There number of possible outcomes is equal to six times six, or thirty-six. The sum of the two dice must be either 2, 3, 5, 7, or 11. There are 15 out of the 36 outcomes that would result in a sum that is a prime number:

[1,1], [1,2], [1,4], [1,6], [2,1], [2,3], [2,5], [3,2], [3,4], [4,1], [4,3], [5,2], [5,6], [6,1], [6,5]

The simplest way to solve this is to find the number of seating arrangements in which Richard and Hang are seated together and then subtract those in which Thomas and Lily are also seated together. Consider Richard and Hang as a unit. This pair can be arranged with the other six speakers in ₇P₇ ways. For each of these ways, Hang could be either on Richard’s left or his right. Thus, there are ₇P₇ × 2 = 10 080 arrangements in which Richard and Hang are seated together. Now also consider Lily and Thomas as a unit. The two pairs can be arranged with the remaining four speakers in ₆P₆ ways, and the total number of arrangements with each of the pairs together is ₆P₆ × 2 × 2 = 2880.

Therefore, the number of seating arrangements in which Richard and Hang are adjacent but Thomas and Lily are not is 10 080 − 2880 = 7 200.

First, find the numbers that in the set that are divisible by 4. 12, 24, 60, 100, 260, 480, and 1000 are all divisible by 4. Now find the numbers that are divisible by 6. 12, 24, 60, 480, 606 are all divisible by 6. The numbers that are divisible by both 4 and 6 are 12, 24, 60, and 480, or 4 total numbers from the set. So 4 out of the 10 numbers are divisible by 4 and 6.  The probablility is 4/10, which reduces to 2/5.  The correct answer is 2/5.

If Sam will celebrate his seventh birthday in 2014, then he was born in 2007. So Mike was born 23 years before in 1984 and Jacob was born 27 years before that in 1957.

If you answered 1964, then you did not factor that Sam was 7 years old at the time of calculation.

If you answered 1984, then you found the year that Mike was born, not Jacob.

If you answered 2007, then you found the year that Sam was born, not Jacob.

If you answered 1987, then you just subtracted 27 years old from the day of the party in 2014.

Probability = what you want ÷ total number

A standard deck of playing cards has 52 cards, with 4 suits and 13 cards in each suit

Choosing two red cards = 26 * 25 = 650

Choosing two cards = 52 * 51 = 2652

So the probabiulity of choosing 2 red cards is 650/2652 = 25/102

If replacement is allowed, then the probability of choosing 2 red cards becomes 676/2704 = 1/4

120 is 40% of 300, so x = 40.

175% of 40 is equal to 70.

There are 9 possible ways for this outcome to be determined. They are contained in this set 

{BB, BR, BY, RB, RR, RY, YB, YR, YY}.

Only one of these outcomes is desired, so the value is 1/9

Another way to solve this is to multiply the probability by the number of instances: thus two consecutive instances of "blue" will have the probability of one instance (1/3) multiplied by one instance (1/3), or 1/9.

Two independent events that each have a probability of 1/2 of occuring.

1/2 * 1/2 = 1/4

Probability = # of Tickets You Bought / # of Tickets Sold

Probability = 8/526 = 0.0152

Probability of each event = 1 side of die / # of sides = 1/6

Probability for multiple events = P1 * P2 * P3

1/6 x 1/6 x 1/6 = 1/216