Baskin Robbins has 2 types of sizes, 31 flavors. and 6 toppings. How many differ

Question

Baskin Robbins has 2 types of sizes, 31 flavors. and 6 toppings. How many different ways are possible If you choose one from each category? How many different ways can 8 looks be arranged in a display? How many different 3·digit combinations are possible if the last digit must be odd and no repetition is allowed? How many different ways can 4 books chosen from a group of 7 be arranged In a display? A bag contains 17 marbles: 10 purple and 7 green. The purple marbles are numbered 1 through 10 and the green ones are numbered 11 through 17. If one marble is chosen, what is the probability that it is purple? If one marble is chosen, what is the probability that it is green? If one marble is chosen, what is the probability that it is both purple and even? If one marble is chosen, what is the probability that it is NOT both purple and even? If one marble is chosen, what is the probability that it is green or even? If one marble is chosen, what is the probability it is even, given that it is green? If two marble are chosen with replacement, what is the probability both are odd? If two marble are chosen without replacement what is the probability that the first is purple and the second is green? Give an example of two events that are mutually exclusive when one marble is chosen. If 89 marbles are chosen at random without replacement, what is the probability that at least one is purple?

Explanation / Answer

(According to Chegg policy only four subquestions will be answered. Please post the remaining in a separate question)

1. The number of ways of choosing size = 2

So we can choose a size in 2 ways

The number of ways of choosing flavors = 31

So we can choose a flavor in 31 ways

The number of ways of choosing toppings = 6

So we can choose a topping in 6 ways.

Therefore the number of ways in which we can choose one of each = 2*31*6 = 372.

2. We shall consider the number of 3 digit combinations without repetition first.

The first digit can be chosen in 10 ways

The second digit can be chosen in 9 ways (all except the first one chosen)

The third digit can be chosen in 8 ways (two digits are taken already)

So the number of 3 digit numbers with no repetition = 10*9*8

Exactly half of these will be odd and rest even. The last digit determines whether the combination is odd or even. So the number of 3 digit non-repeating numbers (including with leading zeros) where last digit is odd is 10*9*8/2

= 360.

3. The first position in the display can be taken by any of the 8 books.

The second position can be taken by 7 other books and so on.

So the total number of ways to arrange them.

4. Since there are 7 books and we need to choose 4 books out of each, this can be done in

7C4 ways

= 7! / [(7-4)!4!]

= 7*6*5/3*2*1

= 35

Now once chosen, the 4 books can be arranged in 4! or

4*3*2*1= 24 ways.

So the total number of ways = 35*24 = 840 ways.

= 8*7*6*5*4*3*2*1

= 40320.

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