Math SL Day 66 Probability Practice [196 marks]

Transcription

1 Math SL Day 66 Probability Practice [96 marks] Events A and B are independent with P(A B) = 0.2 and P(A B) = 0.6. a. Find P(B). valid interpretation (may be seen on a Venn diagram) P(A B) + P(A B), P(B) = 0.8 b. Find P(A B). valid attempt to find P(A) correct working for P(A) correct working for P(A B) P(A B) = P(A) P(B), 0.8 A = , () N () , P(A B) = 0.85 The following Venn diagram shows the events A and B, where P(A) = 0., P(A B) = 0.8 and P(A B) = 0.. The values p and q are probabilities. 2a. (i) Write down the value of q. (ii) Find the value of p. [ marks]

2 (i) q = 0. N (ii) appropriate approach P(A) q, p = 0. [ marks] 2b. Find P(B). [ marks] valid approach P(A B) = P(A) + P(B) P(A B), P(A B) + P(B A ) correct values () 0.8 = 0. + P(B) 0., P(B) = 0.5 [ marks] In a class of 2 students, 2 own a laptop, 0 own a tablet, and own neither. The following Venn diagram shows the events own a laptop and own a tablet. The values p, q, r and s represent numbers of students. a. (i) Write down the value of p. (ii) Find the value of q. (iii) Write down the value of r and of s. [5 marks] (i) p = N (ii) valid approach ( ) 2, 22 8 q = (iii) r = 8, s = 6

3 A student is selected at random from the class. b. (i) (ii) Write down the probability that this student owns a laptop. Find the probability that this student owns a laptop or a tablet but not both. (i) (= ) A2 (ii) valid approach 8 + 6, r + s 2 2 (= ) Two students are randomly selected from the class. Let L be the event a student owns a laptop. c. (i) Copy and complete the following tree diagram. (Do not write on this page.) (ii) Write down the probability that the second student owns a laptop given that the first owns a laptop. (i) N (ii) 20 N Let C and D be independent events, with P(C) = 2k and P(D) = k 2, where 0 < k < 0.5. Write down an expression for P(C D) in terms of a. k.

4 P(C D) = 2k k 2 () P(C D) = 6k b. Find P(C D). [ marks] METHOD finding their P(C D) (seen anywhere) () , ,0.08 correct substitution into conditional probability formula () METHOD 2 recognizing P(C D) = P(C ) finding their P(C ) = P(C) (only if first line seen) () P(C P(C D) = D), 0.27 P(C D) = 0. 2k, 0.6 P(C D) = 0. [ marks] Total [7 marks] ( 2k)(k 2 ) k2 Ann and Bob play a game where they each have an eight-sided die. Ann s die has three green faces and five red faces; Bob s die has four green faces and four red faces. They take turns rolling their own die and note what colour faces up. The first player to roll green wins. Ann rolls first. Part of a tree diagram of the game is shown below. 5a. Find the probability that Ann wins on her first roll. recognizing Ann rolls green 8 P(G) Find the probability that Ann wins the game.

5 5b. recognize the probability is an infinite sum [7 marks] Ann wins on her st roll or 2 nd roll or rd roll, S recognizing GP u = 8 r = 20 6 (seen anywhere) (seen anywhere) correct substitution into infinite sum of GP , ( ), 5 ( ) correct working () 8 6 8, 6 P (Ann wins) = (= ) [7 marks] Total [5 marks] N Adam travels to school by car ( C) or by bicycle ( B). On any particular day he is equally likely to travel by car or by bicycle. The probability of being late ( L) for school is if he travels by car. 6 The probability of being late for school is if he travels by bicycle. This information is represented by the following tree diagram. 6a. Find the value of p. correct working () 6 p = 5 6 Find the probability that Adam will travel by car and be late for school. 6b.

6 multiplying along correct branches () 2 6 P(C L) = 2 6c. Find the probability that Adam will be late for school. multiplying along the other branch adding probabilities of their 2 mutually exclusive paths correct working () P(L) = (= ) N 6d. Given that Adam is late for school, find the probability that he travelled by car. [ marks] recognizing conditional probability (seen anywhere) P(C L) correct substitution of their values into formula () 2 2 P(C L) = [ marks] Adam will go to school three times next week. 6e. Find the probability that Adam will be late exactly once.

7 valid approach X B (, ), ( ) ( ) 2, (, three ways it could happen ) correct substitution () ( ) ( ) ( ) 2, + + correct working () ( ) ( ), Total [5 marks] The following cumulative frequency graph shows the monthly income, I dollars, of 2000 families. 7a. Find the median monthly income.

