Note: Again, don't listen to the idiot across the table, he can screw you up in your thought process. In this one he actually screwed me up because he said, "keep the water pouring." And that is the key: don't keep the water pouring.
Pour water from the 5-gallon bucket into the 3-gallon bucket. This results in 2 gallons of water being left in the 5-gallon bucket. Pour away the water in the 3-gallon bucket and transfer the 2 gallons of water into the 3-gallon bucket. Now the 3-gallon bucket has room for 1 more gallon of water. Fill up the 5-gallon bucket again and pour water into the 2/3 full 3-gallon bucket until it fills up. 4 gallons of water remains in the 5-gallon bucket!
Solve it in two minutes. Note: Don't listen to the idiot across the table, he can screw you up in your thought process.
Take 6 of the 8 balls and put 3 on each side of the scale. If the heavy ball isn't in the group of 6, you know it's one of the remaining 2 and so you put those two in the scale and determine which one. If the heavy ball is in the 6, you have narrowed it down to 3. Of those 3, pick any 2 and put them on the scale. If the heavy ball is in that group of 2, you know which one it is. If both balls are of equal weight, then the heavy ball is the one you sat to the side.
I toss the coin n = 2 times, and got two heads, then E[n] = 2, since the result was simply HH I toss the coin n = 3 times, and got at least two heads, then E[n] = 1/4 [2 + 2 + 2 + 3] = 2.25 since the potential outcomes would be: THH HTH HHT HHH in this case, tossing 3 times produced 3 heads Now to generalize this... ...
Show More Responses
Assume the expected number of times you have to toss to get 2 heads is E_n If the first time you flip, you get one tail, then the exp time you get two heads is: 1/2 * (1+ E_n) If the first time you flip, you get one head, if you get another head the 2nd time, you're done, if you get a tail for the 2nd flip, you start all over again: 1/2 * 1/2 * 2 + 1/2 * 1/2 * (E_n+2) By conditional probability: E_n = 1/2 * (1+ E_n) + 1/4 *(2 + E_n + 2) -> E_n = 6 So you have to flip coin 6 times on average
Assuming his initial q meant below: what is the expected number of flips needed to get two heads in a row. ans) 6 How? By recursive problem construction. Let E = expected #of flips to get HH 1) if we get T => we need to start all over again. need to flip +1 to get H. 2) if we get HT => we need to start all over again. need to flip at least 2 more flips to achieve HH 3) if we get HH => we are done. we used 2 flips to get HH idea: (for having T, HT, HH) E = 0.5 * (E + 1) + (0.5 * 0.5) * (E + 2) + (0.5 * 0.5) * 2 if we solve above recursion, the answer is 6.
7 Race 5 groups of 5. Race the leaders. Let the horses be A1 .. A5 B1 .. B5 C1 .. C5 D1 .. D5 E1 .. E5 where X1 ... X5 was the order the horses came in, in the "first round" races, and where A1 > B1 > C1 > D1 > E1 in the leaders race. Lastly, race A2 A3 B1 B2 C1. Top two in that race, along with A1, are the top 3.
Why not just six races? Race each horse once, 25 horses, 5 per race -> 5 races. Then take the 5 horses with the best time independent of the race they were in.
Problem statement only really refers to ranking, ordering, not obtaining a measure of time for each run.
I am programmed to complete, meaning I am not satisfied until the job is completed
Perhaps you could find a way to perform the given task better or potentially even automate it. This would turn the "menial task" into a skill builder, a way to impress your supervisor, and work more efficiently