person A’s ten attempts (1,3,5,7,9,11,13, 15,17, and 19), we get a total probability value of 0.53122688. Similarly, if we add up the probabilities of Person B’s ten attempts (2,4,6,8,10,12,14, 16, 18, and 20), we get a total probability value of 0.442689067. Therefore, by comparing the two values, we can see that it would clearly be an advantage to take the second turn in Thumb-Tack Russian Roulette. I created the graph below to visual represent this.
Image of page 11

You've reached the end of your free preview.

Want to read all 11 pages?

  • Fall '12
  • jp
  • Probability, Probability theory, Binomial distribution, Geometric distribution, Hypergeometric Distribution

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture