Probability and combinatorics #3 - Questions and Answers

Question 1 A dresser drawer contains one pair of socks with each of the following colors: blue, brown, red, white and black. Each pair is folded together in a matching set. You reach into the sock drawer and choose a pair of socks without looking. You replace this pair and then choose another pair of socks. What is the probability that you will choose the red pair of socks both times? Answer Probabilities: P(red) = 1/5 P(red and red) = P(red) * P(red) = (1/5) * (1/5) = 1/25 Question 2 Given the following: P(A) = 0.3 P(B) = 0.2 P(A or B) = 0.5 P(A and B) = 0 Which of the following is true? A and B are disjoint. A and B are neither disjoint nor independent. A and B are independent. A and B are disjoint and independent. Answer Given: P(A) = 0.3 P(B) = 0.2 P(A or B) = 0.5 P(A and B) = 0 Two events are disjoint, if the events cannot occur at the same time. Since P(A and B) = 0, the events A and B cannot occur at the same time and thus events A and B are disjoint. Two events are independent, if the probability that one event occur in no way affects the probability of the other event ocuring. Moreover, two events are intependent if and only if P(A and B) = P(A) * P(B). P(A) * P(B) = 0.3 * 0.2 = 0.06 P(A and B) = 0 Since 0.06 ≠ 0, the events A and B are not independent. Thus the answer is then: A and B are disjoint. Question 3 Can two events with nonzero probabilities be both independent and mutually exclusive? Explain your reasoning. Answer 1 If two events cannot occur at the same moment, they are mutually exclusive. Two occurrences are considered independent if the occurrence of one of them has no effect on the occurrence of the other. Thus, two events cannot occur simultaneously because if one of the occurrences occurs, we know that the other event does not occur, and thus the second event is influenced by the first event occurring. Result: No Answer 2 No, two events with nonzero probabilities cannot be both independent and mutually exclusive. To see why, let A and B be two events with nonzero probabilities such that they are independent and mutually exclusive. Then, by definition: P(A ∩ B) = P(A) * P(B) (independence) and P(A ∩ B) = 0 (mutual exclusivity) However, these two statements are contradictory since P(A ∩ B) = 0 means that the events A and B have no common outcomes, whereas P(A).P(B) > 0 means that the events are related in some way. Therefore, it is impossible for two events with nonzero probabilities to be both independent and mutually exclusive. Answer 3 Explanation: In probability theory, two events A and B are said to be independent if the occurrence of A does not affect the probability of B, and vice versa. In other words, P(A ∩ B) = P(A)P(B). On the other hand, two events A and B are mutually exclusive (or disjoint) if they cannot occur at the same time, i.e., P(A ∩ B) = 0. Now, let's consider whether two events with nonzero probabilities can be both independent and mutually exclusive. Suppose that A and B are two events with nonzero probabilities such that they are both independent and mutually exclusive. Then, we have P(A∩ B) = 0 and P(A)P(B) ≠ 0. Since P(A ∩ B) = 0, we can write P(AUB) = P(A) + P(B) by the addition rule for probability. Moreover, since A and B are independent, we have P(A ∩ B) = P(A)P(B). Combining these two equations, we get: P(AUB) = P(A) + P(B) = P(A) + P(B) + 2P(A)P(B) 2P(A)P(B) = 0 This implies that either P(A) = 0 or P(B) = 0, which contradicts the assumption that both events have nonzero probabilities. Therefore, two events with nonzero probabilities cannot be both independent and mutually exclusive. Answer 4 We know that two events A and B are said to be independent if the occurrence of A does not affect the occurrence of B, and vice versa. Mathematically, P(A ∩ B) = P(A)P(B). On the other hand, two events are mutually exclusive if they cannot occur simultaneously, i.e., P(A ∩ B) = 0. Now, let's assume that two events A and B have nonzero probabilities, i.e., P(A) > 0 and P(B) > 0. If A and B are independent, then P(A ∩ B) = P(A)P(B) > 0, which means that the intersection of A and B has a nonzero probability. But if A and B are mutually exclusive, then P(A ∩ B) = 0, which contradicts our previous result. Therefore, it is impossible for two events with nonzero probabilities to be both independent and mutually exclusive. Question 4 What is the probability that Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively, in a drawing if 50 people enter a contest and a) no one can win more than one prize. b) winning more than one prize is allowed. Answer Definitions Definition permutation (importance of order): No repetition allowed: P(n,r) = n!/(n-r)! Repetition allowed = n^r Combination definition (order is not crucial): No repetition allowed: C(n,r) = n! / (r!(n-r)!) Repetition allowed: C(n+r - 1,r) = (n+r-1)! / (r!(n-1)!) with n! = n * (n-1) * ... * 2 * 1 Solution: Numerous methods for choosing the winners Since the first, second, and third prizes are different, the order of the winners is important and we must thus utilize the definition of a permutation. Out of the 50 participants, 4 winners must be chosen. n = 50 r = 4 Since repetition of winners is not allowed: P(50,4) = 50! / (50-4)! = 50! / 46! = 50 * 49 * 48 * 47 = 5,527,200 Probability Bo, Collen, Jeff, and Rohini will each receive the first, second, third, and fourth prizes in 1 of the 5,527,200 possible combinations. The probability is calculated by dividing the number of likely outcomes by the total number of outcomes. P(Bo, Colleen, Jeff and Rohini) = # of favorable outcomes / # of possible outcomes = 1 / 5,527,200 ≈ 0.0000001809 = 1.809 * 10^-7 b) Number of methods used to choose the winners We require the concept of a permutation since the order of the winners matters (because the first, second, and third prizes are different). Out of the 50 participants, 4 winners must be chosen. n = 50 r = 4 Since winners may be selected more than once: n^r = 50^4 = 6,250,000 Probability Bo, Collen, Jeff, and Rohini will each receive the first, second, third, and fourth prizes in 1 of the 6,250,000 possible combinations. The probability is calculated by dividing the number of likely outcomes by the total number of outcomes. P(Bo, Collen, Jeff and Rohini) = # of favorable outcomes / # of possible outcomes = 1/6,250,000 ≈ 0.00000016 = 1.6 * 10^-7 Result: a) 1/5,527,200 b) 1/6,250,000 Question 5 Consider a roulette wheel consisting of 38 numbers 1 through 36, 0, and double 0. If Smith always bets that the outcome will be one of the numbers 1 through 12, what is the probability that (a) Smith will lose his first 5 bets; (b) his first win will occur on his fourth bet? Answer a) Observe that the probability that he loses a single bet is simply 26/38 so, because of the independence, the probability that he loses all his five betting is (26/38)^5 ≈ 0.15 b) Define random variable Y that marks the first time that he wins a bet. From the nature of the problem we know that Y ~ Geom(12/38). So, the requred probability is simplify P(Y = 4) = (26/38)^3 * (12/38) ≈ 0.101 a) 0.15, b) 0.101