Solution:

2nd

When the cylinder is spun for the first and last time, the following six outcomes are possible:

First:
First chamber is fired first. The first player loses.

Second:
Second chamber is fired first. The first player loses.

Third:
Third chamber is fired first. The first player loses.

Fourth:
Fourth chamber is fired first. The second player loses. (First shot, player 1, chamber 4 empty. Second shot player 2, chamber 5, empty. Third shot player 1, chamber 6 empty. Fourth shot player 2, chamber 1 not empty.)

Fifth:
Fifth chamber is fired first. The first player loses. (First shot, player 1, chamber 5 empty. Second shot player 2, chamber 6, empty. Third shot player 1, chamber 1 not empty.)

Sixth:
Sixth chamber is fired first. The second player loses. (First shot, player 1, chamber 6 empty. Second shot, player 2, chamber 1, not empty)

Thus, if you analyze, you will find that the second player has a 4/6 chance of winning. Therefore if you want to win the game, going second gives you better shot at it.

Solution:

Tilt the Mug until the coke barely touches the lip of the Mug. If the bottom of the Mug is visible then it is less than half full. If the Mug bottom is still completly coverd by the coke, then it is more than half full.

Solution:

60%

Explanation:
In all such cases where you have to find out the probability of anything with an already given situation, you have to use the formula for conditional probability.

In this case, you have to find out the probability that the second test was passed by a student given that he passed the first test already.

Therefore, P(Second/First) = P (First and Second)/ P (First)
= 0.25/0.42
=0.60
=60 %

Solution:

A) Fill 5 ml gallon ( 5mlGallon - 5, 3mlGallon - 0)
B) Transfer to 3 ml gallon (5mlGallon - 2, 3mlGallon - 3)
C) Empty 3 ml gallon ( 5mlGallon - 2, 3mlGallon - 0)
D) Transfer 2 ml from 5 ml gallon to 3 ml gallon (5mlGallon - 0, 3mlGallon - 2)
E) Fill 5 ml gallon(5mlGallon - 5, 3mlGallon - 2)
F) Transfer 1 ml from 5 ml gallon to 3 ml gallons(5mlGallon - 4, 3mlGallon - 3)

Solution:

Suppose if there is just one unfaithful husband, then his wife will kill her husband as she must be knowing that all other men are faithful but the queen declared that one of them all is unfaithful.

If there are two unfaithful husbands, then both their wives will believe there is just one unfaithful husband, that too the other one whom she knows. Thus both of them will expect the above case to happen at midnight. But when no one is shot, they will realize that there are two unfaithful husbands and since they know about everyone, they will know that the other unfaithful man is their own husband on the next day.

If there are three unfaithful husbands, each of their wives will be knowing about two other unfaithful men so they will be expecting the above case and will be waiting for the gunshots on the second day. When that does not happens, they will realize that there are more than two and since they all will know about the rest, they will realize that their own husband is unfaithful and they will kill him on the third day.

Thus if we talk about general terms and suppose that there are n unfaithful husbands, their wives will believe that there is n-1 unfaithful husbands and will expect the gunshot on n-1 day. When they don’t hear that, they will realize their own husband is the nth.