Nuevo examen - 0 de Febrero de 0 [0 marks] 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. [ marks] (i) attempt to find P(red) P(red),, P(none green) = (= ) (ii) attempt to find P(red) P(green),, recognizing two ways to get one red, one green P(R) P(G), +, P(exactly one green) = 0 (= ) [ marks] Find the expected number of green marbles drawn from the jar. b. [ marks]
P(both green) = 0 (seen anywhere) correct substitution into formula for E(X) 0 0 + + + 0 0 0 expected number of green marbles is 0 (= ), () [ marks] Jar B contains six red marbles and two green marbles. A fair six-sided die is tossed. If the score is or, 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) = (= ) N (ii) P(red jar B) = (= ) N Given that the marble is red, find the probability that it was drawn from jar A. d. [ marks]
recognizing conditional probability P(A R), P(jar A and red) P(red), tree diagram attempt to multiply along either branch (may be seen on diagram) P(jar A and red) = (= ) attempt to multiply along other branch P(jar B and red) = (= ) adding the probabilities of two mutually exclusive paths P(red) = + correct substitution P(jar A red) = P(jar A red) = +, N () [ 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.
attempt to find p 0 0, 0 + 0 + x = 0 p = 0 attempt to find q 0 0, 00 0 0 q = 0 A girl is chosen at random. b. (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 minutes. [ marks] (i) 0 00 0 (= ) N (ii) valid approach 0 + 0, 00 0 0 00 (= ) [ marks] A girl is selected for the competition if she takes less than c. x minutes to complete the race. Given that 0% of the girls are not selected, (i) find the number of girls who are not selected; (ii) find x.
(i) attempt to find number of girls 0., 00 0 00 0 are not selected (ii) 0 are selected () x = 0 Girls who are not selected, but took less than d. 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.
(i) 0 given second chance N (ii) 0 took less than 0 minutes () attempt to find their selected total (may be seen in % calculation) 0 + 0 (= 0), 0+ their answer from (i) 0 ( %) N Bill and Andrea play two games of tennis. The probability that Bill wins the first game is. If Bill wins the first game, the probability that he wins the second game is. If Bill loses the first game, the probability that he wins the second game is. a. Copy and complete the following tree diagram. (Do not write on this page.) [ marks]
N Note: Award for each correct bold probability. [ marks] Find the probability that Bill wins the first game and Andrea wins the second game. b. multiplying along the branches (may be seen on diagram) 0 ( ) c. Find the probability that Bill wins at least one game.
c. METHOD multiplying along the branches (may be seen on diagram) adding their probabilities of three mutually exclusive paths correct simplification 0 0 0 METHOD N () recognizing Bill wins at least one is complement of Andrea wins finding P (Andrea wins ) P (Andrea wins both) = () evidence of complement,, N + +, + + +, + + 0 (= ) p, (R) Given that Bill wins at least one game, find the probability that he wins both games. d. [ marks] P (B wins both) evidence of recognizing conditional probability correct substitution (= ) [ marks] N (A) (R) P(A B), P (Bill wins both Bill wins at least one), tree diagram 0 (= ) Let A and B be independent events, where P(A) = 0. and P(B) = 0.. a. Find P(A B).