8 recognizing that the median is at half the total frequency m = 2500 (dollars) 7b. (i) Write down the number of families who have a monthly income of 2000 dollars or less. (ii) Find the number of families who have a monthly income of more than 000 dollars. (i) 500 families have a monthly income less than 2000 N (ii) 850 correct cumulative frequency, () subtracting their cumulative frequency from families have a monthly income of more than 000 dollars Note: If working shown, award M for = 50, using the table. 7c. The 2000 families live in two different types of housing. The following table gives information about the number of families living in each type of housing and their monthly income I. Find the value of p. correct calculation () 2000 ( ), () p = 585 7d. A family is chosen at random. (i) Find the probability that this family lives in an apartment. (ii) Find the probability that this family lives in an apartment, given that its monthly income is greater than 000 dollars.

9 (i) correct working () 0.65 (exact) (ii) correct working/probability for number of families () , 0.65 [0.6, 0.65] ,, (= ), 0.87 [0.86, 0.87] 7e. Estimate the mean monthly income for families living in a villa. evidence of using correct mid-interval values ( 500,000,500) () attempt to substitute into fx f p [0, 20] (dollars) [ marks] Total [5 marks] 8. Celeste wishes to hire a taxicab from a company which has a large number of taxicabs. The taxicabs are randomly assigned by the company. The probability that a taxicab is yellow is 0.. The probability that a taxicab is a Fiat is 0.. The probability that a taxicab is yellow or a Fiat is 0.6. Find the probability that the taxicab hired by Celeste is not a yellow Fiat. [6 marks]

10 recognize need for intersection of Y and F (R) P(Y F), valid approach to find P(Y F) P(Y ) + P(F) P(Y F), Venn diagram correct working (may be seen in Venn diagram) () P(Y F) = 0. recognize need for complement of Y F P(Y F), 0. P ((Y F) ) = 0.9 N [6 marks] Bill and Andrea play two games of tennis. The probability that Bill wins the first game is. 5 If Bill wins the first game, the probability that he wins the second game is 5. 6 If Bill loses the first game, the probability that he wins the second game is 2. 9a. Copy and complete the following tree diagram. (Do not write on this page.) [ marks]

11 N Note: Award for each correct bold probability. [ marks] 9b. Find the probability that Bill wins the first game and Andrea wins the second game. multiplying along the branches (may be seen on diagram) ( ) 0 5 9c. Find the probability that Bill wins at least one game.

12 METHOD multiplying along the branches (may be seen on diagram) 5 2,, adding their probabilities of three mutually exclusive paths , correct simplification () , (= ) N 0 5 METHOD 2 recognizing Bill wins at least one is complement of Andrea wins 2 (R) finding P (Andrea wins 2) P (Andrea wins both) = () 5 evidence of complement p, 5 N 5 9d. Given that Bill wins at least one game, find the probability that he wins both games. [5 marks] P (B wins both) 5 evidence of recognizing conditional probability correct substitution 5 6 (= ) [5 marks] 2 N (A2) (R) P(A B ), P (Bill wins both Bill wins at least one), tree diagram (= ) Let A and B be independent events, where P(A) = 0. and P(B) = a. Find P(A B).

13 correct substitution P(A B) = 0.8 () 0b. Find P(A B). correct substitution () P(A B) = P(A B) = c. On the following Venn diagram, shade the rion that represents A B. [ mark] N 0d. Find P(A B ). appropriate approach , P(A) P(B ) P(A B ) = 0.2 (may be seen in Venn diagram)

14 Samantha goes to school five days a week. When it rains, the probability that she goes to school by bus is 0.5. When it does not rain, the probability that she goes to school by bus is 0.. The probability that it rains on any given day is 0.2. a. On a randomly selected school day, find the probability that Samantha goes to school by bus. appropriate approach P(R B) + P(R B), tree diagram, one correct multiplication () , 0.2 correct working () , P(bus) = 0.(exact) N b. Given that Samantha went to school by bus on Monday, find the probability that it was raining. [ marks] recognizing conditional probability P(A B) = P(A B) P(B) correct working P(R B) =, [ marks] (R) c. In a randomly chosen school week, find the probability that Samantha goes to school by bus on exactly three days. recognizing binomial probability X B(n, p), ( 5 ) (R) (0.), (0.) ( 0.) 2 P(X = ) = 0.7 d. After n school days, the probability that Samantha goes to school by bus at least once is greater than Find the smallest value of n. [5 marks]

15 METHOD evidence of using complement (seen anywhere) P (none), 0.95 valid approach P (none) > 0.95, P (none) < 0.05, P (none) = 0.95 correct inequality (accept equation) (0.66) n > 0.95, (0.66) n = 0.05 n > (accept n = 7.209) () n = 8 N METHOD 2 valid approach using guess and check/trial and error finding P(X ) for various values of n seeing the cross over values for the probabilities n = 7, P(X ) = 0.95, n = 8, P(X ) = 0.99 recognising > 0.95 (R) n = 8 N [5 marks] Two events A and B are such that P(A) = 0.2 and P(A B) = a. Given that A and B are mutually exclusive, find P(B). correct approach () 0.5 = P(B), P(A B) = 0 P(B) = 0. 2b. Given that A and B are independent, find P(B).

