Question

Suppose you roll an eight-sided die two times hoping to get two numbers whose sum is even. What is the sample space? How many favorable outcomes are there?

Answer 1

@jagr2713

Answer 2

1) Here the experiment is - "A six-sided dice is thrown two times"

2) ==> The number of elements in sample space is = 36
[Since, in one throw there will be any six results as (1,2,3,4,5,6). Any one in the first can combine with any one in the second. Thus it is 6 x 6 = 36]

3) Hence the elements in the sample space = (1,1), (1,2), (1,3), (1,4), (1,5),1,6), (2,1) .... (2,6), (3,1), ..........(3,6), (4,1), ..... (4,6), (5,1), .... (5,6), (6,1) .... (6,6)

4) Required event is: "Sum of the two numbers is even"

5) ==> Favorable is = [(1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3),(3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4) AND (6,6)}; thus number of elements = 18

Answer 3

can you do one more

Answer 4

Sure, If I can.

Answer 5

Darien has raw data that shows how many hot sandwiches, cold sandwiches, wraps, and salads he sold in his sandwich shop in one day. He wants to compare the number of each type of meal sold during the day. Should he use a bar graph, a double bar graph, or a circle graph? Explain.

Answer 6

It would be a double-bar graph because you are comparing large with small for each item, as well as every item to each other.

Answer 7

can you correct me

Answer 8

Pablo rolled a standard die 60 times. He got a 1 twelve times. How does this result compare to the expected results? Explain.

Answer 9

Since a die has 6 sides, one expects each number to be rolled 1/6 of the time. For 60 rolls, one would expect 60(1/6) = 10 rolls to result in 1. 12 rolls of 1 is 2 more than expected.

Answer 10

1.) The probability of rolling a 1 is 1/6. So if the die is rolled 60 times what is the expected number of times you would get a 1. 10 times is the expected value. If the die was rolled hundreds of times and the number of times a 1 came up was recorded the quotient (number of 1s)/(number of rolls) will come closer and closer to 1/6.

2.) For any roll, the probability of getting a 1 is 1/6, so for 60 rolls, the expected number of 1's is 60 x 1/6 = 10. So the result he got was slightly above what would be expected.