Three fair coins are tossed in the air and they land with heads up. Can you calculate the chances that when they are tossed again, two coins will again land with heads up?
50%
The vents in this case are independent. If you form the possible outcomes, they will be as follows
HHH
HTT
HHT
HTH
TTT
TTH
THT
THH
Now among these outcomes, four will give the required results. Thus the probability is 50%.
There is a dressing drawer which contains the following colored socks in pairs: Purple, Magenta, Crimson, White, Yellow and Turquoise. Now, the socks are paired and each pair is together in the matching set. There is no light in the room and you open the drawer and pick up a pair. Then, without noticing any color, you keep them back and pick up again.
Can you calculate the probability that the pair of socks was Yellow both the times?
Whenever two events are independent, we calculate the probability of both occurring as:
P(A and B) = P(A) * P(B)
P(Yellow) = 1/6
P(Yellow and Yellow) = P(Yellow) * P(Yellow)
= 1/6*1/6
= 1/36
You are stuck with a gangster who likes to play it rough. The only way to survive is to accept his invitation to play Russian Roulette.
He presents a revolver in which, three bullets have been placed consecutively. Now he spins the chamber cylinder of the gun. The cylinder wont be spun again. The gun will be passed between both of you till the gun fires and one of you is dead.
Will you prefer to go first or second if you are given a choice ?
2/3
Let us label the chambers for our convenience as C1, C2... C6.
Now, when the cylinder is spun, there can be the following six outcomes.
1. If C1 is fired first: Player 1 dies.
2. If C2 is fired first: Player 1 dies.
3. If C3 is fired first: Player 1 dies.
4. If C4 is fired first: Player 2 dies (First shot, player 1, C4 empty. Second shot player 2, C5, empty. Third shot player 1, C6 empty. Fourth shot player 2, C1 not empty.)
5. C5 is fired first: Player 1 dies (First shot, player 1, C5 empty. Second shot player 2, C6, empty. Third shot player 1, C1 not empty.)
6. C6 is fired first: Player 2 dies (First shot, player 1, C6 empty. Second shot, player 2, C1, not empty)
Therefore, if you choose to go second, you have a 4/6 or 2/3 chance of winning.
Two men play a dice game involving roll of two standard dice. Man X says that a 12 will be rolled first. Man Y says that two consecutive 7s will be rolled first. The men keep rolling until one of them wins.
What is the probability that X will win ?
36/71
Can you calculate the probability of getting a sum of six when a dice is thrown twice?
Event = {(1,5) , (2,4), (3,3), (4,2), (5,1)}
Therefore, P (getting sum = 6) = 5/36
Chances of Mr. Button winning the local lottery in 8%. All participants lined up and Mr. Button is 3rd in the row. The first two participants lose the lottery.
What is the chance of Samuel Mr. Button now?
8%
The winning chance probability is still 8% as the outcome of Mr.Bitton winning the lottery is a separate event from rest losing the lottery.
What is the probability of rolling a sum of 7 with two fair six-sided dice?
0.1667 or 16.67%
When rolling two fair six-sided dice, there are a total of 36 possible outcomes (6 outcomes for the first die multiplied by 6 outcomes for the second die).
To calculate the probability of rolling a sum of 7, we need to determine the number of combinations that result in a sum of 7. These combinations are:
1 + 6
2 + 5
3 + 4
4 + 3
5 + 2
6 + 1
So, there are 6 favorable outcomes that result in a sum of 7.
The probability of rolling a sum of 7 is calculated by dividing the number of favorable outcomes by the total possible outcomes:
Probability = (Number of Favorable Outcomes) / (Total Possible Outcomes) = 6 / 36 = 1/6 ≈ 0.1667
Therefore, the probability of rolling a sum of 7 with two fair six-sided dice is approximately 0.1667 or 16.67%.
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!
A 52% bias toss for head using the 51% tail bias coin was done to obtain a fair result.
Can you find how bias is the floor in this case?
First let us assume that all other condition are fair here.
Now the toss will generate a 52:48 distribution in the favour of heads. Therefore, the toss bias factor for heads is 52/48.
In the same manner, the coin will be generating 49:51 distribution in favour of tails which makes the coin bias factor for heads in this case to be 49/51.
So, we have a combined bias factor of (52 * 49) / (51 * 48) = 2548 / 2448 which will be cancelled by a 2448 / 2548 floor factor.
The floor will be generating a distribution of 2448 / (2548 + 2448) : 2548 / (2548 + 2448) in the favour of tails which amounts to 51.00080064% tails approximately.
A fresh card pile is taken out of a box (the pile has 54 cards including 2 jokers). One joker is taken out and then the cards are shuffled for a good amount of times. After shuffling, two piles are made by dividing that one pile.
What is the possibility that one of the piles will have a card sequence from A to K in order?
For x number of cards, the probability to obtain a precise sequence will be 1/x!
Here, we want 13 cards in sequence, which is why, x = 13
Therefore the possibility of obtaining a sequence from A to K in order = 1/6227020800