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 | 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

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

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