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SIG ER Associate test ACTUAL UPDATED QUESTIONS AND CORRECT ANSWERS

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SIG ER Associate test ACTUAL UPDATED QUESTIONS AND CORRECT ANSWERS You have a 6-face die and a 10-face die. What is the expected value of the sum of the two? - CORRECT ANSWERSE[X1 + X2] = E[X1] + E[X2] = 3.5 + 5.5 = 9 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 ANSWERSE_{B|G} = 0.4*1 + 0.6*(1 + E_{B|G}) which leads to E_{B|G} = 2.5. E_{B|B} = 0.7*1 + 0.3*(1 + E_{B|G}) = 1 + 0.3*2.5 = 1.75 You flip a weighted coin that comes up H with probability 0.4 and T with probability 0.6. If you flip the coin 5 times, what is the probability that you see at least 3 tails? - CORRECT ANSWERSP = (5 choose 3) * 0.6^3 * 0.4^2 + (5 choose 4) * 0.6^4 * 0.4^1 + (5 choose 5) * 0.6^5 = 0.68256 ~= 0.68 A can 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 ANSWERSA and B working together would finish the job in 1/(1/100 + 1/120) min = 600/11 min. A, B and 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 not have to start from zero, we can say that they have to work only for a fraction of the previous time. The fraction left to work is 1 - 40/(600/11). 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 you get two numbers the same, you get 5 $. If the numbers are all different, you lose 2 $. What is the expected win? - CORRECT ANSWERSTotal possible outcomes is 6^3 = 216. There are 6 cases in which the numbers are the same. There are 6C2 * 3! = 90 cases in which exactly two numbers are the same. There are 6*5*4 = 120 cases in which all numbers are different. Check: 6 + 90 + 120 = 216. Ok! So the expected win is (6*10$ + 90*5$ - 120*2$)/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

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SIG ER Associate test ACTUAL
UPDATED QUESTIONS AND CORRECT
ANSWERS
You have a 6-face die and a 10-face die. What is the expected value of the sum of the two? -
CORRECT ANSWERS✅✅E[X1 + X2] = E[X1] + E[X2] = 3.5 + 5.5 = 9


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 ANSWERS✅✅E_{B|G} = 0.4*1 + 0.6*(1 + E_{B|G}) which leads to
E_{B|G} = 2.5.
E_{B|B} = 0.7*1 + 0.3*(1 + E_{B|G}) = 1 + 0.3*2.5 = 1.75


You flip a weighted coin that comes up H with probability 0.4 and T with probability 0.6. If
you flip the coin 5 times, what is the probability that you see at least 3 tails? - CORRECT
ANSWERS✅✅P = (5 choose 3) * 0.6^3 * 0.4^2 + (5 choose 4) * 0.6^4 * 0.4^1 + (5 choose
5) * 0.6^5 = 0.68256 ~= 0.68


A can 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
ANSWERS✅✅A and B working together would finish the job in 1/(1/100 + 1/120) min =
600/11 min. A, B and 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 not have to start from
zero, we can say that they have to work only for a fraction of the previous time. The fraction
left to work is 1 - 40/(600/11). 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 you get two numbers the same,
you get 5 $. If the numbers are all different, you lose 2 $. What is the expected win? -
CORRECT ANSWERS✅✅Total possible outcomes is 6^3 = 216. There are 6 cases in
which the numbers are the same. There are 6C2 * 3! = 90 cases in which exactly two
numbers are the same. There are 6*5*4 = 120 cases in which all numbers are different.
Check: 6 + 90 + 120 = 216. Ok! So the expected win is (6*10$ + 90*5$ - 120*2$)/216 = 1.25
$

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