correct substitution 0. 0. P(A B) = 0. () b. Find P(A B). correct substitution () P(A B) = 0. + 0. 0. P(A B) = 0. c. On the following Venn diagram, shade the rion that represents A B. [ mark] N d. Find P(A B ). appropriate approach 0. 0., P(A) P( B ) P(A B ) = 0. (may be seen in Venn diagram)
Two events A and B are such that P(A) = 0. and P(A B) = 0.. a. Given that A and B are mutually exclusive, find P(B). correct approach () 0. = 0. + P(B), P(A B) = 0 P(B) = 0. b. Given that A and B are independent, find P(B). Correct expression for P(A B) (seen anywhere) P(A B) = 0.P(B), 0.x attempt to substitute into correct formula for P(A B) correct working P(A B) = 0. + P(B) P(A B), P(A B) = 0. + x 0.x () 0. = 0. + P(B) 0.P(B), 0.x = 0. P(B) = (= 0., exact) N Samantha goes to school five days a week. When it rains, the probability that she goes to school by bus is 0.. 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.. 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. 0., 0. correct working () 0. 0. + 0. 0., 0. + 0. 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 0. 0. 0. [ marks] P(R B) =, 0.9 (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), ( ) (R) (0.), (0.) ( 0.) P(X = ) = 0. d. After n school days, the probability that Samantha goes to school by bus at least once is greater than 0.9. Find the smallest value of n. [ marks]
METHOD evidence of using complement (seen anywhere) P (none), 0.9 valid approach correct inequality (accept equation) METHOD N () 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 recognising [ marks] N (R) P (none) > 0.9, P (none) < 0.0, P (none) = 0.9 (0.) n > 0.9, (0.) n = 0.0 n >.09 (accept n =.09) n = n =, P(X ) = 0.9, n =, P(X ) = 0.99 0.99 > 0.9 n = Adam travels to school by car ( ) or by bicycle ( ). 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. The probability of being late for school is C if he travels by bicycle. This information is represented by the following tree diagram. B p Find the value of. a. correct working () p =
Find the probability that Adam will travel by car and be late for school. b. multiplying along correct branches P(C L) = () Find the probability that Adam will be late for school. c. multiplying along the other branch adding probabilities of their mutually exclusive paths correct working + + P(L) = (= ) () N Given that Adam is late for school, find the probability that he travelled by car. d. [ marks] recognizing conditional probability (seen anywhere) P(C L) correct substitution of their values into formula () P(C L) = [ marks] Adam will go to school three times next week. e. Find the probability that Adam will be late exactly once.
valid approach correct substitution correct working Total [ marks] X B (, ), ( ) ( ), ( ) () (), three ways it could happen ( ) ( ), + + ( ) 9 9 9 ( ) ( ), + + 9 Let and C D P(C) = k P(D) = k 0 < k < 0. be independent events, with and, where. Write down an expression for a.. k P(C D) in terms of P(C D) = k k P(C D) = k () b. Find P( C D). [ marks] METHOD finding their P( C D) (seen anywhere) () 0. 0., 0. 0., 0.0 correct substitution into conditional probability formula () METHOD recognizing finding their P( C ) = P(C) (only if first line seen) () P( D) =, [ marks] C P( D) 0. P( C D) = 0. Total [ marks] C P( C D) = P( C ) k, 0. P( C D) = 0. ( k)( k ) k
At a large school, students are required to learn at least one language, Spanish or French. It is known that 9a. % of the students learn Spanish, and 0% learn French. Find the percentage of students who learn both Spanish and French. valid approach Venn diagram with intersection, union formula, P(S F) = 0. + 0.0 (accept % ) At a large school, students are required to learn at least one language, Spanish or French. It is known that 9b. % of the students learn Spanish, and 0% learn French. Find the percentage of students who learn Spanish, but not French. valid approach involving subtraction Venn diagram, 0 (accept 0% ) 9c. At a large school, students are required to learn at least one language, Spanish or French. It is known that [ marks] % of the students learn Spanish, and 0% learn French. At this school, % of the students are girls, and % 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.
