Aptitude
This can be solved using Linear Programming.
Let
d1, d2, . . . dn is the denomination of coins.
c1, c2, . . . cn is the final count of coins
Mary and Tom park their car in an empty parking lot with n >= 2 consecutive parking spaces (i.e. n spaces in a row, where only one car fits in each space). Mary and Tom pick parking spaces at random; of course, they must each choose a different space. (All pairs of distinct parking spaces are equally likely.) What is the probability that there is at most one empty space between them?
Solution
1st car can be placed in n ways
2nd car can be placed in (n-1) ways
Therefore total possibilities are n(n-1)
For n = 3 the probability is always 1
For n = 4,
If fist car is parked in 1st or 4th slot, 2nd car can be placed in 2 ways to satisfy the condition = 1 (2) + 1 (2)
If fist car is parked in 2nd or 3rd slot (last but one), 2nd car can be placed in 3 ways to satisfy the condition = 1 (3) + 1 (3)
P = (4 + 6) / (4 * 3) = 10/12
Generalizing n >= 4 First and Last slot = 2 ways Last but one = 3 ways For every other slot (n - 4) * 4
P = (2 + 2 + 3 + 3 + (n-4)4) / (n(n-1)) = (4n - 6) / (n(n-1))
Four ants are in the corners of a square, one in each corner. Each have two possible direction in which they can walk. What is the probability that ants won't collide with each other?
Each ant have two possible actions, which leads to a total of 16 (4^2) possible actions. If all ants walk in same direction, they won't collide, which can happen in two ways (cw or ccw). So probability of not colliding is 2/16
- It is common that most of the ML experts are PhDs
- In industries ML experts hold various degrees
- You are meeting a Scholar with PhD.
What is the probability that the Scholar is ML expert? Use variables when necessary.
P(ML) Probability of a person being ML Expert
P(PhD) Probability of a person having PhD
According to 1st statement P(PhD | ML) is close to one
Need to Compute P(ML | PhD)
Using bayes theorem P(ML | PhD) = P (PhD | ML) * P(ML) / P(PhD)