Question

11.)  

An experiment consists of rolling two fair dice and adding the dots on the two sides facing up. Assuming each simple event is as likely as any​ other, find the probability that the sum of the dots is greater than 2.

The probability that the sum of the dots is greater than 2 is

12.)

An experiment consists of rolling two fair dice and adding the dots on the two sides facing up. Find the probability of the sum of the dots indicated.

Getting a sum of 1

The probability of getting a sum equal to 1 is

13.)

An experiment consists of tossing 4 fair​ (not weighted)​ coins, except one of the 4 coins has a head on both sides. Compute the probability of obtaining exactly 1 headhead.

The probability of obtaining exactly 1 headhead is

15.)

An experiment consists of rolling two fair​ (not weighted) dice and adding the dots on the two sides facing up. Each die has the number 1 on two opposite​ faces, the number 2 on two opposite​ faces, and the number 3 on two opposite faces. Compute the probability of obtaining the indicated sum.

Sum of 8

The probability of getting a sum of 8 is nothing

16.)

An experiment consists of dealing 7 cards from a standard​ 52-card deck. What is the probability of being dealt exactly 1 ace​?

The probability of being dealt exactly1 ace is approximately

Answer 1

Sample space for rolling of two dice is

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

Sum	2	3	4	5	6	7	8	9	10	11	12
Frequency	1	2	3	4	5	6	5	4	3	2	1

Total outcomes = 36

11. Probability that the sum of the dots is greater than 2, P(X >2) = 1- P(X 2)

12. Probability of getting a sum equal to 1, P(X = 1) = 0/ 36 = 0

13. As one coin has head on both sides so it will always show head.

when there is exactly one head then that mean other 3 coins will have tail.

So, probability of obtaining exactly 1 head = 0.5 *0.5 *0.5 * 1 = 0.125

15.

Sum	2	3	4	5	6
Frequency	4	8	12	8	4

Probability of getting a sum of 8 = 0/36 = 0

16. No. of aces = 4

Probability of being dealt exactly1 ace =