There are two even numbers and three odd numbers, so P (even) = 2/5 and P (odd) = 3/5.

If a die is rolled twice, there are 6 * 6 = 36 possible outcomes. 

Each number is equally probable in a fair die. Thus you only need to count the number of outcomes that fulfill the requirement of adding to 7 or including a 3. These include:

1 6

2 5

3 4

4 3

5 2

6 1

3 1

3 2

3 3

3 5 

3 6

1 3

2 3

5 3

6 3

This is 15 possibilities. Thus the probability is 15/36 = 5/12.

Note that drawing balls from each box are independent events. Thus their probabilities can be combined with multiplication. 

Probability of drawing green from A:

10/18 = 5/9

Probability of drawing green from B:

The only probabilites that we know from this is that P(only A) + P(only B) + P (A and B) = 0.88, and that P(neither) = 0.12. We cannot calculate the probability of P(A) unless we know two of the probabilites that add up to 0.88.

Since any of the first set can be summed with any of the second set, the addition sign in the equation works like a conjunction. As such, there are 4 * 6 = 24 possible combinations of a and b. Only one of these combinations, 11 + 16 = 27, works. Thus the probability is 1/24, or about 0.04.

Answer: .005
Explanation: The probability of two consecutive draws without replacement from a deck of cards is calculated as the number of possible successes over the number of possible outcomes, multiplied together for each case. Thus, for the first ace, there is a 4/52 probability and for the second there is a 3/51 probability. The probability of drawing both aces without replacement is thus 4/52*3/51, or approximately .005.

Calculate the chance of drawing either 2 reds, two greens, or two blues.  Then, subtract this from 1 (100%) to calculate the possibility of drawing a pair of different colors. 

The combined probability of RR, GG, and BB is: (10 * 9) / (37 * 36) + (15 * 14) / (37 * 36)  + (12 * 11) / (37 * 36)

This simplifies to: (90 + 210 + 132) / 1332 = 432 / 1332

Subtract from 1: 1 - 432 / 1332 = (1332 - 432) / 1332 = approx. 0.6757 or 67.57%

There are 52 cards in a standard deck, 13 of which are hearts

13/52 X 12/51 =

1/4 X 12/51 =

12/ 204 = 3/51  = 1/17

First we need to find out how many marbles are in the bag in total. 5 + 8 + 3 = 16. He removes a green marble so now there are only 15 in total. When he removes the red one there are then 14 marbles in the bag. 14 is your denominator. The odds of picking a green one are 5 – 1 or 4 because there are only 4 green marbles left in the bag; therefore, the odds of picking another green marble is 4/14

There are 18 students in the class, and 2 must be selected from the 9 girls and 9 boys. Key to this question is noting that the 2 students must be unique: ie once a student is selected to lead the class, he or she cannot be chosen to be the back-up.

Since these are independent events, the probability of each event is found, and the events are multiplied by each other to find the total.

Quantity A:

P(boy leader) = 9/18 = 1/2 as there are 9 boys out of a possible 18 students. 

Once the boy has been chosen, there are 8 boys and 17 students total from which to choose the second student. 

P(girl back-up) = 9/17 because there are 9 girls and 17 students left.

P(Quantity A) = (1/2)(9/17) = 9/34

Quantity B:

P(boy leader) = 9/18 = 1/2 as there are 9 boys out of a possible 18 students. 

Once the boy has been chosen, there are 8 boys and 17 students total from which to choose the second student. 

P(boy back-up) = 8/17 because there are 8 boys and 17 students left.

P(Quantity B) = (1/2)(8/17) = 8 / 34 = 4/17