16 Correct expression for P(A B) (seen anywhere) P(A B) = 0.2P(B), 0.2x attempt to substitute into correct formula for P(A B) P(A B) = P(B) P(A B), P(A B) = x 0.2x correct working () 0.5 = P(B) 0.2P(B), 0.8x = 0. P(B) = (= 0.75, exact) 8 N Jar A contains three red marbles and five green marbles. Two marbles are drawn from the jar, one after the other, without replacement. a. Find the probability that (i) none of the marbles are green; (ii) exactly one marble is green. [5 marks] (i) attempt to find P(red) P(red) 2, 8 7, P(none green) = 6 56 (= ) 28 (ii) attempt to find P(red) P(green) 5, 8 7 5, recognizing two ways to get one red, one green 2P(R) P(G), 5 5 +, P(exactly one green) = 0 (= ) [5 marks]

17 b. Find the expected number of green marbles drawn from the jar. [ marks] P(both green) = (seen anywhere) correct substitution into formula for E(X) expected number of green marbles is (= 5 ), () [ marks] Jar B contains six red marbles and two green marbles. A fair six-sided die is tossed. If the score is or 2, a marble is drawn from jar A. Otherwise, a marble is drawn from jar B. c. (i) Write down the probability that the marble is drawn from jar B. (ii) Given that the marble was drawn from jar B, write down the probability that it is red. (i) P(jar B) = 6 (= 2 ) N (ii) P(red jar B) = 6 8 (= ) N d. Given that the marble is red, find the probability that it was drawn from jar A. [6 marks]

18 recognizing conditional probability P(A R), P(jar A and red), tree diagram P(red) attempt to multiply along either branch (may be seen on diagram) P(jar A and red) = 8 (= ) 8 attempt to multiply along other branch P(jar B and red) = (= ) 2 adding the probabilities of two mutually exclusive paths 2 6 P(red) = correct substitution P(jar A red) = P(jar A red) = , N () [6 marks] A running club organizes a race to select girls to represent the club in a competition. The times taken by the group of girls to complete the race are shown in the table below. a. Find the value of p and of. q

19 attempt to find p 20 70, x = 20 p = 50 attempt to find q 80 20, q = 60 b. A girl is chosen at random. (i) Find the probability that the time she takes is less than minutes. (ii) Find the probability that the time she takes is at least 26 minutes. [ marks] (i) (= ) 20 N (ii) valid approach , (= ) 5 [ marks] c. A girl is selected for the competition if she takes less than minutes to complete the race. x Given that 0% of the girls are not selected, (i) find the number of girls who are not selected; (ii) find x.

20 (i) attempt to find number of girls 0., are not selected (ii) 20 are selected () x = 20 d. Girls who are not selected, but took less than 25 minutes to complete the race, are allowed another chance to be selected. The new times taken by these girls are shown in the cumulative frequency diagram below. (i) (ii) Write down the number of girls who were allowed another chance. Find the percentage of the whole group who were selected.

21 (i) 0 given second chance N (ii) 20 took less than 20 minutes () attempt to find their selected total (may be seen in % calculation) (= 0), 20+ their answer from (i) 70 ( %) N A Ferris wheel with diameter 22 metres rotates clockwise at a constant speed. The wheel completes 2. rotations every hour. The bottom of the wheel is metres above the ground. A seat starts at the bottom of the wheel. 5a. Find the maximum height above the ground of the seat. valid approach + diameter, + 22 maximum height = 5 (m)

22 After t minutes, the height h metres above the ground of the seat is given by h = 7 + a cos bt. 5b. (i) Show that the period of h is 25 minutes. (ii) Write down the exact value of b. (i) period = period = 25 minutes AG N0 (ii) b = 2π 25 (= 0.08π) N 5c. Find the value of. a [ marks] METHOD valid approach max 7, a = 5, 2 7 a = 6 (accept a = 6 ) () a = 6 METHOD 2 attempt to substitute valid point into equation for h 2π = 7 + a cos( ) 25 correct equation () 5 = 7 + a cos(π), = 7 + a a = 6 [ marks] 5d. Sketch the graph of h, for. 0 t 50