(i) valid approach tree diagram, multiplying probabilities, P(S G) P(G) correct calculation 0. 0. () P(G S) = 0. (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., 0.9 0. G and S are not independent AG N0 [ marks] R At a large school, students are required to learn at least one language, Spanish or French. It is known that 9d. % of the students learn Spanish, and 0% learn French. At this school, % of the students are girls, and % of the girls learn Spanish. A boy is chosen at random. Find the probability that he learns Spanish. [ marks]
METHOD % are boys (seen anywhere) P(B) = 0. 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), 0.0 correct substitution 0. + 0.x = 0., 0.x = 0.0 correct manipulation P(S B) = 0.0 0. () () P(Spanish boy) = 0. 0. P(Spanish boy) = 0. [0., 0.] N [ marks] METHOD % are boys (seen anywhere) 0. 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 0. + 0.x = 0. correct manipulation 0.x = 0.0, P(S B) = 0.0 0. [ marks] () () P(Spanish boy) = 0. 0. P(Spanish boy) = 0. [0., 0.],
Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement. 0a. (i) Copy and complete the following tree diagram. [ marks] (ii) Find the probability that two white balls are chosen. (i), and (, and ) (ii) multiplying along the correct branches (may be seen on diagram) (= ) [ marks] N () Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement. 0b. 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. [ marks] A standard die is rolled. If or 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.
P(bagA) = (= ), P(bagB) = (= ) (seen anywhere) appropriate approach P(WW A) + P(WW B) ()() correct calculation + + [ marks] 0 P(W) = (= ), N Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement. 0c. 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. A standard die is rolled. If or 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. recognizing conditional probability P(A B), P(B) P(A WW) = P(WW A) P(WW) correct numerator () P(A WW) =, correct denominator, probability (= ) 0 N ()
Events A and B are such that P(A) = 0., P(B) = 0. and P(A B) = 0.. The values q, r, s and t represent probabilities. Write down the value of t. a. [ mark] t = 0. N [ mark] (i) Show that b. r = 0.. (ii) Write down the value of q and of s. [ marks] (i) correct values 0. + 0. 0., 0.9 0. r = 0. (ii) q = 0., s = 0. [ marks] AG N0 (i) Write down c. P( B ). (ii) Find P(A B ). [ marks] (i) 0. N (ii) P(A ) = A B [ marks]
A box contains six red marbles and two blue marbles. Anna selects a marble from the box. She replaces the marble and then selects a second marble. a. Write down the probability that the first marble Anna selects is red. [ mark] Note: In this question, method marks may be awarded for selecting without replacement, as noted in the examples. P(R) = (= ) [ mark] N Find the probability that Anna selects two red marbles. b. attempt to find P(Red) P(Red) P(R) P(R),, P(R) = (= ) 9 Find the probability that one marble is red and one marble is blue. c. [ marks]
METHOD attempt to find P(Red) P(Blue) P(R) P(B),, recognizing two ways to get one red, one blue P(RB) + P(BR), ( ), + [ marks] METHOD recognizing that P(R, B) is P(B) P(R) attempt to find P(R) and P(B) P(R) = ; P(B) = [ marks] P(R,B) = (= ),, P(R,B) = (= ) Let f(x) = + kx +, where x k Z. Find the values of k such that a. f(x) = 0 has two equal roots.
METHOD evidence of discriminant b ac, discriminant = 0 correct substitution into discriminant k, k = 0 k = ± METHOD N recognizing that equal roots means perfect square attempt to complete the square, ( + kx + ) x correct working (x + k) = k k = ±, N (R) Each value of k is equally likely for b. k. Find the probability that f(x) = 0 has no roots. evidence of appropriate approach b ac < 0 correct working for k < k <, k <, list all correct values of k p = A N
The diagram below shows the probabilities for events A and B, with P( A ) = p. Write down the value of p. a. [ mark] p = N [ mark] Find b. P(B). [ marks] multiplying along the branches 0, adding products of probabilities of two mutually exclusive paths + + 0 0 [ marks] P(B) = 0 (= ) 0, Find c. P( A B). [ marks]
appropriate approach which must include A (may be seen on diagram) P( A B) P(B) P(A B) P(B) ) (do not accept A 0 P( B) = P( B) = (= ) A () [ marks] Consider the events A and B, where P(A) = 0., P(B) = 0. and P(A B) = 0.. The Venn diagram below shows the events A and B, and the probabilities p, q and r. Write down the value of a. (i) p ; (ii) q ; (iii) r. [ marks] (i) p = 0. N (ii) q = 0. N (iii) r = 0. N [ marks] Find the value of b. P(A B ).
