# Lesson: Probability of Event A AND Event B

## Comment on Probability of Event A AND Event B

### P(A OR B) = P(A) + P(B) - P(A

P(A OR B) = P(A) + P(B) - P(A AND B)
which would give
P(A AND B) = P(A) + P(B) - P(A OR B)

How do we know that this formula shouldn't be used here? ### Great question!

Great question!

That formula will work, but it will take much MUCH longer.

Let event A = 1st selection is green
Let event B = 2nd selection is green

So, P(A AND B) = P(1st is green AND 2nd is green)
= P(1st green) + P(2nd green) - P(1st green OR 2nd green)
= 4/7 + 4/7 - P(1st green OR 2nd green)

At this point, if we use the formula P(A OR B) = P(A) + P(B) - P(A AND B) then we find ourselves trying to determine the value of P(A and B) anyway.

However, we don't need to do this.

That is, P(1st green OR 2nd green) = 1 - P(neither is green)

P(neither is green) = P(1st red AND 2nd red)
= P(1st red) x P(2nd red)
= 3/7 x 2/6
= 6/42
= 1/7

So, P(1st green OR 2nd green) = 1 - 1/7 = 6/7

Now that we know the value of P(1st green OR 2nd green), we can write:

P(A AND B) = P(1st is green AND 2nd is green)
= P(1st green) + P(2nd green) - P(1st green OR 2nd green)
= 4/7 + 4/7 - 6/7
= 2/7

### I found this question.

I found this question.
Q: Find the probability of getting 2 Jacks and 1 Queen when 3 cards are selected from a standard deck of cards.

Can you give me the solution for this. ### P(2 Jacks and 1 Queen) = (#

P(2 Jacks and 1 Queen) = (# of ways to get 2 Jacks and 1 Queen)/(TOTAL # of ways to select 3 cards)

DENOMINATOR: TOTAL # of ways to select 3 cards
Since the order in which we select the cards doesn't matter, we can use COMBINATIONS.
We can select 3 cards from 52 cards in 52C3 ways
52C3 = (52)(51)(50)/(3)(2)(1)
= (52)(17)(25)
= 22,100

NUMERATOR: # of ways to get 2 Jacks and 1 Queen

STAGE 1: Select 2 Jacks from 4 possible Jacks
Since the order in which we select the 2 Jacks doesn't matter, we can use COMBINATIONS.
We can select 2 Jacks from 4 Jacks in 4C2 ways (= 6 ways)

STAGE 2: Select 1 Queen from 4 possible Queen
We can complete this stage in 4 ways

Applying the Fundamental Counting Principle, we can complete the two stages in (6)(4) ways
= 24 ways

So.....
P(2 Jacks and 1 Queen) = 24/22,100
= 6/5525

Cheers,
Brent

### A jar contains 4 marbles: 2

A jar contains 4 marbles: 2 red and 2 white. 2 marbles are chosen at random.

Quantity A: The probability that the marbles chosen are the same colour
Quantity B: The probability that the marbles chosen are different colours.

My Approach
For the above question P(Same color) [used AND formula] : 2/4 * 1/3 = 1/6
P(Different color) : 1- P(Same Color) : 5/6
and the answer I got was B
But other site sugget P(Same color) : 1/3
and P(Different color) : 2/3 and answer = B
Kindly correct where I went wrong Question link (with my solution): https://greprepclub.com/forum/a-jar-contains-4-marbles-2-red-and-2-white...

There's a problem with the first line of your solution.

P(same color on both selections) = P(1st marble is ANY color AND 2nd marble matches 1st marble)
= P(1st marble is ANY color) x P(2nd marble matches 1st marble)
= 1 x 1/3
= 1/3

Does that help?

Cheers,
Brent

### can you please tell me why it

can you please tell me why it is 1? IGOT THE SAME SOULTION 1/6 ### Are you referring to the part

Are you referring to the part where I say P(1st marble is ANY color) = 1?

If so, then consider this way of looking at it:
The jar has 4 marbles: 2 red and 2 white
If you randomly select 1 marble, what is the probability that the marble is ANY color?
The probability is 1, since you are GUARANTEED to choose a marble of ANY color.

ASIDE: Here's another way to calculate P(same color on both selections)

P(same color on both selections) = P(1st pick is red and 2nd pick is red OR 1st pick is white and 2nd pick is white)
= P(1st pick is red and 2nd pick is red) + P(1st pick is white and 2nd pick is white)
= (2/4 x 1/3) + (2/4 x 1/3)
= 1/6 + 1/6
= 1/3

Does that help?

Cheers,
Brent

### GRE practice question

GRE practice question (difficulty level: 160 to 170) – Greenlight Test Prep

would you elaborate why P(pick ANY sock 1st) = 1? Isn't it 1/16 instead? In this question, the color of the first sock is irrelevant. In fact, it can be ANY color. Once we've selected the first sock, what matters is whether or not the second sock matches the first sock.

Since the first sock can be ANY color, P(pick ANY sock 1st) = 1

Does that help?

Cheers,
Brent

### Thank you very much for the

Thank you very much for the quick response.

it's not crystal-clear at this point. However, I think that I kind of get the idea. ### Here's another way to put it:

Here's another way to put it:

Let's say that you are randomly selecting a sock from the drawer, and I will give you \$100 if you select ANY sock. What is the probability that you win the \$100?

