Easy | Probability |

We have a weighted coin which shows a Head with probability p, (0.5<p<1). How do we get a fair toss from this? That is, how do we toss this coin in such a way that we can have probability of winning = loosing = 50%?

Hint

Clearly we cannot have a fair toss in a single flip of this coin. But by tossing this coin 2 times, we can assign the outputs to "win" or "loose", such that we have a 50% chance for both.

Answer

Toss 2 times, mapping HT to "win", TH to "loose" and repeat the process otherwise.

Solution

Toss the coin twice. If consecutive Heads-Tails appears (HT), we "win". In case of (TH), we "loose". If (TT) or (HH) appears, repeat the process. Probability of infinite repetition (p*p*.....) + (1-p)*(1-p)*.... = 0 + 0 = 0;

Probability of H-T and T-H is equal hence it's fair now.

I proposed a faster method, "lets keep tossing the coin to form a sequence of H's & T's . I win if HT appears before TH" . Was I bluffing?

Actually yes, the probability of HT before TH is just p in second game.

Probability of H-T and T-H is equal hence it's fair now.

I proposed a faster method, "lets keep tossing the coin to form a sequence of H's & T's . I win if HT appears before TH" . Was I bluffing?

Actually yes, the probability of HT before TH is just p in second game.

Source: Common

Enable Like and Comment Easy | Probability |

Spiderman has two girlfriends, Mary Jane & Gwen Stacy. After every mission, he rushes to the central subway. Since spidey is a nice man, much impartial, he takes which-ever train arrives first. From subway, one series go towards Mary's place, and another series move towards Stacy. Trains from either series appear every 10 minutes. Also, Peter Parker sticks with the train which arrives first.

But somehow, he notices that he is spending 9 times more dates with Mary Jane than Stacy. Can you explain why?

But somehow, he notices that he is spending 9 times more dates with Mary Jane than Stacy. Can you explain why?

Hint

Do you think ratio should be 50:50? Do you think the gap between each train's arrival is 5 minutes? Can a train arrive always earlier than the other and still be at every 10 minutes gap?

Solution

Train to Mary Jane's place comes at say times 0 and 10, while the train to Stacy's place comes at times 1 and 11. So 9/10 times, train to Mary Jane's place is appearing earlier.

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

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