Please add good ones you’ve heard of…I need to practice.

okay, try this one. This is the best one i have come across. There are 12 identical looking balls. And one ball in this lot does not have the same weight as the rest of them. That is, 11 of these balls are exactly the same weight and 1 of them is either heavier or lighter than the rest. You have a balance. You need to identify the problem ball by using the balance ONLY 3 times. How will you do it?

^ I have this particular one down pat since I’ve encountered it before. It is an interesting one, but I don’t want to redo a problem I’ve encountered.

There are three dudes and sublimitys mom in the room, what is probability that exactly two of them will get on said mom at the same time

what is the probability that you will get a sequence of least 6 consecutive heads or tails after flipping a fair coing 200 times

Mobius Striptease Wrote: ------------------------------------------------------- > what is the probability that you will get a > sequence of least 6 consecutive heads or tails > after flipping a fair coing 200 times Hmmm, let’s take the base case of “6”. There are are 200-5 slots for this to occur, each with a probability of (1/2)^6. So for: k = 6, (200-5)/2^6 Since you introduced the term “at least”, that means we will have to consider everything from k = 6 to k = 200. For k = 7, (200-6)/2^7 … For general k: define f(k) = (200 - (k -1))/(2^k) Add these up, we have P = SUM(k = 6, 7, …, 200) f(k)

rus1bus Wrote: ------------------------------------------------------- > okay, try this one. This is the best one i have > come across. > > There are 12 identical looking balls. And one ball > in this lot does not have the same weight as the > rest of them. That is, 11 of these balls are > exactly the same weight and 1 of them is either > heavier or lighter than the rest. > > You have a balance. You need to identify the > problem ball by using the balance ONLY 3 times. > How will you do it? take 6 and balance 3vs3. if they balance, move on to the next set of 3vs3. one side has to weigh more than the other, so you’ve narrowed it down to 3 balls. pick two and balance them. if they balance, the third is the heavy one. is there a different way?

6 vs. 6 discard ligher. take the heavier side and go 3 vs 3 discard ligher. take the heavier side and go heavier side 1 vs 1 either you get the heavier ball, or you have the one that’s left out as the heavier one.

You can also start out 5 vs 5

You are assuming the faulty ball is heavy, so you are discarding balls on the lower side of balance. That is not appropriate. Dont assume the faulty ball to be heavy. It could be lighter than the rest too. Faulty ball in the problem is given as either heavier or lighter than the rest. But whether it is heavy or light is not given.

^You can start out 5 vs 5: 5 vs 5 - if balance then 1 vs 1 = 2 Steps 5 vs 5 - if not balance then 5 balls from heavier side and do 2 vs 2, if balance then remaining ball = 2 steps 5 vs 5 - if not balance then 5 balls from heavier side and do 2 vs 2 as before, but if not balance, then 1 vs 1 = 3 steps. Oh, just read it could be lighter than the rest… tricky.

An oldy but goody. The monte hall three doors problem. Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to switch your pick door No. 2?” Is it to your advantage to switch your choice?

Sanka’s Mom Wrote: ------------------------------------------------------- > An oldy but goody. You are leaving yourself open to a MILF Hunter comment, lol. Anyways, I know this problem so will let others do it.

depends is the goat hot?

sublimity Wrote: ------------------------------------------------------- > … > For general k: define f(k) = (200 - (k -1))/(2^k) > > Add these up, we have P = SUM(k = 6, 7, …, 200) > f(k) in general, when you produce a formula and claim it represents the probability of something, it should be less than 1

^ ? that is greater than 1?

Just looking at it but if K=6 then it would be 195/64? If I am looking at that correctly

i meant to divide everything by 1/(2^200), ooops

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i had something like that on a recent interview, but it was pr(getting more than 60 heads) from 100 flips of a fair coin. Remember that this coin flipping is a binomial distribution with mean = n*p and variance = n*p*(1-p). You can apply the central limit theorem to assume that the binomial method converges to the normal distribution for such a large number of trials… then mean = 50 (n=100, p = 1/2), variance = 100*0.0*0.5 = 25, so stdev = sqrt(variance) = 5. Let x = number of heads, then x = 60 is mean + 2 standard deviations, Pr(x>= 60 ) = 1 - pr(x<=60) = 1 - F(x<=60)… in the end it came out to .025