Easy | Probability |

Two bullets are put into a gun's round barrel consecutively, which has capacity of 6. The gun is shot once, but no bullet is fired. Does rolling the barrel (shuffling) before next shot increase the probability of firing a bullet?

Hint

Since one empty slot has been used, there must be more probability of shooting a bullet in next chance, than shooting after a shuffle.

Answer

Yes!

Solution

Since there are 6 total slots with 2 bullets, the probabilty of firing a bullet after shuffle is 2/6.

The probability of getting bullet without shuffle is 1/4. To understand this, look at the image, and imagine that one of the empty slots (3,4,5,6) just got shot. Now assumming the barrel turns clockwise, next shot will be 4 ,5 ,6 or 1. and out of these four, only (1) has bullet. Hence probability of firing a bullet is 1/4. Note that this same case stands in anticlockwise direction.

With shuffle its 2/6=1/3. Hence yes, shuffling does increase probability of shooting bullet (from 1/4 to 1/3)

The probability of getting bullet without shuffle is 1/4. To understand this, look at the image, and imagine that one of the empty slots (3,4,5,6) just got shot. Now assumming the barrel turns clockwise, next shot will be 4 ,5 ,6 or 1. and out of these four, only (1) has bullet. Hence probability of firing a bullet is 1/4. Note that this same case stands in anticlockwise direction.

With shuffle its 2/6=1/3. Hence yes, shuffling does increase probability of shooting bullet (from 1/4 to 1/3)

Source: Quant Interview

Enable Like and Comment Easy | Probability |

How do you place 50 good candies and 50 rotten candies in two boxes such that if you choose a box at random and take out a candy at random, it better be good!

That means probability of choosing a good candy should be highest.

That means probability of choosing a good candy should be highest.

Hint

Placing all bad in 1 box and all good in another will give probability of choosing good just half. But placing only one type of candy in one box makes sure that good candy has a probability of at-least 1/2

Answer

Put 1 good candy in one box and all other (49 good and 50 rotten candies) in second box

Solution

Put 1 good candy in one box and all other (49 good and 50 rotten candies) in second box. This will give a probability of (1/2)*1 + (1/2)*(49/100) = 74.5%

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

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