### #1 - Monday Probability Problem

What is the probability of getting five Monday in a 31-days month ?

If a 31day month starts on a Saturday, Sunday or Monday it will have five Monday, if not it will have 4 Monday.

Probability is 3/7.

What is the probability of getting five Monday in a 31-days month ?

Probability is 3/7.

kukki and fukki are a married couple (dont ask me who he is and who she is) :)

They have two kids, one of them is a girl. Assume safely that the probability of each gender is 1/2.

What is the probability that the other kid is also a girl?

Hint: It is not 1/2 as you would first think.

1/3

This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.

The following are possible combinations of two children that form a sample space in any earthly family:

Girl - Girl

Boy - Girl

Boy - Boy

Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample space.

This leaves only three possibilities, one of which is two girls. Hence the probability is 1/3

I throw two dice simultaneously.

What is the probability of getting sum as 9 of the two numbers shown ?

1/12

Possible cases = 6*6 = 36

Desired cases = [(3,6), (4,5), (6,3), (5,4)] = 4

=> 4/36 = 1/12

What is the probability of choosing the correct answer at random from the options below.

a) 1/4

b) 1/2

c) 1

d) 1/4

If the answer is 1/4, then because 2 out of 4 answer choices are '1/4', the answer must actually be 1/2. This is a contradiction. So the answer cannot be 1/4.

If the answer is 1/2 (or 1), then because 1/2 (or 1) is 1 out of 4 answer choices, the answer must be 1/4. This is again a contradiction. So the answer cannot be 1/2 (nor 1).

So none of the provided answer choices are correct. Therefore the probability of choosing the correct answer is 0%

The host of a game show, offers the guest a choice of three doors. Behind one is a expensive car, but behind the other two are goats.

After you have chosen one door, he reveals one of the other two doors behind which is a goat (he wouldn't reveal a car).

Now he gives you the chance to switch to the other unrevealed door or stay at your initial choice. You will then get what is behind that door.

You cannot hear the goats from behind the doors, or in any way know which door has the prize.

Should you stay, or switch, or doesn't it matter?

You better switch!

Your first choice has a 1/3 chance of having the car, and that does not change.

The other two doors HAD a combined chance of 2/3, but now a Goat has ben revealed behind one, all the 2/3 chance is with the other door.

Henry has been caught stealing cattle, and is brought into town for justice. The judge is his ex-wife Gretchen, who wants to show him some sympathy, but the law clearly calls for two shots to be taken at Henry from close range. To make things a little better for Henry, Gretchen tells him she will place two bullets into a six-chambered revolver in successive order. She will spin the chamber, close it, and take one shot. If Henry is still alive, she will then either take another shot, or spin the chamber again before shooting.

Henry is a bit incredulous that his own ex-wife would carry out the punishment, and a bit sad that she was always such a rule follower. He steels himself as Gretchen loads the chambers, spins the revolver, and pulls the trigger. Whew! It was blank. Then Gretchen asks, 'Do you want me to pull the trigger again, or should I spin the chamber a second time before pulling the trigger?'

What should Henry choose?

Henry should have Gretchen pull the trigger again without spinning.

We know that the first chamber Gretchen fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Henry has Gretchen pull the trigger again, the probability that a bullet will be fired is 1/4.

If Gretchen spins the chamber again, the probability that she shoots Henry would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position.

Bruna was first to arrive at a 100 seat theater.

She forgot her seat number and picks a random seat for herself.

After this, every single person who get to the theater sits on his seat if its available else chooses any available seat at random.

Neymar is last to enter the theater and 99 seats were occupied.

Whats the probability what Neymar gets to sit in his own seat ?

1/2

one of two is the possibility

1. If any of the first 99 people sit in neymar seat, neymar will not get to sit in his own seat.

2. If any of the first 99 people sit in Bruna's seat, neymar will get to sit in his seat.

Mr. Black, Mr. Gray, and Mr. White are fighting in a truel. They each get a gun and take turns shooting at each other until only one person is left. Mr. Black, who hits his shot 1/3 of the time, gets to shoot first. Mr. Gray, who hits his shot 2/3 of the time, gets to shoot next, assuming he is still alive. Mr. White, who hits his shot all the time, shoots next, assuming he is also alive. The cycle repeats. If you are Mr. Black, where should you shoot first for the highest chance of survival?

He should shoot at the ground.

If Mr. Black shoots the ground, it is Mr. Gray's turn. Mr. Gray would rather shoot at Mr. White than Mr. Black, because he is better. If Mr. Gray kills Mr. White, it is just Mr. Black and Mr. Gray left, giving Mr. Black a fair chance of winning. If Mr. Gray does not kill Mr. White, it is Mr. White's turn. He would rather shoot at Mr. Gray and will definitely kill him. Even though it is now Mr. Black against Mr. White, Mr. Black has a better chance of winning than before.

How many people must be gathered together in a room, before you can be certain that there is a greater than 50/50 chance that at least two of them have the same birthday?

Only twenty-three people need be in the room, a surprisingly small number. The probability that there will not be two matching birthdays is then, ignoring leap years, 365x364x363x...x343/365 over 23 which is approximately 0.493. this is less than half, and therefore the probability that a pair occurs is greater than 50-50. With as few as fourteen people in the room the chances are better than 50-50 that a pair will have birthdays on the same day or on consecutive days.

Three people enter a room and have a green or blue hat placed on their head. They cannot see their own hat, but can see the other hats.

The color of each hat is purely random. They could all be green, or blue, or any combination of green and blue.

They need to guess their own hat color by writing it on a piece of paper, or they can write 'pass'.

They cannot communicate with each other in any way once the game starts. But they can have a strategy meeting before the game.

If at least one of them guesses correctly they win $50,000 each, but if anyone guess incorrectly they all get nothing.

What is the best strategy?

Simple strategy: Elect one person to be the guesser, the other two pass. The guesser chooses randomly 'green' or 'blue'. This gives them a 50% chance of winning.

Better strategy: If you see two blue or two green hats, then write down the opposite color, otherwise write down 'pass'.

It works like this ('-' means 'pass'):

Hats: GGG, Guess: BBB, Result: Lose

Hats: GGB, Guess: --B, Result: Win

Hats: GBG, Guess: -B-, Result: Win

Hats: GBB, Guess: G--, Result: Win

Hats: BGG, Guess: B--, Result: Win

Hats: BGB, Guess: -G-, Result: Win

Hats: BBG, Guess: --G, Result: Win

Hats: BBB, Guess: GGG, Result: Lose

Result: 75% chance of winning!