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sagnik242
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sagnik2422

Great explanation, what about the 4th question I had?

Now that you have understood the first 3, you should be easily able to deduce the 4th on your own!
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While going through the OG math review, i don't understand why the probability of both A and C happening is "at least" 0.08???

I am understanding the rest of the things in the paragraph but not the highlighted segment...
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While going through the OG math review, i don't understand why the probability of both A and C happening is "at least" 0.08???

I am understanding the rest of the things in the paragraph but not the highlighted segment...

Check Karishma's reply here: gmat-og-13-probability-questions-173028.html#p1376868

Hope it helps.
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sagnik242
Consider an experiment with events A, B, and C for which P(A) = 0.23 P(B) = 0.40 and P(C) = 0.85. Also, suppose that A and B are mutually exclusive and B and C are independent.

In the GMAT book they explain: Note that P(A or C) cannot be determined using the info given. But it can be determined that A and C are not mutually exclusive since P(A) + P(C) = 1.08, which is greater than 1, and therefore cannot equal P (A or C) ; from this it follows that P (A and C >= (Greater than or equal to) 0.08. One can also deduce that P(A and C) <= (less than or equal to) 0.23 since A ∩ C is a subset of A, and that P (A or C) >= P(C) = 0.85 since C is a subset of A ∪ C. Thus one can conclude that 0.85<= P(A or C) <= 1 and 0.08 <= P (A and C) <= 0.23
MY QUESTIONS:
1. Where is the 0.08 coming from in this statement from above : Note that P(A or C) cannot be determined using the info given. But it can be determined that A and C are not mutually exclusive since P(A) + P(C) = 1.08, which is greater than 1, and therefore cannot equal P (A or C) ; from this it follows that P (A and C >= (Greater than or equal to) 0.08
2. How do I know that C is a subset of A or C?
3. How do I know that A nd C is a subset of A?
4. How are these two inequalities obtained exactly ? Thus one can conclude that 0.85<= P(A or C) <= 1 and 0.08 <= P (A and C) <= 0.23

What are mutually exclusive events? They are those which have nothing in common e.g.
A = "It will rain tomorrow" and
B= "It will not rain tomorrow"
are mutually exclusive events. If one happens, the other cannot happen. P(A and B) = 0
Also, in any given day, either it will rain or not so P(A or B) = 1. A and B cover the entire range of possibilities.

Similarly
A = I will get an A+ on my Math test and
B = I will get a B on my Math test
are two mutually exclusive events. If you get an A+, you cannot get a B and vice versa. So P(A and B) = 0
But you could also get a C or a D so P(A or B) < 1

With mutually exclusive events, P(A or B) = P(A) + P(B) - This must be less than or equal to 1.

What about independent events? They are not connected to each other e.g.
A = I will get A+ on my Math test
B = I will get B on my English test.
It is possible that both happen. A doesn't depend on B - they are independent.

In given question, P(A) = 0.23 P(B) = 0.40 and P(C) = 0.85
A and B are mutually exclusive so P(A and B) = 0 and P (A or B) = .23+.4 = .63

Also, B and C are independent.

Since P(A) + P(C) = .23 + .85 = 1.08, we can say that A and C are not mutually exclusive. (See the highlighted part above)
So A and C must have something common i.e. P(A and C) must be greater than 1.

We know the general formula P(A or C) = P(A) + P(C) - P(A and C) (similar to sets formula)
Since a probability is always <= 1, P(A or C) <= 1
P(A) + P(C) - P(A and C) <= 1
1.08 - P(A and C) <= 1
P(A and C) >= 0.8

2.
P(A or C) includes all cases where A happens or C happens or both happen. So it includes all cases where C happens (irrespective or whether A happens or not). So P(C) is a subset of P(A or C).

3.
P(A and C) includes those cases where both A and C happen. This will be a subset of all cases where A happens (irrespective or whether C happens or not). Say A happens in a total of 20 cases. A and C both will happen in less than or equal to 20 cases, right? A and C both cannot happen in more than 20 cases since A happens in 20 cases only. Out of these 20 cases, C may also happen in some cases.

Great explanation.

The text marked in RED seems to me as typographical errors.

1. P(A or C) must be greater than 1.

2. P(A and C) >= 0.08
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The text marked in RED seems to me as typographical errors.

1. P(A or C) must be greater than 1.

2. P(A and C) >= 0.08

Yes. Edited. Thanks!

Note that P(A or C) cannot be greater than 1. It was just P(A) + P(C) > 1. Note that any one probability cannot be more than 1.

P(A or C) = P(A) + P(C) - P(A and C)
Whatever is common in A and C, needs to get subtracted. P(A or C) will have a maximum value of 1.
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arunspanda


The text marked in RED seems to me as typographical errors.

1. P(A or C) must be greater than 1.

2. P(A and C) >= 0.08

Yes. Edited. Thanks!

Note that P(A or C) cannot be greater than 1. It was just P(A) + P(C) > 1. Note that any one probability cannot be more than 1.

P(A or C) = P(A) + P(C) - P(A and C)
Whatever is common in A and C, needs to get subtracted. P(A or C) will have a maximum value of 1.


Similarly, can we deduce the following from the given information.

1. Events B and C are not mutually exclusive as P (B and C) is non-zero..
P(B) + P(C) = 0.4 + 0.85 = 1.25 or, events B and C are overlapping sets with common outcomes.
2. 0.25 =< P(B and C) =< P(B)
3. P(C) =< P(B or C) = < 1
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VeritasPrepKarishma
arunspanda


The text marked in RED seems to me as typographical errors.

1. P(A or C) must be greater than 1.

2. P(A and C) >= 0.08

Yes. Edited. Thanks!

Note that P(A or C) cannot be greater than 1. It was just P(A) + P(C) > 1. Note that any one probability cannot be more than 1.

P(A or C) = P(A) + P(C) - P(A and C)
Whatever is common in A and C, needs to get subtracted. P(A or C) will have a maximum value of 1.


Similarly, can we deduce the following from the given information.

1. Events B and C are not mutually exclusive as P (B and C) is non-zero..
P(B) + P(C) = 0.4 + 0.85 = 1.25 or, events B and C are overlapping sets with common outcomes.
2. 0.25 =< P(B and C) =< P(B)
3. P(C) =< P(B or C) = < 1

Yes, you are right on all three counts.



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