Easy | Probability |

A father claims about snowfall last night. First daughter tells that the probability of snowfall on a particular night is 1/8. Second daughter tells that 5 out of 6 times the father is lying! What is the probability that there actually was a snowfall?

Hint

Conditional Probability or Baye's Theorem

Solution

Let S = Snowfall occurred, and C = Claim

Probability of (Snowfall given Claim) = P(S | C) = P(C|S)*P(S)/P(C)

Now, P(C|S) = 1/6, P(S) = 1/8

P(C ) = P(true claim) + P(False Claim) = P(C|S)*P(S) + P( false claim|no snow)*P(no snow)

This is same as [1/6*1/8]/[ 1/6*1/8 + 7/8*5/6] = 1/36

Probability of (Snowfall given Claim) = P(S | C) = P(C|S)*P(S)/P(C)

Now, P(C|S) = 1/6, P(S) = 1/8

P(C ) = P(true claim) + P(False Claim) = P(C|S)*P(S) + P( false claim|no snow)*P(no snow)

This is same as [1/6*1/8]/[ 1/6*1/8 + 7/8*5/6] = 1/36

Source: Written Test

Enable Like and Comment Easy | General |

You are in a game against devil, on a perfectly round table and with an infinite pile of pennies. He says, "OK, we'll take turns putting one penny down, no overlapping allowed, and the pennies must rest flat on the table surface. The first guy who can't put a penny down loses." You can go first. How will you guarantee victory?

Hint

Try the problem if coins have unit radii, and table has radius 1 & 3.

Solution

You place a penny right in the center of the table. After that, whenever the devil places a penny on the table, mimic his placement by placing a penny diametrically opposite and at same distance from center. If he has a place to place a penny, so will you. The devil will run out of places to put a quarter before you do.

Easy | Probability |

Two witches make a nightly visit to an all-night coffee shop. Each arrives at a random time between 0:00 and 1:00. Each one of them stays for exactly 30 minutes. On any one given night, what is the probability that the witches will meet at the coffee shop?

Hint

Graph Shading

Answer

3/4

Solution

Plot the graph of X and Y, where X and Y denote the time each witch arrives. This will form a 60x60 square for total feasible region. The probability that they meet is |X-Y|<= 30. Draw this as favorable region, by joining line (30,0) with (60,30) and (0,30) with (30,60). Clearly the interior of this region has area 3/4 th of total. Hence the probability = 3/4 = 0.75

Easy | Strategy |

So there's this king. Someone breaks into his wine cellar where he stores 1000 bottles of wine. This person proceeds to poison one of the 1000 bottles, but gets away too quickly for the king's guard to see which one he poisoned or to catch him.

The king needs the remaining 999 safe bottles for his party in 4 weeks. The king has 10 prisoners who deserve execution. The poison takes about 3 weeks to take effect, and any amount of it will kill whoever drinks it. How can he figure out which bottle was poisoned in time for the party?

The king needs the remaining 999 safe bottles for his party in 4 weeks. The king has 10 prisoners who deserve execution. The poison takes about 3 weeks to take effect, and any amount of it will kill whoever drinks it. How can he figure out which bottle was poisoned in time for the party?

Hint

Convert to binary strings?

Solution

The king has to mix wine in order to isolate the single poisoned one. 2) There are 10 servants. After about 3 weeks, each one can be either dead or alive, meaning that there are 2^10 = 1024 possible outcomes. Since 1024 > 1000, it's actually possible for some scheme to work.

Here's the scheme: The king assigns each servant a number from 1-10. The king assigns each bottle a number from 0-999. When he labels them, though, he writes the number on the bottle in binary with ten digits, like this: 0: 000000000 1: 000000001 2: 000000010 3: 000000011 4: 000000100 5: 000000101 ... 999: 1111100111 and so on.

Now the strategy is simple: Pick bottle 1, write its binary form, look at the positions where '1' appears (here: 1st only). Make the corresponding prisoner drink small quantitiy of that wine. (ie prisoner 1 takes a sip of wine #1). Continue this process upto 1000th wine. After 3 weeks, suppose only prisoner number 4 & 7 die. This means the binary representation of poisoned wine has '1' at position 4 & 7, and rest all zeros. Convert this binary number to decimal and that gives the poisoned wine.

Here's the scheme: The king assigns each servant a number from 1-10. The king assigns each bottle a number from 0-999. When he labels them, though, he writes the number on the bottle in binary with ten digits, like this: 0: 000000000 1: 000000001 2: 000000010 3: 000000011 4: 000000100 5: 000000101 ... 999: 1111100111 and so on.

Now the strategy is simple: Pick bottle 1, write its binary form, look at the positions where '1' appears (here: 1st only). Make the corresponding prisoner drink small quantitiy of that wine. (ie prisoner 1 takes a sip of wine #1). Continue this process upto 1000th wine. After 3 weeks, suppose only prisoner number 4 & 7 die. This means the binary representation of poisoned wine has '1' at position 4 & 7, and rest all zeros. Convert this binary number to decimal and that gives the poisoned wine.

Source: Common

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