23 N Note: Award for approximately correct domain, for approximately correct range, for approximately correct sinusoidal shape with 2 cycles. Only if this last awarded, award for max/min in approximately correct positions. 5e. In one rotation of the wheel, find the probability that a randomly selected seat is at least metres above the ground. 05 [5 marks] setting up inequality (accept equation) h > 05, 05 = 7 + a cos bt, sketch of graph with line y = 05 any two correct values for t (seen anywhere) t = 8.7, t = 6.628, t =.7, t =.628 valid approach , t t , 2( ) 25 p = 0.0 [5 marks], M 6a. At a large school, students are required to learn at least one language, Spanish or French. It is known that 75% of the students learn Spanish, and learn French. 0% Find the percentage of students who learn both Spanish and French.

24 valid approach Venn diagram with intersection, union formula, P(S F) = (accept 5% ) 6b. At a large school, students are required to learn at least one language, Spanish or French. It is known that 75% of the students learn Spanish, and learn French. 0% Find the percentage of students who learn Spanish, but not French. valid approach involving subtraction Venn diagram, (accept 60% ) 6c. At a large school, students are required to learn at least one language, Spanish or French. It is known that 75% of the students learn Spanish, and learn French. 0% At this school, 52% of the students are girls, and 85% of the girls learn Spanish. A student is chosen at random. Let G be the event that the student is a girl, and let S be the event that the student learns Spanish. (i) Find P(G S). (ii) Show that G and S are not independent. [5 marks]

25 (i) valid approach tree diagram, multiplying probabilities, P(S G) P(G) correct calculation () P(G S) = 0.2 (exact) N (ii) valid reasoning, with words, symbols or numbers (seen anywhere) P(G) P(S) P(G S), P(S G) P(S), not equal, one correct value P(G) P(S) = 0.9, P(S G) = 0.85, G and S are not independent AG N0 [5 marks] R 6d. At a large school, students are required to learn at least one language, Spanish or French. It is known that 75% of the students learn Spanish, and learn French. 0% At this school, 52% of the students are girls, and 85% of the girls learn Spanish. A boy is chosen at random. Find the probability that he learns Spanish. [6 marks]

26 METHOD 8% are boys (seen anywhere) P(B) = 0.8 appropriate approach P(girl and Spanish) + P(boy and Spanish) = P(Spanish) correct approach to find P(boy and Spanish) P(B S)= P(S) P(G S), P(B S)= P(S B) P(B), 0.08 correct substitution () x = 0.75, 0.8x = 0.08 correct manipulation P(S B) = () P(Spanish boy) = P(Spanish boy) = 0.62 [0.6, 0.62] N [6 marks] METHOD 2 8% are boys (seen anywhere) 0.8 used in tree diagram appropriate approach tree diagram, correctly labelled branches on tree diagram () () first branches are boy/girl, second branches are Spanish/not Spanish correct substitution x = 0.75 correct manipulation 0.8x = 0.08, P(S B) = [6 marks] () () P(Spanish boy) = P(Spanish boy) = 0.62 [0.6, 0.62],

27 Events A and B are such that P(A) = 0., P(B) = 0.6 and P(A B) = 0.7. The values q, r, s and t represent probabilities. 7a. Write down the value of t. [ mark] t = 0. N [ mark] 7b. (i) Show that r = 0.2. (ii) Write down the value of q and of s. [ marks] (i) correct values , r = 0.2 AG (ii) q = 0., s = 0. [ marks] N0 7c. (i) Write down P(B ). (ii) Find P(A B ). [ marks] (i) 0. N (ii) P(A B ) = A2 [ marks]

28 8a. Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement. (i) Copy and complete the following tree diagram. [5 marks] (ii) Find the probability that two white balls are chosen. (i) , and (, and ) N (ii) multiplying along the correct branches (may be seen on diagram) (= ) () [5 marks] 8b. Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement. [5 marks] Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, the probability that they are both white is 2. 7 A standard die is rolled. If or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bag B. Find the probability that the two balls are white.

29 2 P(bagA) = (= ), 6 2 P(bagB) = (= ) (seen anywhere) 6 ()() appropriate approach P(WW A) + P(WW B) correct calculation [5 marks] , 5 2 P(2W) = (= ) N 8c. Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement. Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, the probability that they are both white is 2. 7 A standard die is rolled. If or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bag B. Given that both balls are white, find the probability that they were chosen from bag A.

30 recognizing conditional probability P(A B), P(B) P(A WW) = P(WW A) P(WW) correct numerator correct denominator probability 8 (= ) 20 5 N () 6 2 P(A WW) =, , () International Baccalaureate Organization 208 International Baccalaureate - Baccalauréat International - Bachillerato Internacional Printed for North Hills Preparatory