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## Comment on

Independent Events## Hello Brent,

How does the answer differ if the second set has the following numbers, {3, 11,13,17,18}?

## You're referring to the

You're referring to the question that starts at 3:05 in the video.

So, the question BECOMES:

If a number is randomly selected from the set {2, 3, 6, 8}, and a number is randomly selected from the set {3, 11, 13, 17, 18} what is the probability that both selected numbers will be odd?

P(both odd) = P(1st number selected is odd AND 2nd number selected is odd)

= P(1st number selected is odd) x P(2nd number selected is odd)

= 1/4 x 4/5

= 1/5

Cheers,

Brent

## Thank you, Brent.

## Can we use the dependent

## You're referring to this

You're referring to this question: https://www.greenlighttestprep.com/module/gre-probability/video/744

Let's name the two sets as follows:

Set A {1, 2, 4, 6, 7}

Set B: {2, 3, 4, 5, 6, 7, 8, 9}

We must select 1 number from each set, and we want P(numbers add to 11)

Notice that, if we select 1 from Set A, then there is NO number in Set B that will give us a sum of 11.

So, P(numbers add to 11) = P(number from Set A is 2, 4, 6 or 7 AND number from Set B adds to 1st number to get a sum of 11)

= P(number from Set A is 2, 4, 6 or 7) x P(number from Set B adds to 1st number to get a sum of 11)

= 4/5 x 1/8

= 1/10

Does that help?

Cheers,

Brent

## why 1/8 when we have 4

## Once we have selected a 2, 4,

Once we have selected a 2, 4, 6 or 7 from Set A, we must then select a number from Set B.

At this point, there are 8 numbers in Set B to choose from, and only ONE of those values will pair with the number from Set A to get a sum of 11.

Does that help?

Cheers,

Brent

## We can also show that the

We can also show that the answer is 1/10 by listing all possible outcomes.

Set A {1, 2, 4, 6, 7}

Set B: {2, 3, 4, 5, 6, 7, 8, 9}

1) Set A: 1 and Set B: 2

2) Set A: 1 and Set B: 3

3) Set A: 1 and Set B: 4

4) Set A: 1 and Set B: 5

5) Set A: 1 and Set B: 6

6) Set A: 1 and Set B: 7

7) Set A: 1 and Set B: 8

8) Set A: 1 and Set B: 9

9) Set A: 2 and Set B: 2

10) Set A: 2 and Set B: 3

11) Set A: 2 and Set B: 4

12) Set A: 2 and Set B: 5

13) Set A: 2 and Set B: 6

14) Set A: 2 and Set B: 7

15) Set A: 2 and Set B: 8

16) Set A: 2 and Set B: 9 (SUM = 11)

17) Set A: 4 and Set B: 2

18) Set A: 4 and Set B: 3

19) Set A: 4 and Set B: 4

20) Set A: 4 and Set B: 5

21) Set A: 4 and Set B: 6

22) Set A: 4 and Set B: 7 (SUM = 11)

23) Set A: 4 and Set B: 8

24) Set A: 4 and Set B: 9

25) Set A: 6 and Set B: 2

26) Set A: 6 and Set B: 3

27) Set A: 6 and Set B: 4

28) Set A: 6 and Set B: 5 (SUM = 11)

29) Set A: 6 and Set B: 6

30) Set A: 6 and Set B: 7

31) Set A: 6 and Set B: 8

32) Set A: 6 and Set B: 9

33) Set A: 7 and Set B: 2

34) Set A: 7 and Set B: 3

35) Set A: 7 and Set B: 4 (SUM = 11)

36) Set A: 7 and Set B: 5

37) Set A: 7 and Set B: 6

38) Set A: 7 and Set B: 7

39) Set A: 7 and Set B: 8

40) Set A: 7 and Set B: 9

As you can see, there are 40 possible outcomes, and 4 of them add to 11

So, P(sum = 11) = 4/40 = 1/10

Cheers,

Brent

## and 'being that' 2 and 9 had

## Sorry, but I'm not sure what

Sorry, but I'm not sure what you mean. Can you please rephrase your question?

Cheers,

Brent

## We looked at the formula that

## I'm not sure what you mean by

I'm not sure what you mean by "So we have 2+9 = 11, this event has already occurred."

We are selecting one number from EACH set of numbers. So, the fact that we have a 2 in Set A and a 2 in Set B does not change the correct answer.

So, if we select a 2 from Set A, then among the 8 numbers in Set B, only ONE is the needed 9.

Does that help?

## Hi Mr Hanneson,

I still don't understand the " with replacement" and " without replacement " concept in any of your questions. Can you please explain it to me.

