Math 10B: Worksheet 4 Solutions

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1 Math 10B: Worksheet 4 Solutions February In a superlottery, a player selects numbers out of the first 100 positive integers. What is the probability that a person wins the grand prize by picking numbers that are among the 18 numbers selected at random by a computer? Let A be the event that the chosen numbers are out of the 18 that will fetch you the prize. We then have P(A) = A Ω = ( 18 ( 100 ) ) Suppose that 100 people enter a contest and that different winners are selected at random for first, second, and third prizes. What is the probability that at least one of Phoebe, Chandler, and Monica wins a prize if each has entered the contest? Let W be the event that at least one of Phoebe, Chandler and Monica wins a prize. Note then that W = Ω W c = P (100, P (97, so that P(W ) = W Ω = P (100, P (97, P (100, = In a group of international referees, 5 are from Europe, 4 from Asia and 3 from North America. To officiate at a tournament, 3 referees are chosen at random from the group. Find the probability that (a) a referee is chosen from each continent, Observe first that there are a total of Ω = ( ) 3 = 220 outcomes. Out of these, only ( )( 1)( 1) = 60 outcomes have a referee from each continent so we have P(a referee is chosen from each continent) = =

2 (b) exactly two referees are chosen from Asia, Note that there are ( 4 2)( 8 1) = 48 ways of achieving this so we get P(exactly two referees are chosen from Asia) = = 55. (c) the three referees are chosen from the same continent. Observe that there are ( 5 + ( 4 + ( 3 = = 15 ways of accomplishing this so we have P(the three referees are chosen from the same continent) = = What is the probability of these events when we randomly select a permutation of the 26 lowercase letters of the English alphabet? (a) z is the first letter of the permutation. Observe first that there are Ω = 26! possible permutations. Out of these, only 25! have z as the first letter (place z up front and permute the remaining 25) so we have P(z is the first letter of the permutation) = 25! 26! = (b) z precedes a in the permutation. Note that for any permutation in which z precedes a, we have another permutation in which a precedes z (just switch z and a). Thus, precisely half the permutations have z before a so P(z precedes a) = 1 2. (c) a immediately precedes m, which immediately precedes z in the permutation. For such permutations, we consider amz as one joint block; we therefore need to permute = 24 objects, of which there are 24! ways so that P(amz is a block) = 24! 26! = (d) m, n, and o are in their original places in the permutation. Leaving m, n, and o are in their original places, we can shuffle the remaining 23 characters in the 23 places so we get 23! valid permutations. We therefore have P(m, n, and o are in their original places) = 23! 26! =

3 5. In a large group of people, it is known that 20% have a hot breakfast, 40% have a hot lunch and 50% have a hot breakfast or a hot lunch (or both). Find the probability that a person chosen at random from this group (a) has a hot breakfast and a hot lunch, Let B and L be the events that a randomly chosen person has a hot breakfast and a hot lunch respectively. We then have P(B L) = P(B) + P(L) P(B L) = = 0.1. (b) has a hot lunch, given that the person chose a hot breakfast. P(L B) = P(L B) P(B) = = In a group of 100 people, 40 own a dog, 25 own a cat and 15 own a dog and a cat. Find the probability that a randomly chosen person does not own a cat, given that she owns a dog. Let C and D be the events that a randomly chosen person owns a cat and a dog respectively. We then have P(C c D) = 1 P(C D) = 1 P(C D) P(D) = 1 15/100 40/100 = A team needs to win at least two of its remaining three games to secure the championship. Experts assess that the probabilities of winning the games are 0.6, 0.7 and 0.8 respectively. Assuming that the results of the games are independent, what is the probability that the team will win the championship? Let W i be the event that the team wins game i, i = 1, 2, 3. We then have P(win at least two games) = P(win exactly two games) + P(win all three games) = (0.6)(0.7)(0.2) + (0.6)(0.(0.8) + (0.4)(0.7)(0.8) + (0.6)(0.7)(0.8) ( results are indpendent) = = Two events A and B are such that P(A) = 8, P(B) = 1 and P(A B) = 1. Find the probability that 3

4 (a) exactly one of the events occurs, P(exactly one of the events occurs) = P(A B) P(A B) = P(A) + P(B) 2P(A B) = ( ) 3 = = (b) neither event occurs. Note that P(A B) = P(A) + P(B) P(A B) = 15 = 4 5 so that P(A c B c ) = P((A B) c ) = = In a restaurant, 40% of the customers choose steak as their main course. If steak is indeed chosen, there is a 60% likelihood that it will be followed by ice-cream while if steak is not chosen, the likelihood of ice-cream goes down to 20%. Find the probability that a randomly chosen customer will choose (a) steak and ice-cream, Let S and I be the events that steak and ice-cream are chosen respectively. We then have P(S I) = P(I S)P(S) = (0.6)(0.4) = (b) ice-cream. P(I) = P(S I) + P(S c I) = P(I S)P(S) + P(I S c )P(S c ) = (0.2)(0.6) = Each day in March is classified as either rainy or fine. If a day is rainy, there is a 60% chance that the next day will also be rainy while if a day is fine, there is a 70% chance that the next day will be fine. Assume that there is a 30% chance that March 1 is rainy. (a) Find the probability that March 3 is fine. Let R i and F i be events that March i is rainy and fine respectively, i = 1, 2, 3. The given values are represented in Figure 1. 4

5 R 1 F F 2 R 2 R 2 F Figure 1 We then have P( ) = (0.(0.6)(0.6) + (0.(0.4)(0. + (0.7)(0.(0.6) + (0.7)(0.7)(0. = (b) Given that March 3 is rainy, find the probability that March 1 is fine. P(F 1 ) = P(F 1 ) P( ) (0.7)(0.(0.6) + (0.7)(0.7)(0. = =