Easy | Probability |

There is a regular die and a special invisible die. You know that regular die has integers 1 to 6, but don't know what's on the invisible dice. After tossing, I speak the sum of outcome of both die. It so happens that the outcome is an interger between 1 to 12, with equal probability (1/12 each). Can you guess what are the numbers printed on special invisible dice?

Hint

For the sum to be 1, we need atleast 0, and for sum to be 12, what do we need? Can you guess the digits of invisible dice now?

Answer

0 0 0 6 6 6

Solution

Faces of cube have numbers: 0 0 0 6 6 6 . If we compute conditional probability, we get 0 with probabilty (1/2) and 6 with probability (1/2). Hence the sum is 1,2....12 with probability of (1/12) each

Source: Technical Interview

Enable Like and Comment Easy | Probability |

Sheldon says "Suppose I have two children. Younger one is a girl". What is the probability that both children are girls?

"Forget all that, and suppose I have two children, and atleast one of them is a boy". Find probability of two boys.

"Suppose I have two kids, lets call them Bouba and Kiki", says Dr. Cooper, "and suppose Bouba is a girl !" What is the probability that I have two daughters?"

"Forget all that, and suppose I have two children, and atleast one of them is a boy". Find probability of two boys.

"Suppose I have two kids, lets call them Bouba and Kiki", says Dr. Cooper, "and suppose Bouba is a girl !" What is the probability that I have two daughters?"

Answer

1/2, 1/3, 1/3

Solution

First and second are un-ambiguous. Third is debatable as per the interpretation. I believed 1/3, as the names Cooper used were anonymous, and should have given no extra information. But it turns out that Cooper did provide some extra information. He had actually fixed a person (A). Also, calculating the probability in part 3 using conditionals will give 1/2.

Source: Written Test

Enable Like and Comment 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

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