Question

HELP QUICK!!!!!!! I REALLY NEED IT!!!

To win the game, Elena has to roll an even number first and a number less than 3 second. Her probability of winning is 6/36.
Marta has a lower probability of winning than Elena has. Which could be the outcome that Marta needs to win the game? Check all that apply.
rolling a sum of 7
rolling a sum of 6
rolling a sum of 2 or a sum of 9
rolling a sum that is greater than 9
rolling a sum that is greater than 2 but less than 5

Answer 1

Your answer would be, Rolling Sum of Six, Rolling a sum of two, or a sum of nine, and Rolling a sum that is greater than two, but less than five.





HOpe that helps!!!

Answer 2

Probability of an event is the of chance of its occurrence. Martha can get probability of 6/36 by choosing

• Option 1 : E = Rolling a sum of 7
• Option 4: E = rolling a sum that is greater than 9

How to calculate the probability of an event?

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

$P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}}$

Where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

Using the above definitions to find the probabilities of all the listed events:

Two dice are rolled and there are $6 \times 6 = 36$ elementary events(calculated by rule of product as for each outcome of 6 outcomes of first die, there can be 6 outcomes of other die) in total and all equally likely (assumed dice were fair).

• Case 1 : E = Rolling a sum of 7

Sum of 7 can come as (1,6), (2,5), (3,4), (4,3), (5,2) , (6,1) (where (x,y) means x came on first die and y came on second die)

Thus, we have 6 favorable cases, thus,

P(E) = 6/36

• Case 2: E =  Rolling a sum of 6

Sum of 6 can come as (1,5) , (2,4), (3,3) , (4,2), (5,1) (total 5 favorable events)

Thus, we have

P(E) = 5/36

• Case 3: Rolling a sum of 2 or a sum of 9

A = Sum of 2 can come as (1,1)

P(A) = 1/36

B = Sum of 9 can come as  (3,6), (4,5), (5,4), (6,3) (total 4 favorable events)

P(B) = 4/36

$P(A \cup B) = P(A) + P(B) = \dfrac{1}{36} + \dfrac{4}{36} = \dfrac{5}{36}$

• Case 4: E = rolling a sum that is greater than 9

Sum greater than 9 come as (4,6), (5,5), (6,4), (5,6), (6,5), (6,6) (total 6 favorable cases)

Thus,

P(E) = 6/36

• Case 5: E = rolling a sum that is greater than 2 but less than 5

That means sum should be 3, or 4

They come as (1,2), (2,1), (2,2), (1,3), (3,1) (total  5 favorable cases)

Thus, we get P(E) = 5/36

Thus, Martha can get probability of 6/36 (highest for the given options)

by choosing

• Option 1 : E = Rolling a sum of 7
• Option 4: E = rolling a sum that is greater than 9

Learn more about probability here:

brainly.com/question/1210781