Probability Practice Exam Questions and Solutions, Exams of Quantitative Techniques

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SIG Quantitative Evaluation – Probability Practice Exams [2025]

If it is a good day (G) there are 60% chances tomorrow will be G and 40% chances tomorrow will be bad (B). If it is a B day, there 30% chances tomorrow will be G and 70% chances tomorrow will be B. If today is B, what is the expected number of days before seeing another B?- CORRECT ANSWER-E_{B|G} =0.4*1 + 0.6*(1 + E_{B|G}) which leads to E_{BIG} =2.5 E_{BIB} =0.7*1 + 0.3% +E_{BlG}) = 1+ 0.3*2.5 = 1.75 So the final answor is 1.75 days. You flip a weighted coin thal comes up H with probability 0.4 and T with probability 0.6. If you Acan finish a job in 100 min, B can finish the same job in 120 min. A and B work together on this Job, but after 40 min C comes to help them and they finish the job in additional 10 min. How long would it take to C to finish the job by himself?- CORRECT ANSWER-A and B working together would finish the job in 1/1/1090 + 1/120) min = 600/11 min. A, Band C working together (starting from zero) would finish the job in 1/(1/100 + 1/120 + 1/x) min = 1/(11/600 + 1/x) min Because they do nal have to start from zero, we can say that thoy have lo wark only for a fraction of the previous time. The fraction left to work is 1 - 404660/11). Hence we can write [1 - 40/(600/11) 1/100 + 1/120 + 1/x) min = 10 min, and solving for x gives 120 min. You roll 3 dice. If you get the same number, you earn 10. If vou get two numbers the same, you pected win?- CORRECT get oS. Ifthe numbers are all different, you lose 2 $. What is the e: ANSWER1.258 Total possible outcomes is 6°3 = 216. There are 6 cases in which the numbers are the same. There are 6C2 * 3! = 90 cases in which exactly two numbers are the same. There are 6*5*4 = 120 cases in which all numbers are different. Check: 6 + 90 + 120 = 216. Ok! So the expected win is (6*10$ + 90*5S - 120*28)/216 = 1.25 $ There are two racks, X and Y. X contains red (R) socks with probability 0.4 and black (B) socks with probability 0.6. Y contains R with probability 0.7 and B with probability 0.3. I pick a rack randomly. From that rack, ] pick 2R. What is the probability that those 2R come from rack X?- ER-246% chance Use total probability A and B play the game rocks (R), scissors (S), paper (P). You know that they never drew. Morcover, A played 3R, 1P and 6S, whercas B played 2R, 4P. How many times did A win? - Since the table is circular it doesn’t matter where the youngest sits. Fixing the youngest person in seat 1, there are 2 seatings that work - the oldest person in seat 2 and the oldest person in seat. 3. The total possible number of seatings is 4*3*2 = 24, so the probability is 2/24 = 1/12. The: sa 91% chance of seeing a shooting star in the next hour, what is the probability of seeing Pino star in half an hour)*2 = 0.09 Pino star in half an hour) = 0.3 P(star in half an hour} =1- 0.3 =0.7 = 70% Suppose you are given a cube, you break the cube in 3X3X3 pieces. However, before doing so, all 6 faces of the cube were colored green. All the other faces not colored green will be white after breaking it up. You randomly pick asmall cube and see one face which is green. What is the probability thal the cube is an edge (not corner!). For a 3x3x3 cube consisting of 27 smaller cubes, there will be 8 corner cubes, 12 edge cubes, 6 face cubes, and 1 center cube. All cubes except the center will have paint on it. Now apply Bayes’ law. Piedge | green side) = p(green side | edge) pfedge) / p(green side) p(green side | edge) = 2/6 pledge) = 12/27 * 1/27 + 1/5 * 6/27 + 2/6 * 12/27 + 3/6*8/27 = 1/3 (this could also be found by observing thal 6*9=34 sides aro painted, with a total of 27*6=3 p(green side) = 0/ 54 Lotal sides existing) Therefore, pledge | green side) = 12/27 how many combinations can you roll an 8 on a dice - CORR: