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Margaret [11]
3 years ago
5

A batch of cookie dough will be sliced up into 100 cookies and then baked. 400 raisins have been included in the batch of dough,

and the dough has been thoroughly mixed so as to randomize the ingredients. What is the chance that, despite these precautions, one or more cookies in the batch will contain no raisins at all?
a. How many raisins should be put in the batch of dough to be 99% sure that no cookie comes out with no raisins in it?
Mathematics
1 answer:
Lesechka [4]3 years ago
6 0

Answer:

if there is gonna be 100 cookies an 400 raisins in the batch there should be a 1% chance that no 1 or more cookie will come out with out raisins.  You have more raisins than the number of cookies which if u ever bake cookies when adding in things like that you should add in as much raisin as there is cookie.

Step-by-step explanation:

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The Eco Pulse survey from the marketing communications firm Shelton Group asked individuals to indicate things they do that make
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Answer:

a) There is a probability of 42% that the person will feel guilty for only one of those things.

b)There is a probability of 46% that a randomly selected person will not feel guilty for either of these reasons

Step-by-step explanation:

This probability problem can be solved by building a Venn like diagram for each probability.

I say that we have two sets:

-Set A, for those people that will feel guilty about wasting food.

-Set B, for those people that will feel guilty about leaving lights on when not in a room.

The most important information is that there is a .12 probability that a randomly selected person will feel guilty for both of these reasons. It means that P(A \cap B) = .12.

The problem also states that there is a .39 probability that a randomly selected person will feel guilty about wasting food. It means that P(A) = 0.39. The probability of a person feeling guilty for only wasting food is PO(A) = .39-.12 = .27.

Also, there is a .27 probability that a randomly selected person will feel guilty about leaving lights on when not in a room. So, the probability of a person feeling guilty for only leaving the lights on is PO(B) = 0.27-0.12 = 0.15.

a) What is the probability that a randomly selected person will feel guilty for either wasting food or leaving lights on when not in a room?

This is the probability that the person feels guilt for only one of those things, so:

P = PO(A) + PO(B) = 0.27 + 0.15 = 0.42 = 42%

b) What is the probability that a randomly selected person will not feel guilty for either of these reasons

The sum of all the probabilities is always 1. In this problem, we have the following probabilies

- The person will not feel guilty for either of these reasons: P

- The person will feel guilty for only one of those things:  PO(A) + PO(B) = 0.42

- The person will feel guilty for both reasons: PB = 0.12

So

`P + 0.42 + 0.12 = 1

P = 1-0.54

P = 0.46

There is a probability of 46% that a randomly selected person will not feel guilty for either of these reasons

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2 years ago
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