Lesson: Independent Events

Comment on Independent Events

A={2,3,4,5}
B={4,5,6,7,8}

Two integers will be randomly selected from the sets
above, one integer from set A and one integer from
set B. What is the probability that the sum of the two
integers will equal 9 ?

Hi can u pls explain this que. Im getting confused
gmat-admin's picture

The best/fastest way to solve this question is to apply the basic probability formula.

So, P(sum is 9) = (number of pairs that add to 9)/(total number of pairs)

As always, start with the denominator: total number of pairs
We can select a number from set A in 4 ways, and we can select a number from set B in 5 ways. So, the TOTAL number of pairs = 4 x 5 = 20

Now the numerator: number of pairs that add to 9
Let's list all possible pair:
- select 2 from set A, and 7 from set B
- select 3 from set A, and 6 from set B
- select 4 from set A, and 5 from set B
- select 5 from set A, and 4 from set B
So, there are 4 pairs that add to 9

So, P(sum is 9) = 4/20 = 1/5

For more information, watch https://www.gmatprepnow.com/module/gmat-probability/video/742

Thanks Brent. Understood now.

Hi Brent,
How do I calculate the first example (selecting balls without replacement) if what I wanted to know was the probability of the second ball being green?

I mean, I understand that the two events are dependent, however, how can I know what is the probability of event B occurring if I don't know whether or not event A occurred.

What I got so far is:
If the first ball is green, then the probability of the second ball being green is: P(B|A) = 3/6 = 1/3

If the first ball is not green, then the probability of the second ball being green is: P(B|A) = 4/6 = 2/3

How do I reconcile these two probabilities into one?
gmat-admin's picture

Sorry, I'm not 100% sure what probability you are trying to calculate.

If you want to calculate P(2nd ball is green), then we can say...

P(2nd ball is green) = P(1st ball is green and 2nd ball is green OR 1st ball is red and 2nd ball is green)

= P(1st ball is green AND 2nd ball is green) + P(1st ball is red AND 2nd ball is green)

= [P(1st ball is green) x P(2nd ball is green)] + [P(1st ball is red) x P(2nd ball is green)]

= [4/7 x 3/6] + [3/7 x 4/6]

= 12/42 + 12/42

= 24/42

= 4/7

ASIDE: Notice that P(2nd ball is green) is equal to P(1st ball is green). For more on this phenomenon see my post (7 posts down) at: http://www.beatthegmat.com/beat-this-probability-qs-t185719.html

Does that help?

Hi Brent,
What I want to calculate is the probability of the second ball being green (similarly to the question on the post you mentioned).

It's clear now. I see where I went wrong.
Thanks

Can you please help me with this question?

Gordon buys 5 dolls for his 5 nieces. The gifts include two identical S beach dolls, one E, one G, one T doll. If the youngest niece doesn't want the G doll, in how many different ways can he give the gifts?
gmat-admin's picture

Glad to help.

You'll find my step-by-step solution here: http://www.beatthegmat.com/p-amp-c-problem-t269505.html

Cheers,
Brent

While preparing notes for this module, I ran into a confusion. In an earlier lesson we saw that P(A and B) = 0 when we have mutually exclusive events; and in this lesson we different formulae for P(A and B) for dependent and independent scenarios.
Am I mixing two different concepts here? If yes, how so? If no, why aren't the formulae inconsistent?
gmat-admin's picture

If P(A and B) = 0, then the two events cannot BOTH occur.
When this is the case, we say that events A and B are mutually exclusive.

This property has an effect on OR probabilities. Here's why:
We know that P(A OR B) = P(A) + P(B) - P(A and B)
So, if we know that events A and B are mutually exclusive, we can write: P(A OR B) = P(A) + P(B), since we would know that P(A and B) = 0

Here's an example: A box contains 2 red balls, 1 green ball, and 4 yellow balls. If we randomly select 1 ball from the box.
Let's say A = the event that a red ball is selected, and B = the event that a green ball is selected.
In this case, events A and B are mutually exclusive, because we can't select a ball that is both red and green.
In other words, P(A and B) = 0
In this case, P(A or B) = P(A) + P(B)
= P(selecting a red ball) + P(selecting a green ball)
= 2/7 + 1/7
= 3/7
So, that's the deal with mutually exclusive events, as they affect OR probabilities.

Now let's examine dependent events...
We say that events A and B are dependent if the occurrence of one event effects the probability of the other event.
For example, let's randomly select one number from the following set of values: {3, 5, 9, 21, 24}
Let A = the selected number is prime
Let B = the selected number is odd
We can see that P(A) = 2/5
And we can see that P(B) = 4/5

If I select a number at random and tell you that the selected number is prime, we can be certain that the number is also odd.
We know this because, if the selected number is prime, then the selected number must be either 3 or 5 (both of which are odd).
In other words, if we know event A has occurred, then P(B) = 1
However, if we don't know whether or not event A has occurred, then P(B) = 4/5

So, knowing whether event A occurs or not affects the likelihood of event B.
Therefore, we say events A and B are dependent.

To summarize.....
- If events A and B cannot both occur, then events A and B are mutually exclusive. In other words, if P(A and B) = 0, then the two events are mutually exclusive.
- If the occurrence of one event effects the probability of the other event, but then the two events are dependent.

Does that help?

Thanks for the detailed explanation. It helped.

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