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Change the reference to independence to mutually exclusive. Do you understand why the two events are not mutually exclusive?so would that be the answer

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Change the reference to independence to mutually exclusive. Do you understand why the two events are not mutually exclusive?so would that be the answer

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This is not possible as the formula for probabilities is P(A or B) = P(A) + P(B) - P(A and B) and you can't add separate probabilities because the two events are not independent, probabilities also cannot be above one. is that correct?

If \(\displaystyle A\) is the event the pair multiply to an even number and \(\displaystyle B\) is the event that a pair multiply to more than thirty then those events are neither independent nor mutually exclusive. then the number \(\displaystyle \|A\|=144-36=108\). The reason being that there are \(\displaystyle 36\) pairs in which both entries are odd, meaning the product is odd. Now how pairs have a product greater than thirty \(\displaystyle \|B\|=?\)I think you meanmutually exclusive, right? This amounts to the second possible answer I suggested. It's not at all clear how much the student is expected to explain, but this more extended answer can't hurt!

Also what is \(\displaystyle \|A\cap B\|\;?\).

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Is the answer below correct?

This is not possible as the formula for probabilities is P(A or B) = P(A) + P(B) - P(A and B) and you can't add separate probabilities because the two events are not mutually exclusive, probabilities also cannot be above one.

This is not possible as the formula for probabilities is P(A or B) = P(A) + P(B) - P(A and B) and you can't add separate probabilities because the two events are not mutually exclusive, probabilities also cannot be above one.

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Dr. Peterson, back in response 11 (we are up to 20 now) told you that "P(A or B)= P(A)+ P(B)- P(A and B)". The probability that the two dice come up "an even number"

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Correct.Is the answer below correct?

Note that mutually exclusive means that P(A and B) = 0, which is why you

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This is the very reason that for years I would begin each probability theory class with the statement 'if you have not had the foundations class or if you do not know set operations you need to leave right now'. It is a fundamental property of counting that \(\displaystyle \|A\cup B\|=\|A\|+\| B\|-\|A\cap B\|\) That is, the number of elements in at least one of sets \(\displaystyle A\text{ or }B\) equals the number in \(\displaystyle A\) plus the number in \(\displaystyle B\) minus the number common to both, (the last term accounts the over count). In probability it becomes \(\displaystyle \mathcal{P}(A\cup B)=\mathcal{P}(A)+\mathcal{P}(B)-\mathcal{P}(A\cap B)\).This is not possible as the formula for probabilities is P(A or B) = P(A) + P(B) - P(A and B) and you can't add separate probabilities because the two events are not independent, probabilities also cannot be above one. is that correct?

We have agreed to use \(\displaystyle A\) to be the event the product of the outcomes on the pair is even and \(\displaystyle B\) is product is over thirty.

There are a total of \(\displaystyle 144\) possible pairs. Of those \(\displaystyle 108\) pairs have at least one that is even so that the product is even.

The set \(\displaystyle \{1,3,5,7,9,11\}\times\{1,3,5,7,9,11\}\) contains \(\displaystyle 36\) pairs of odd entries so the products are odd; \(\displaystyle 144-36=108\).

Now as Prof. Ivy points out, this is tedious. There are seventy five pairs, that yield a product that is more than \(\displaystyle 30\,.\) Of those \(\displaystyle 48\) are also even.

So here goes: \(\displaystyle \begin{align*}\|A\cup B\|&=\|A\|+\| B\|-\|A\cap B\| \\&=108+75-48\\&=135 \end{align*}\)

Ah, the probability the product is even

Anotheron any roll of a set of two fair 12 sided dice, the probability of obtaining a product that is an even number or a product of greater than 30, adds to a number larger than 1. Given we know the probability of an event must lie between 0 and 1 explain in your own words if this is possible

The problem appears to say that the probability of an even product or a product greater than 30

Probably, however, the problem instead says that the the sum of the probability of an even product and the probability of a product greater than 30 exceeds 1. The question then asks for an explanation of how that is possible. That is a reasonable problem.

There is a reason why we ask that problems be quoted exactly and completely. When the problem given makes no sense, we should ask the student to start by making sure that we have the full and exact statement of the problem. Frequently, students have trouble because they form an incorrect paraphrase of the actual problem.