The probability = 1.

### Hi Brent ,

Hi Brent ,
Unknowingly I had approached the question in video this way
total balls=7
total outcomes when tw must be selected - > 6*7=42
total green balls->4
when two green balls should be selected ->4*3=12

so P(green and green)=12/42

It yields the answer but is it a correct approach

Thanks You're using counting techniques (versus using probability rules), which is fine for many probability questions (for more on this, watch: https://www.greenlighttestprep.com/module/gre-probability/video/759)

In your approach, you're saying that the green balls are all unique, and the red balls are all unique.
We might say G1, G2, G3 and G4 represent the 4 green balls.
And say R1, R2, and R3 represent the 3 red balls.

We have 7 unique balls in total
So, there are 7 ways to select the first ball
And there are 6 ways to select the second ball
So, the TOTAL number of outcomes = (7)(6) = 42

In how many different ways can we select two green balls?
There are 4 unique balls
So, there are 4 ways to select the first green ball
And there are 3 ways to select the second green ball
So, the total number of ways to get two green balls = (4)(3) = 12

So, P(both balls are green) = 12/42 = 2/7

Cheers,
Brent

### Hello Brent,

Hello Brent,

What is it to do with the probability of unfair toss value. I know you have already explained two such solution. but can you explain me verbally.

for example: for the following question
The probability is 0.6 that an “unfair” coin will turn up tails on any given toss. If the coin is tossed 3 times, what is the probability that at least 1 of the tosses will turn up tails?
the solution provided you follows:
We want P(select at least 1 tails)
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 tails) = 1 - P(not getting at least 1 tails)
What does it mean to not get at least 1 tails? It means getting zero tails.
So, we can write: P(getting at least 1 tails) = 1 - P(getting zero tails)
-------------------

P(getting zero tails)
P(getting zero tails) = P(heads on 1st toss AND heads on 2nd toss AND heads on 3rd toss)
= 0.4 x 0.4 x 0.4
= 0.064

So, P(getting at least 1 tails) = 1 - 0.064 = 0.936

where do we actually use the the probability of unfair coin toss. what's the purpose of the value? ### Given: P(unfair coin turns up

Given: P(unfair coin turns up TAILS) = 0.6
This means that: P(unfair coin turns up HEADS) = 1 - 0.6 = 0.4

So, when we get to: P(getting zero tails) = P(HEADS on 1st toss) x P(HEADS on 2nd toss) x P(HEADS on 3rd toss)
We can substitute values to get: P(getting zero tails) = 0.4 x 0.4 x 0.4

Does that help?

### Of the 700 members of a

Of the 700 members of a certain organization, 120 are lawyers. Two members of the organization will be selected at random. Which of the following is closest to the probability that neither of the members selected will be a lawyer?
A. 0.5
B. 0.6
C. 0.7
D. 0.8
E. 0.9

can you please explain should we consider this as with replacement or without replacement. Also for the total number of outcomes is it 700 or 580. can you please clarify these along with solution.

Best Regards,
Revathi Hi Revathi,

Cheers,
Brent

### Hey Brent! Can you help with

Hey Brent! Can you help with this one: Flipping a fair two-sided coin, the probability of getting heads is equal to the probability of getting tails. If the coin is flipped three times consecutively, what is the probability of getting heads at least twice?

Here's what I did:

Complement is: P(not getting head at least twice) = p(getting head only once)

P(getting head one time AND getting tail the second time AND getting tail the third time)= (1/2)*(1/2)*(1/2) = 1/8

Hence, getting a head atleast two times = 1 - 1/8 = 7/8

Where did I go wrong here? ### The mistake occurs here: P

The mistake occurs here: P(NOT getting heads at least twice) = p(getting head only once)

First recognize that, if we flip a coin 3 times, there are four possible cases:
i) 0 heads and 3 tails
ii) 1 heads and 2 tails
iii) 2 heads and 1 tails
iv) 3 heads and 0 tails

In cases iii and iv, we get heads AT LEAST twice.
This means, in cases i and ii, we DON'T get heads at least twice.

Does that help?

### Mind showing the full solve,

Mind showing the full solve, I'm still not getting the right answer when I account for the 0 heads. ### APPROACH #1: Listing and

APPROACH #1: Listing and Counting
In my opinion, the fastest approach is to list all 8 possible outcomes, and then count how many outcomes have at least two heads.
Here are all possible outcomes:
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT

As you can see, 4 of the 8 outcomes have at least two heads.
So, P(at least 2 heads) = 4/8 = 1/2
----------------------------

APPROACH #2: Probability Rules
= P(2 heads and 1 tails) + P(3 heads and 0 tails)

P(2 heads and 1 tails) = P(HHT or HTH or THH)
= P(HHT) + P(HTH) + P(THH)
= (1/2)(1/2)(1/2) + (1/2)(1/2)(1/2) + (1/2)(1/2)(1/2)
= 1/8 + 1/8 + 1/8
= 3/8

P(3 heads and 0 tails) = P(HHH)
= (1/2)(1/2)(1/2)
= 1/8

So, P(at least 2 heads) = 3/8 + 1/8 = 4/8 = 1/2