Question

In this activity, you'll calculate a probability and use it to predict the result of repeating a simple chance-based trial many times.
You are in charge of the casino night game area at a fundraising event. In one of the games, called Odd Odds, the player rolls two six-sided dice. The player gets points if the product of the two numbers rolled is odd. So, success in the game depends on the chances of getting an odd number for the result.

1)Find the number of outcomes in the sample space, n(S), of the trial of this game.

2)List and count all the outcomes for event E, in which the product of the two numbers rolled is odd.

3)Find the probability of getting an odd number. In this case, you will calculate the probability, P(E), of event E, in which the product of the two numbers rolled is odd. Write the probability as a fraction reduced to lowest terms and as a decimal correct to two places.

Answer 1

Answer:

Step-by-step explanation:

They are saying the player gets points if the PRODUCT of the two numbers rolled is odd not the SUM  So the answer should be

(1,3) (1,5) (3,1) (3,3) (3,5) (5,1)(5,3)(5,5)  which is 9 out of 36 which would be 1/4 or 0.25

Answer 2

1) The outcomes for rolling two dice, the sample space, is as follows:
(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)
There are 36 outcomes in the sample space.
2)  The ways to roll an odd sum when rolling two dice are:
(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5).  There are 18 outcomes in this event.
3)  The probability of rolling an odd sum is 18/36 = 1/2 = 0.5