Questions and Answers #6 Probability

Questions and Answers Sheet 6 Probability Question #1 If the mean, median, and mode are all equal for the set {3, 4, 5, 8, 𝑥}, what is the value of x? Answer: 1st question to ask would be, " Are the values in order?" Clearly they can't be because then x would have to be equal to 8 in order to have a mode of 8. But while 8 could be the mode, it is not the median. 2nd question " Where must x lie?" The mode will have to be equal to the median. (the middle value). This would lead to {3, 4, 5, 𝑥, 8} ⇒ [3, 4, 5, 5, 8} What about the mean? If x=5, the sum of the numbers =25 25 Mean = = 5 5 Question #2 The probability of drawing a green candy from a jar of 20 candies is be added to the jar in order to reduce the probability to 1 6 1 4 . How many yellow candies should ? Answer: 10 yellow candies should be added to the jar in order to reduce the probability to 1 6 . Explanation: 1 As the probability of drawing a green candy from a jar of 20 candies is , number of green candies in the 4 1 jar are 20 × = 5. 4 Now let us assume we add x yellow candies. So total number of candies become 20+x and probability of drawing a green candy from a jar becomes As we desire the probability to be 5 20+𝑥 = 1 6 1 5 20+𝑥 6 or (20 + 𝑥) = 5 × 6 = 30 and x=30-20=10 1 Hence, 10 yellow candies should be added to the jar in order to reduce the probability to . 6 Question #3 A study shows that approximately 25% of all businessmen who wear ties wear them so tightly that they reduce blood flow to the brain. At a board meeting of 15 businessmen, all of whom wear ties, what's the probability at least one tie is too tight? Answer: We are told that: 25 1 𝑝(tight tie) = = 100 4 1 3 4 4 𝑃(not a tight tie) = 1 − = And so: 𝑃(at least one too tight) = 1 − 𝑃(none are too tight) 3 15 =1=( ) 4 = 0.986636 ⋯ Question #4 Sociologists say that 95% of married women claim that their husband's mother is the biggest bone of contention in their marriages. Suppose that six married women are having coffee together. What's the probability all of them dislike their mother-in-law? Answer: The probability that one of woman disliking her mother-in-law is: 𝑝1 = 0.95 The probability that two women dislike their mothers-in-law is (probability rules): 𝑝2 = 𝑝1 × 𝑝1 = 𝑝12 = 0.952 = 0.9025 By extension it is simple to see that the case of all 6 disliking their mothers-in-law would be 𝑝6 = 𝑝16 = 0.956 ≅ 0.735 Therefore the probability that all 6 women dislike their mothers-in-law is 73.5%. Question #5 At a local university, 54.3% of incoming first year students have a PC. If 3 students are selected randomly, what are the probabilities that at least one has a computer? Answer: Let X denote number of students with computer in n=3 independent students, where each has a probability p=0.543 We denote this by: 𝑋 ∼ Binomial(𝑛 = 3, 𝑝 = 0.543) Probability of x number of students with computer is: 𝑃(𝑋 = 𝑥) = ((𝑛), (𝑥))𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 where ((n),(x)) is combinatorial expression: 𝑛! ((𝑛), (𝑥)) = 𝑥!(𝑛−𝑥)! 𝑃(𝑋 ≥ 1), 𝑛 = 3, 𝑝 = 0.543 𝑃(𝑋 ≥ 1) = 1 − 𝑃(𝑋 < 1) We first calculate: 𝑃(𝑋 < 1) = 𝑃(0 ≤ 𝑋 < 1) = 𝑃(𝑋 = 0) 𝑃(𝑋 = 0) = ((3), (0))(0.543)0 (1 − 0.543)3 = 1(0.543)0 (0.457)3 ≈ 0.095444 𝑃(𝑋 = 0) = 0.095444 ≈ 0.095 𝑃(𝑋 < 1) = 0.0954 𝑃(𝑋 ≥ 1) = 1 − 𝑃(𝑋 < 1) = 1 − 0.0954 = 0.9046 Question #6 1 5 3 12 Suppose S and T are mutually exclusive events. Find P(S or T) if 𝑃(𝑆) = and 𝑃(𝑇) = Answer: We use the following fundamental definition from Probability and Set Theory: 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) And if we apply this to our problem we have: 𝑃(𝑆 ∪ 𝑇) = 𝑃(𝑆) + 𝑃(𝑇) − 𝑃(𝑆 ∩ 𝑇) 1 5 3 3 12 = + − 𝑃(𝑆 ∩ 𝑇) = − 𝑃(𝑆 ∩ 𝑇) 4 ? Now we are told that S and T are mutually exclusive, and so: 𝑃(𝑆 ∩ 𝑇) = 0 3 Hence, 𝑃(𝑆 ∪ 𝑇) = 4 Question #7 How do you find the probability of an event given the odds of the event occurring is 2:1 ? Answer: Probabilities always range between 0 and 1. The odds on the other hand are defined as the probability that the event will occur divided by the probability that the event will not occur. As odds of the event occurring is 2:1 . the probability that the event will occur is twice the probability that the event will not occur. But if probability of happening of event is p, probability of event not happening is 1-p 𝑝 2 2 Hence = 2 or 𝑝 = 2 − 2𝑝 or 3𝑝 = 𝑖. 𝑒. 𝑝 = and desired probability is . 1−𝑝 3 3 Question #8 Shironda rolls a number cube and chooses a card from a standard deck of cards. What is the probability that Shironda rolls a 5 on the number cube and then chooses a 5 from the deck? 1) 2) 3) 4) 1 78 1 24 5 78 10 39 Answer: The probability that one rows a 5 on a standard six-sided cube is 1 6 since there is 1 number on the dice that has the number 5 out of a total of 6 possibilities. On a deck of 52 cards, there are four 5's (for each suit). Thus, the probability is 4 1 = . 52 13 These events are independent. Thus, to calculate their P(A and B), we multiply their probabilities. 1 78 . Question #9 How do you find the odds of an event occurring given its probability 5 8 Answer: Answer: 5 3 Explanation: Odds is the ratio of favourable outcomes to non-favourable outcomes: If P is the probability of an event occurring, then: 1-P is the probability of the event not occurring. The odds will then be: ? 1 6 ⋅ 1 13 = 𝑃 ⇒ 1−𝑃 5 8 5 1−( ) 8 = 5 8 3 8 = 5 3 To make this clearer: The probability of throwing a 6 with a standard die is The odds of throwing a 6 are 𝑃 = 1−𝑃 1 6 1 6 1−( ) = 1 6 5 6 = 1 1 6 5 1 5 Question #10 If 𝑃(𝐴) = 1 2 and 𝑃(𝐵) = 8 9 what is P(A and B)? Answer: Without additional information t is not possible to answer the question. However, if we assume that the events A and B are independent events then we can, as, in that case: 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵) 18 = 29 Hence, = 4 9 Question #11 The experimental probability that Kristen will hit the ball when she is at bat is a season, how many times can Kristen expect to hit the ball? Answer: Number of times she is expected to hit the ball = 𝑃 × Total times she bat 3 = × 80 5 = 16 3  80 5 = 3 × 16 = 48 𝑡𝑖𝑚𝑒𝑠 3 5 . If she is at bat 80 times in