advertisement

BU7527 Example sheet — 1 Mike Peardon — mjp@maths.tcd.ie School of Mathematics, TCD Monday, 28th September Try to answer these questions before tomorrow’s lectures. We will go through solutions in class. 1. Two dice are rolled. What is the probability the biggest number showing on either of the dice was more than four given the smallest number was more than two? Deﬁne two events, A ≡ “biggest number > 4” and B ≡ “smallest number > 2”. Now the theorem of conditional probability tells us P (A|B) = 12/36 3 P (A ∩ B) = = P (B) 16/36 4 2. A group of friends, three men and six women are going out and call a taxi. A random selection of four friends get into the taxi. Find the probability the taxi contains a. Two men and two women. b. Four women. c. At least two men. The possible combinations are (M, W ) ∈ {(3, 1), (2, 2), (1, 3), (0, 4)} and their probabilities are 1 = 21 (3, 1): P (3, 1) = 1 C4 × 39 · 82 · 17 (2, 2): P (2, 2) = 2 C4 × ( 93 · 82 ) × ( 67 × 56 ) = (1, 3): P (1, 3) = 1 C4 × ( 93 ) × ( 67 × 56 × 54 ) = (0, 4): P (0, 4) = 0 C4 × ( 96 × 85 × 47 × 36 ) = 5 14 10 21 5 42 So the three probabilities are: a. Two men and two women has probability 5/6 b. Four women has probability 5/42 c. At least two men has probability 1/21 + 5/14 = 17/42 1 ı9em In a game, a player takes turns to roll a die and keeps a count of N , the number of turns taken. If the number showing is less than or equal to N , the game stops. Find P (N = n) at the end of the game for all possible values of n. n P (N = n) 1 2 3 4 5 6 1 6 5 6 5 6 5 6 5 6 5 6 = · 2 6 4 6 4 6 4 6 4 6 · · = 3 6 3 6 3 6 3 6 · · · = 4 6 2 6 2 6 = · · · · · · · · 5 6 1 6 = = 1 6 5 18 5 18 5 27 50 648 10 648 = 16.7% = 27.8% = 27.8% = 18.5% = 7.7% = 1.5% 3. Two fair dice are rolled. If we deﬁne two events A and B such that A is said to have occurred if the ﬁrst roll is a six and B is said to have occured if the sum of the two rolls is seven then are A and B independent? What if the sum of the rolls is ten? Deﬁne events A ≡“ﬁrst roll is 6” and B ≡“sum of rolls is 7” then we see P (A) = 1/6, P (B) = 1/6 and P (A ∩ B) = 1/36 so since P (A) × P (B) = 1/36 = P (A ∩ B) the events are independent. 4. Bayes theorem: The probability your friend ﬁnds a mathematics lecture interesting is 95% and mathematics lectures make up 10% of their modules this semester. The probability they ﬁnd a lecture for any other course interesting is 5%. You overhead your friend saying they have just enjoyed an interesting lecture. What is the probability they have just attended a mathematics lecture? Bayes theorem tells us P (A) P (B|A). P (B) Deﬁne A as the event “friend goes to maths lecture” and B is the event “friend goes to an interesting lecture”. So we have P (A) = 0.1. Then we also have P (B|A) = 0.95 and P (B|Ac ) = 0.05. We can compute the probability of attending an interesting lecture as P (A|B) = P (B) = P (B|A)P (A) + P (B|Ac )P (Ac ) = 0.95 × 0.1 + 0.05 × 0.9 = 0.14 Finally, using Bayes theorem gives P (A|B) = P (A) 0.1 P (B|A) = × 0.95 = 67.9%... P (B) 0.14 2 5. Binomial experiment: To improve proﬁts, the airline BinomialAir sells more tickets for each ﬂight than there are seats on its planes. Each plane has 20 seats and the airline always sells 24 tickets. On average, 19 passengers actually check in. What is the probability that on a particular ﬂight, more passengers check in than can be accommodated on the plane? Need to consider the probability either 24,23,22 or 21 passengers turn up. This is a . So binomial experiment, with p = 19 24 24! 24! 0! 24! P (23) = 23! 1! 24! P (22) = 22! 2! 24! P (21) = 21! 3! P (24) = p24 q 0 = 0.37% p23 q 1 = 2.32% p22 q 2 = 7.02% p21 q 3 = 13.5% The sum P (24) + P (23) + P (22) + P (21) is then 23.3%. 3