Medium | Probability |

A line of 100 airline passengers is waiting to board a plane. They each hold a ticket to one of the 100 seats on that flight. For convenience, let's say that the nth passenger in line has a ticket for the seat number 'n'. Being drunk, the first person in line picks a random seat (equally likely for each seat). All of the other passengers are sober, and will go to their proper seats unless it is already occupied; If it is occupied, they will then find a free seat to sit in, at random.

What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)?

What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)?

Hint

Can last passenger arrive at any other seat than 1 or 100?

Answer

1/2

Solution

Notice that the last passenger can only take seat #1 or #100. If any passenger takes seat #1, the cycle stops, and all the subsequent passengers take their own seats (including last). Otherwise, if #100 seat is taken before #1, the cycle is paused, i.e., the subsequent passengers do take their own seats, but the last passenger would take seat #1. Now for any passenger from 1st to 99th, who is picking random vacancy, will choose between #1, #100 or any other seat equally likely. Thus, by symmetry, #1 or #100, any one will be taken first - with equal probability. Hence last person ends up at his seat with probability 0.5

Medium | Probability |

A stick is broken into 3 parts, by choosing 2 points randomly along its length. With what probability can it form a triangle?

Hint

All three broken parts must satisfy the triangle inequality. Or rather, each of the broken part must be less than half of stick's length.

Answer

1/4

Solution

All 3 sides have to have lengths less than half the length of the stick. the conditions are min{ x.y}<= 0.5; max{x,y}>=0.5; |x-y|<=0.5 . looking at the unit square, and dividing into 8 congruent triangles by lines parallel to the axes and y=x line, its easy to see 2 of the 8 triangles satisfy the condition. so the answer is 1/4

Source: Quant Interview

Enable Like and Comment Medium | Probability |

In this gambling game, a player can buy a ticket for Rs 1 on any number from 1 to 6. Three identical and unfair dice are rolled. If the booked number appears on 0, 1, 2 or 3 dice, player wins Rs 0, 1, 2 or 3 respectively, without returning the original Rs 1. What is expected money you can win after buying a ticket for Rs 1?

Hint

Book the full house

Answer

1/2

Solution

We bet Rs 1 on each number 1-6. In any case, we get Rs 3 back. That is 1/2 per ticket. Hence the expected amount of money we can win is 1/2. A tedious way to arrive at this answer is to calculate the probability of getting 1, 2 or 3 faces common to our booking.

Source: 50 puzzles in prob

Enable Like and Comment Medium | Probability |

A and B are in a team called AB, playing against C. If AB team wins it gets Rs 3, nothing otherwise.

Game is: A and B are placed in 2 separate rooms far away. A will toss a coin and B will also toss a coin; A will have to guess outcome of B's toss and B will guess A's. If both guesses are right, team AB wins Rs 3, nothing otherwise.

Should they play the game, by giving Rs 1 in start to C.

Game is: A and B are placed in 2 separate rooms far away. A will toss a coin and B will also toss a coin; A will have to guess outcome of B's toss and B will guess A's. If both guesses are right, team AB wins Rs 3, nothing otherwise.

Should they play the game, by giving Rs 1 in start to C.

Hint

Winning probability in not 1/4. They can make strategy before game.

Answer

1/2

Solution

They will have same coin with probability 1/2. They can speak their own coin's face as the guess of other's. They win game with probability 1/2. Pay off will be positive, and hence they should play!

Source: Top Quant Interview

Enable Like and Comment Medium | Probability |

Snow-particles are falling on the ground one after another. A particular snowflake turns out to be of type "Stellar Dendrite" with probability 'p' if its previous particle was also Stellar Dendrite, and with probability 'q' if previous one was something else. If a snowflake is picked from ground, what is the probability that it is Stellar Dendrite?

PS:Although no two snowflakes are alike, yet there are various crystalline structures to categorize their interesting shapes. The image depicts the most popular shape, called Stellar Dendrites, which means star-like particles with tree-like branches.

PS:Although no two snowflakes are alike, yet there are various crystalline structures to categorize their interesting shapes. The image depicts the most popular shape, called Stellar Dendrites, which means star-like particles with tree-like branches.

Hint

Need to form a recursive equation of conditional probability

Answer

probability is q/(1-p+q)

Solution

Solution by Palak:

Let x be the probability that a snowflake picked from ground is Stellar Dendrite. Thus, when a new snowflake is falling, with prob=x the last snowflake was Stellar Dendrite => prob the new falling snowflake is Stellar Dendrite = x*p + (1-x)*q. But, for the composition of the snowflakes on the ground to remain constant, xp+(1-x)q should be =x => x=1/(1+(1-p)/q)

This is a kind of steady state analysis.

Let x be the probability that a snowflake picked from ground is Stellar Dendrite. Thus, when a new snowflake is falling, with prob=x the last snowflake was Stellar Dendrite => prob the new falling snowflake is Stellar Dendrite = x*p + (1-x)*q. But, for the composition of the snowflakes on the ground to remain constant, xp+(1-x)q should be =x => x=1/(1+(1-p)/q)

This is a kind of steady state analysis.

Source: Self

Enable Like and Comment Latest solved Puzzles

Color Switches Weird Sequences Intersecting Pillars Consecutive sums Scaling a Square Difficulty Level

© BRAINSTELLAR |