Latest Probability Riddles

Let us take this one step at a time.

Let us calculate the probability that Anna and Brian have the same number in their mind.
101/125

Now, there's a 1/5 chance that the numbers will be same and 4/5 chance that the numbers are different.

Let us now include Christy in this data. There can be two cases.
1) Anna and Brian have the same number. In that case, Christy will have only one number to compare.
2) Anna and Brian did not have the same number. In that case, Christy will have two numbers to compare to.

For the first case, the probability will be 5/25. This is if Anna and Brian did have the same numbers.

But if Anna and Brian did not have the same numbers, there is a 2/5 probability that Christy is having the same number (this is because Christy gets to match her number with both Anna and Brian). In that case, we can simply multiply the probabilities.
4/5 * 2/5 = 8/25

Otherwise, if Christy is not having the same number, the probability is 3/5. Now multiplying with the previous chain:
4/5 * 3/5 = 12/25

Now, we can include Drake in our calculations. If we follow the path where Drake's number matches with Anna and Brian, the probability will be 25/125.

Now let us join that with Christy's probability. If Christy's number matches with Anna and Brian and Drakes' number also matches, then the probability will be:
4/5 * 2/5 = 40/125

If Christy's number does not match with Anna and Brian but Drake's number matches with Christy's, the probability will be:
4/5 * 3/5 * 3/5 = 36/125

But if Christy's number does not matches with Anna and Brian and even Drake's does not matches with Christy, then the probability will be:
4/5 * 3/5 * 2/5 = 24/125

Now, we have to tell the probability when all the four friends have same numbers, so we will just add up the probability where all the numbers matches:

25/125 + 40/125 + 36/125 = 101/125.

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.

The probability will be 1/36.

For the first dice, there can be six possibilities. Similarly, for the second dice as well, there can be six possibilities.
Thus the total possibilities is 6 * 6 = 36.
The outcome we need is that the first comes up with a 2 and the second comes up with a 5. That is possible only in one possibility.
Therefore the required probability is 1/36.

If you put a single precious black stone in one bag and all the others in the other bag, it will give you almost three/fourth of the probability of picking up a precious black stone.

Since the probability of having a girl or a boy is same, half of the families will have a boy first and stop. The other half of the families will have a girl and from half of those families, the second born will be a boy and they will stop while the others will again have a girl. This process will continue.

Suppose the number of couples are N, the number of boys will be N.

1/2 have a boy and stop: 0 girls
1/4 have a girl, then a boy: N/4 girls
1/8 have 2 girls, then a boy: 2*N/8 girls
1/16 have 3 girls, then a boy: 3*N/16 girls
1/32 have 4 girls, then a boy: 4*N/32 girls
…
Total: N boys and
1N 2N 3N 4N
– + – + – + — +… = ~N

Therefore the proportion of boys to girl will be quite close to 1:1.

2/3

2 possible scenario are :
case number side shown other side
1 A1 (H) A2 (H)
2 A2 (H) A1 (H)
3 B1 (H) B2 (T)
4 B2 (T) B1 (H)

case 4 is not possible so ans is 2/3

If a 31day month starts on a Saturday, Sunday or Monday it will have five Monday, if not it will have 4 Monday.
Probability is 3/7.

1/2

one of two is the possibility
1. If any of the first 99 people sit in neymar seat, neymar will not get to sit in his own seat.
2. If any of the first 99 people sit in Bruna's seat, neymar will get to sit in his seat.

1/12

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

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