P(A B ) = A Note: Award for an unfinished answer such as. 0. 0. c. Hence, or otherwise, show that the events A and B are not independent. [ mark] valid reason 0., 0. 0. R thus, A and B are not independent AG N0 [ mark] José travels to school on a bus. On any day, the probability that José will miss the bus is. If he misses his bus, the probability that he will be late for school is. If he does not miss his bus, the probability that he will be late is. Let E be the event he misses his bus and F the event he is late for school. The information above is shown on the following tree diagram. Find a. (i) P(E F) ; (ii) P(F).
(i) N (ii) evidence of multiplying along the branches, adding probabilities of two mutually exclusive paths ( ) + ( ), ( ) + ( ) P(F) = Find the probability that b. (i) José misses his bus and is not late for school; (ii) José missed his bus, given that he is late for school. [ marks] (i) () (ii) recognizing this is P(E F) (= ) [ marks] A N c. The cost for each day that José catches the bus is euros. José goes to school on Monday and Tuesday. Copy and complete the probability distribution table. [ marks] A N [ marks]
The cost for each day that José catches the bus is euros. José goes to school on Monday and Tuesday. d. Find the expected cost for José for both days. correct substitution into E(X) formula 0 + + 9 9 + 9 9 E(X) = (euros) 9, In a class of 00 boys, boys play football and boys play rugby. Each boy must play at least one sport from football and rugby. (i) a. (ii) Find the number of boys who play both sports. Write down the number of boys who play only rugby. [ marks] (i) evidence of substituting into n(a B) = n(a) + n(b) n(a B) + 00, Venn diagram 0 (ii) N [ marks] b. One boy is selected at random. (i) Find the probability that he plays only one sport. (ii) Given that the boy selected plays only one sport, find the probability that he plays rugby.
(i) METHOD evidence of using complement, Venn diagram p, 00 0 0 00 (= ) METHOD attempt to find P(only one sport), Venn diagram + 00 0 00 (ii) 0 0 (= ) 0 9 (= ) 00 A Let A be the event that a boy plays football and B be the event that a boy plays rugby. c. Explain why A and B are not mutually exclusive. valid reason in words or symbols (R) P(A B) = 0 if mutually exclusive, P(A B) 0 if not mutually exclusive correct statement in words or symbols P(A B) = 0., P(A B) P(A) + P(B), P(A) + P(B) >, some students play both sports, sets intersect d. Show that A and B are not independent. [ marks] valid reason for independence P(A B) = P(A) P(B), P(B A) = P(B) correct substitution N 0 00 00 0 00 [ marks] 00, (R)
Jan plays a game where she tosses two fair six-sided dice. She wins a prize if the sum of her scores is. a. Jan tosses the two dice once. Find the probability that she wins a prize. [ marks] outcomes (seen anywhere, even in denominator) () valid approach of listing ways to get sum of, showing at least two pairs (, )(, ), (, )(, ), (, )(, ), (, )(, ), lattice diagram P(prize) = (= ) 9 N [ marks] b. Jan tosses the two dice times. Find the probability that she wins prizes. recognizing binomial probability B(, ), binomial pdf, 9 ( ) ( ) 9 ( 9 ) P( prizes) = 0.0 In a group of students, take art and take music. One student takes neither art nor music. The Venn diagram below shows the events art and music. The values p, q, r and s represent numbers of students. (i) 9a. Write down the value of s. (ii) Find the value of q. (iii) Write down the value of p and of r. [ marks]
(i) s = N (ii) evidence of appropriate approach, + q = q = (iii) p =, r = [ marks] 9b. (i) (ii) A student is selected at random. Given that the student takes music, write down the probability the student takes art. Hence, show that taking music and taking art are not independent events. (i) P(art music) = (ii) METHOD P(art) = (= ) A evidence of correct reasoning R the events are not independent AG N0 METHOD P(art) P(music) = 9 (= ) evidence of correct reasoning R the events are not independent AG N0 9c. Two students are selected at random, one after the other. Find the probability that the first student takes only music and the second student takes only art.
