## Answers to Problems

Answers to these problems

999*1001 = 999,999

If a family has five children, the probability that at least three are girls is one half. The family must have at least 3 boys or at least 3 girls, but cannot have both. Probabilities are symmetric substituting boys for girls, so the probability is 50%.

You have two glass balls and are in a 100-story building. There’s a window on each floor. When you drop the ball out a window, it may or may not break, depending on how high you are. There’s a certain floor that is the transition from not breaking to breaking, so that if you drop it from that floor or above, it will break, and below that floor, it will not. This floor could be any of the floors from one to one hundred with equal probability. What is the fewest number of drops you need to make to be sure you accurately locate the transition floor?

14. Instead of the question asked, ask how many stories you can explore with $n$ drops. Drop the ball from story $n$ to begin. If it breaks, you need to drop from story 1, then 2, then 3 possibly up to $n-1$, requiring $n$ total drops. If it does not break, now drop it from $n + (n-1)$. This is the right way because if it breaks on the second drop, we’ll still need a maximum of $n$ total drops. Continuing, we see the highest building we can explore with $n$ drops is $n + (n-1) + (n-2) + ... + 1 = n(n+1)/2$. Set this equal to 100 and we get $n = 14$ (on rounding up).

Your friend puts two balls in a jar. Each ball is either red or green, and your friend chooses the color of each ball with a fair coin flip before putting it in. You come up, open the jar, and without looking can smell that there’s at least one red ball in it (but two red balls smell the same as one). What’s the probability that both balls are red?

There are three ways to explain the fact that you can smell a red ball. The first ball was red and the second green, the first was green and the second red, or both red. All are equally likely. That makes 1/3 chance both are red.

Same as last problem, but this time you reach in and pull out a red ball. What’s the probability that the remaining ball is also red?

Once you can smell are red ball, there are three explanations, as before. Between those three explanations, there are a total of 4 red balls. You pulled one out, and all those red balls were equally likely to be the one you grabbed. 2 of the 4 red balls come from the scenario where both balls are red, so there’s a 50% chance that the next ball pulled will be red.

Suppose that there are two barrels, each containing a number of plastic eggs. In both barrels, some eggs are painted blue and the rest are painted red. In the first barrel, 90% of the eggs contain pearls and 20% of the pearl eggs are painted blue. In the second barrel, 45% of the eggs contain pearls and 60% of the empty eggs are painted red. Would you rather have a blue pearl egg from the first or second barrel?

This was a trick question. They’re the same.

100 prisoners are to be executed, but they are given a chance to save themselves by playing a game. They will all stand in a single file line, so the prisoner in back can see all the other prisoners and the prisoner in front can see no one. The warden will then put a white or black hat on each prisoner’s head, choosing at random as he gets to the prisoner. Then the warden goes to the prisoner at the back of the line (who can see everyone else’s hat, but not his own) and asks him what color his hat is. He can respond only with either “white” or “black”. If he gets it right, he lives. This continues down the line. Each prisoner can hear the responses of all the prisoners who come before him. If the prisoners are allowed to get together and discuss strategy beforehand, how many of the 100 can be saved?

99, with a 50% chance to save the 100th. The first prisoner says “black” if he sees an even number of black hats on the other prisoners’ heads and “white” if he sees an odd number of black hats. He has a 50% chance to live. The next prisoner counts the hats in front and sees if he agrees with what the first prisoner says. If so, his hat is white, if not it’s black. He identifies his hat. The third prisoner does the same, including “seeing” the hat of the guy behind him, who he knows identified his hat color correctly. So on down the line.

There are eight pitchers of wine, one of which is poisoned. You have some lab rats to test the wine on. If a rat drinks any poison wine, it will die some time within the next 24 hours. How many rats do you need to use to design a test that is certain to discover the poisoned bottle in 24 hours?

Each rat is a bit, and distinguishing between eight states requires $log_2(8) = 3$ bits. For example, give the first rat pitchers 1,2,3,4; the second pitchers 1,2,5,6; the third 1,3,5,7.

Prove that there exist numbers x and y that are both irrational, but x^y is rational.

Recall (or prove) that $\sqrt{2}$ is irrational. Consider $\sqrt{2}^{\sqrt{2}}$. Is this rational? If so, we’re done. If it’s irrational, consider
$\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}$
Simplifying algebraically, this is
$\sqrt{2}^2 = 2$
which is rational. So if $\sqrt{2}^{\sqrt{2}}$ is irrational, then we’ve found two irrational numbers that give us a rational. Interestingly we don’t need to know whether $\sqrt{2}^{\sqrt{2}}$ is rational or not to complete the proof.

Suppose you cut a cone out of a sheet of paper. How does the time it takes the cone to fall to the floor when dropped from the ceiling depend on the radius of the cone?

For a reasonable cone made of reasonable paper dropped from a reasonable height, the cone will be at its terminal velocity for most of its fall, so we only need to consider the terminal velocity of the cone.

The weight of the cone scales with the area of the cone, but the wind resistance also scales with the area. The terminal velocity will then remain the same regardless of the radius of the cone. This actually works, btw.

Take a 6*6 chessboard and an 8*8 chessboard. For each, you’re allowed to make one cut through it along the lines between the squares. This will give you four pieces total. How can you make the cuts so those 4 pieces can be rearranged into a 10*10 chessboard? Try the same thing with other Pythagorean triples.

Draw the 10×10 board with the 8×8 superimposed in one corner and the 6×6 in the opposite corner. There’s a 4×4 overlap, which you can cut in half removing half from the 6×6 and half from the 8×8, and they fit in the empty corner of the 10×10.

I think this puzzle is impossible for Pythagorean triples besides multiples of 3-4-5.

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### One Response to “Answers to Problems”

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