Easy | Probability |

In a world where everyone wants a girl child, each family continues having babies till they have a girl. What do you think will the boy to girl ratio be eventually? (Assuming probability of having a boy or a girl is the same)

Answer

1:1

Solution

Suppose there are N couples. First time, N/2 girls and N/2 boys are born (ignoring aberrations). N/2 couples retire, and rest half try another child. Next time, N/4 couples give birth to N/4 girls and rest N/4 boys. Thus, even in second iteration, ratio is 1:1. It can now be seen that this ratio always remain same, no matter how many times people try to give birth to a favored gender.

Easy | Probability |

Probability of accidents on a road in one hour is 3/4. What is the probability of accidents in half hour?

Hint

The probability of no event in two disjoint intervals of half hours is same as probability of no event in one full hour.

Answer

1/2

Solution

Probability of no accident in 1 hour = (prob of no accident in 1/2 hour )*(no accidents in next disjoint 1/2 hour)=p^2

hence 1/4=p^2, hence p=1/2, hence probability of accident in half hour=1/2

Read Palak's detailed solution:

Assuming things to be memoryless, and thus disjoint time intervals to be independent, prob. of no accidents in [0,2t) = prob. of no accidents in [0,t) AND [t,2t) => 1-p(2t) = (1-p(t))^2, where p(t) is the prob. of accident in time interval of length t. Given that p(1h)=3/4 => p(0.5h)=1/2 and p(2h)=15/16.

The way we have defined p(t) above, it is the probability that at least 1 accident happens in time interval t. Thus, as t increases, p(t) increases.

If, instead, p(t) is defined as the fraction of vehicles that met with an accident in time interval t, then again p(2t)=p(t)+(1-p(t))*p(t)=1-(1-p(t))^2

hence 1/4=p^2, hence p=1/2, hence probability of accident in half hour=1/2

Read Palak's detailed solution:

Assuming things to be memoryless, and thus disjoint time intervals to be independent, prob. of no accidents in [0,2t) = prob. of no accidents in [0,t) AND [t,2t) => 1-p(2t) = (1-p(t))^2, where p(t) is the prob. of accident in time interval of length t. Given that p(1h)=3/4 => p(0.5h)=1/2 and p(2h)=15/16.

The way we have defined p(t) above, it is the probability that at least 1 accident happens in time interval t. Thus, as t increases, p(t) increases.

If, instead, p(t) is defined as the fraction of vehicles that met with an accident in time interval t, then again p(2t)=p(t)+(1-p(t))*p(t)=1-(1-p(t))^2

Source: Placement tests

Enable Like and Comment Easy | Probability |

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?". What is the probability to win car if you switch?

Answer

2/3

Solution

Probability that your initial choice did not have car is 2/3. So you should switch to the other gate, and win the car with probability 2/3

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.

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