P(first takes only music) = P(second takes only art) = evidence of valid approach P(music and art) = 0 (= ) 0 (seen anywhere) (seen anywhere) A company uses two machines, A and B, to make boxes. Machine A makes 0% of the boxes. 0% of the boxes made by machine A pass inspection. 90% of the boxes made by machine B pass inspection. A box is selected at random. 0a. Find the probability that it passes inspection. [ marks] evidence of valid approach involving A and B P(A pass) + P(B pass), tree diagram correct expression () P(pass) = 0. 0. + 0. 0.9 P(pass) = 0. [ marks] 0b. The company would like the probability that a box passes inspection to be 0.. Find the percentage of boxes that should be made by machine B to achieve this.
evidence of recognizing complement (seen anywhere) P(B) = x, P(A) = x, P(B), 00 x, x + y = evidence of valid approach 0.( x) + 0.9x, 0.x + 0.9y correct expression 0. = 0.( x) + 0.9x, 0. 0. + 0.9 0. = 0., 0.x + 0.9y = 0. 0% from B The Venn diagram below shows events A and B where P(A) = 0., P(A B) = 0. and P(A B) = 0.. The values m, n, p and q are probabilities. (i) Write down the value of n. a. (ii) Find the value of m, of p, and of q. (i) n = 0. N (ii) m = 0., p = 0., q = 0. N Find b. P( B ).
appropriate approach P( B ) = P(B), m + q, (n + p) P( B ) = 0. Two fair -sided dice, one red and one green, are thrown. For each die, the faces are labelled,,,. The score for each die is the number which lands face down. a. List the pairs of scores that give a sum of. [ marks] three correct pairs N (, ), (, ), (, ), RG, RG, RG [ marks] The probability distribution for the sum of the scores on the two dice is shown below. b. [ marks] Find the value of p, of q, and of r. p =, q =, r = [ marks] N Fred plays a game. He throws two fair -sided dice four times. He wins a prize if the sum is on three or more throws. c. Find the probability that Fred wins a prize. [ marks]
let X be the number of times the sum of the dice is evidence of valid approach X B(n, p), tree diagram, sets of outcomes produce a win one correct parameter n =, p = 0., q = 0. Fred wins prize is P(X ) () () appropriate approach to find probability M complement, summing probabilities, using a CDF function correct substitution () 0.99,, 0.0 + 0.0090, + probability of winning = 0.00 ( ) N [ marks] Let A and B be independent events, where P(A) = 0. and P(B) = x. Write down an expression for a. P(A B). [ mark] P(A B) = P(A) P(B)(= 0.x) [ mark] N b. Given that P(A B) = 0., (i) find x ; (ii) find P(A B).
