Probability Riddles

1/3

Explanation
Assuming D1 is the dress for Sister1, D2 is the dress for Sister2 and D3 is the dress for Sister3.

Therefore the total number of cases are illustrated below.

Sister1 Sister2 Sister3
D1 D2 D3
D1 D3 D2
D2 D1 D3
D2 D3 D1 ..... (1)
D3 D1 D2 .... (2)
D3 D2 D1

In both steps (1) & (2), no one gets the correct Dress.
Therefore probability that no sister gets the correct dress is 2/6 = 1/3

Place 1 white marble in one bowl, and place the rest of the marbles in the other bowl (49 whites, and 50 blacks).

This way you begin with a 50/50 chance of choosing the bowl with just one white marble, therefore life! BUT even if you choose the other bowl, you still have ALMOST a 50/50 chance at picking one of the 49 white marbles.

Only twenty-three people need be in the room, a surprisingly small number. The probability that there will not be two matching birthdays is then, ignoring leap years, 365x364x363x...x343/365 over 23 which is approximately 0.493. this is less than half, and therefore the probability that a pair occurs is greater than 50-50. With as few as fourteen people in the room the chances are better than 50-50 that a pair will have birthdays on the same day or on consecutive days.

1/3

This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.

The following are possible combinations of two children that form a sample space in any earthly family:
Girl - GirlGirl - Boy
Boy - Girl
Boy - Boy

Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample space.
This leaves only three possibilities, one of which is two girls. Hence the probability is 1/3

There are six possible pairings of the two balls withdrawn, RED+RED, RED+WHITE, WHITE+RED, RED+BLUE, BLUE+RED, WHITE+BLUE. We know that the WHITE + BLUE combination has not been drawn. This leaves five possible combinations remaining. Therefore the chances that the RED + RED pairing has been drawn are 1 in 5. Many people cannot accept that the solution is not 1 in 3, and of course it would be, if the balls had been drawn out separately and the color of the first ball announced as red before the second had been drawn out. However, as both balls had been drawn together, and then the color of one of the balls announced, then the above solution, 1 in 5, must be the correct one.

2/13

There 4 kings & 4 queens, which implies there are eight cards in the pack of 52 that would satisfy the conditions.
=>8/52
=> 2/13

1/12

Possible cases = 6*6 = 36
Desired cases = [(3,6), (4,5), (6,3), (5,4)] = 4
=> 4/36 = 1/12

This can be solved in a generic and complex way but let us not go into all that.

There can be four ways through which the pair of dice results in a sum of 5. There can be six ways through which the pair of dice can result in a sum of 7.

Now, we want the probability of the pair of dice resulting in a sum of 5 before a sum of 7. Thus probability = 4/(4+6) = 4/10 or 2/5.

60%

Explanation:
This type of questions call in for the application of conditional probability as the question is asking you to find out the probability that the second test was passed with the given situation that the first one was passed.

The formula for conditional probability:
P (Second/First) = P (First and Second)/ P (First)
= 0.25/0.42
=0.60
=60%

Analyzing thing, the friend has better chances of winning. Let us assume and analyze what can happen at any point of time supposing that we have not already completed the sequence. If the last flip was tails, start at 1 and if it is heads, start at 2.

1) If any one between us flips to get tails, nothing will change. We will still be at step 1 and we both will be needing 3 additional flips to complete. But if either of us gets heads, we continue.

2) If either of us gets heads, we stay on this step. But if we get tails, we continue. At this point, I am 3A and my friend is 3B.

3A) If I get tails now, I will have to start all over again from step 1 till I again get heads and I will have to flip the coin at least 3 times to complete my sequence.

3B) If the friend gets heads, he goes back to the step 2 and try again for tails. He will have to flip the coin at least 2 times to complete his sequence.

Thus, he obviously has better chances of winning.