## Consider this illustrative

Consider this illustrative example:

Let's say we have a bag of 2 red balls and 2 green balls.

We're going to select 1 ball and then another ball, and we want to determine the probability that both balls are red.

If we have a proviso that we select the balls WITH REPLACEMENT, then after the 1st ball is selected, we put that ball back in the bag

(so, the 4 original balls are in the bag), and then select the 2nd ball.

If we have a proviso that we select the balls WITHOUT REPLACEMENT, then after the 1st ball is selected, we DON'T put that ball back in the bag (so, there are now 3 balls remaining in the bag), and then select the 2nd ball.

Does that help?

Cheers,

Brent

## Hey, can you explain the 3rd

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/in-a-probability-experiment-g-and-h-are...

Great question!!

Here's my solution: https://gre.myprepclub.com/forum/in-a-probability-experiment-g-and-h-are...

Cheers,

Brent

## Hi,

Based on the above question https://gre.myprepclub.com/forum/in-a-probability-experiment-g-and-h-are-independent-events-2254.html,

plz see below the question from greprep club::

https://gre.myprepclub.com/forum/the-probability-that-gary-will-eat-eggs-2383.html

Plz let me know shouldn't the approach and the result of both question be the same

## Question #1: https:/

Question #1: https://gre.myprepclub.com/forum/in-a-probability-experiment-g-and-h-are...

Question #2: https://gre.myprepclub.com/forum/the-probability-that-gary-will-eat-eggs...

The two questions are similar since we can use the OR probability formula, P(A or B) = P(A) + P(B) - P(A and B), for both questions.

At the same time, the two questions differ in the following way:

In Question #1, the probability that both events occur is greater than 0. That is, P(G and H) > 0

In Question #2, the probability that both events occur is 0. That is, P(Gary eats eggs and cereal) = 0

Does that help?

Cheers,

Brent

## Hi Brent,

I have a problem in understanding the ques. I have read ur explanation but I couldnot relate why Question 2 ( link 2) give an Option C as an answer.

Plz Let me know, where Iam going wrong

As the first quest : QTY A - The probability that either G will occur or H will occur, but not both : This is definitely less than QTY B, which include probability of r or S or both

As for the 2nd ques.

QTY A :Probability that Gary eats eggs or cereal for breakfast on a particular day: This means the probability of eating egg or cereal but not both -

But the above statement should be less than 1, since the probability cannot be greater than 1 and the probability of 2 events occurring is also 0 hence QTY A < 1.

Plz help me out.

## The big difference is that,

The big difference is that, for question #1, P(G and H) > 0.

For question #2, P(eggs and cereal) = 0.

We get:

P(eggs or cereal) = P(eggs) + P(cereal) - P(eggs and cereal)

= 3/7 + 4/7 - 0

= 7/7

= 1

Does that help?

Cheers,

Brent

## Thanks, got that.

## https://gre.myprepclub.com/forum

This question has got me confused. So mutually exclusive and independence are 2 SEPARATE things?

Mutually exclusive has to do with OR probabilities

Independence has to do with AND probabilities

Is this correct?

## https://gre.myprepclub.com/forum

https://gre.myprepclub.com/forum/in-a-probability-experiment-g-and-h-are...

The OR probability formula: P(A or B) = P(A) + P(B) - P(A and B)

Two events (say event A and event B) are mutually exclusive if they cannot occur together.

This means P(A and B) = 0

So, if event A and event B are mutually exclusive, then we can say P(A or B) = P(A) + P(B)

We can remove the "P(A and B)" part since P(A and B) = 0 when the events are mutually exclusive.

More on mutually exclusive events here: https://www.greenlighttestprep.com/module/gre-probability/video/747

More on the OR probability formula: https://www.greenlighttestprep.com/module/gre-probability/video/748

The AND probability formula: P(A and B) = P(A) x P(B|A)

Two events (say event A and event B) are independent if the occurrence of one event does NOT affect the likelihood of the other event.

So, if event A and event B are independent, then we can say P(A and B) = P(A) x P(B)

Cheers,

Brent

## https://gre.myprepclub.com/forum

for this question for QA I got r+s-rs but in this case since they are independent wouldnt we say that P(G&H) is mutually exclusive? Independent would mean two events are separate from one another cannot happen at the same time, therefor for QA I am getting r+s-0 and therefore A is greater than B. Not sure why this logic fails?

## If events A and B are

If events A and B are possible (that is P(A) > 0 and P(B) > 0), then the two events cannot be both independent and mutually exclusive.

Mutually exclusive events can't both occur at the same time, whereas independent event can both occur at the same time.