(i) evidence of using P(A B) = P(A) + P(B) P(A)P(B) correct substitution 0. = 0. + x 0.x, 0. = 0.x x = 0. (ii) P(A B) = 0. N c. Hence, explain why A and B are not mutually exclusive. [ mark] valid reason, with reference to P(A B) R N P(A B) 0 [ mark] There are 0 students in a classroom. Each student plays only one sport. The table below gives their sport and gender. a. One student is selected at random. (i) Calculate the probability that the student is a male or is a tennis player. (ii) Given that the student selected is female, calculate the probability that the student does not play football. (i) correct calculation 9 + 0 0 +++ 0 (ii) correct calculation, 0 P(male or tennis) = 0 0 + 0, () () P(not football female) = Two students are selected at random. Calculate the probability that neither student plays football. b. [ marks]
b. P(first not football) =, 0 P(second not football) = 0 9 P(neither football) = 0 P(neither football) = 0 0 [ marks] 0 9 N Consider the independent events A and B. Given that. P(B) = P(A), and P(A B) = 0., find P(B). [ marks]
METHOD for independence P(A B) = P(A) P(B) expression for P(A B), indicating P(B) = P(A) () P(A) P(A), x x (R) substituting into P(A B) = P(A) + P(B) P(A B) correct substitution 0. = x + x x, 0. = P(A) + P(A) P(A)P(A) correct solutions to the equation 0.,. (accept the single answer 0.) P(B) = 0. [ marks] METHOD N for independence P(A B) = P(A) P(B) expression for P(A B), indicating P(A) = P(B) () P(B) P(B), x x (R) (A) substituting into P(A B) = P(A) + P(B) P(A B) correct substitution 0. = 0.x + x 0.x, 0. = 0.P(B) + P(B) 0.P(B)P(B) correct solutions to the equation (A) 0.,. (accept the single answer 0.) P(B) = 0. (accept x = 0. if x set up as P(B) ) N [ marks]
The letters of the word PROBABILITY are written on cards as shown below. Two cards are drawn at random without replacement. Let A be the event the first card drawn is the letter A. Let B be the event the second card drawn is the letter B. Find a. P(A). [ mark] P(A) = [ mark] N Find b. P(B A). P(B A) = 0 A Find c. P(A B). [ marks] recognising that P(A B) = P(A) P(B A) correct values P(A B) = [ marks] P(A B) = 0 () 0 N
Two standard six-sided dice are tossed. A diagram representing the sample space is shown below. Let X be the sum of the scores on the two dice. a. (i) Find P(X = ). (ii) Find P(X > ). (iii) Find P(X = X > ). [ marks] (i) number of ways of getting X = is P(X = ) = (ii) number of ways of getting X > is P(X > ) = (= ) (iii) P(X = X > ) = (= ) [ marks] A b. Elena plays a game where she tosses two dice. If the sum is, she wins points. If the sum is greater than, she wins point. If the sum is less than, she loses k points. Find the value of k for which the game is fair. [ marks]
attempt to find P(X < ) M fair game if E(W) = 0 (may be seen anywhere) attempt to substitute into E(X) formula M 0 ( ) + ( ) k( ) correct substitution into E(W) = 0 0 ( ) + ( ) k( ) = 0 work towards solving + 0k = 0 [ marks] P(X < ) = 0 = 0k k = (=.) 0 M N R A four-sided die has three blue faces and one red face. The die is rolled. Let B be the event a blue face lands down, and R be the event a red face lands down. a. Write down (i) P(B); (ii) P(R). (i) P(B) = N (ii) P(R) = N
If the blue face lands down, the die is not rolled again. If the red face lands down, the die is rolled once again. This is b. represented by the following tree diagram, where p, s, t are probabilities. Find the value of p, of s and of t. p = N s =, t = N c. Guiseppi plays a game where he rolls the die. If a blue face lands down, he scores and is finished. If the red face lands down, he scores and rolls one more time. Let X be the total score obtained. (i) Show that P(X = ) =. (ii) Find P(X = ). [ marks] (i) P(X = ) = P (getting and ) = = AG N0 (ii) P(X = ) = + (or ) = [ marks] () (i) Construct a probability distribution table for X. d. (ii) Calculate the expected value of X. [ marks]
(i) A (ii) evidence of using E(X) = xp(x = x) E(X) = ( ) + ( ) () = (= ) [ marks] If the total score is, Guiseppi wins e. $0. If the total score is, Guiseppi gets nothing. Guiseppi plays the game twice. Find the probability that he wins exactly $0. win $0 scores one time, other time P() P() = (seen anywhere) evidence of recognising there are different ways of winning $0 P() P() + P() P(), ( ), + + + 9 P(win $0) = (= ) N International Baccalaureate Organization 0 International Baccalaureate - Baccalauréat International - Bachillerato Internacional Printed for Colio Aleman de Barranquilla