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## Comment on

Guessing Strategies## Hi Brent!

For a certain probability experiment, the probability that event A will occur is 1/2 and the probability that event B will occur is 1/3. Which of the following values could be the probability that the event AUB (that is, the event A or B, or both) will occur?

indicate all such values

a)1/3 b)1/2 c)3/4

for P(A or B) i did like P(A)*P(~A) + P(B)*P(~B)

1/2 * 2/3 + 1/3 * 1/2 for that i got 3/4,

But in ETS quant guide b) and c) both are answers.

Please let me know about P(AUB) (A union B) and how b) is also answer.

you have explained about P(A intersection B), both A and B can occur, in mutually exclusive function.

## We want P(A or B)

We want P(A or B)

P(A or B) = P(A) + P(B) - P(A and B)

= 1/2 + 1/3 - P(A and B)

= 5/6 - P(A and B)

This means P(A or B) must be less than or equal to 5/6. At this point, it comes down to the value of P(A and B)

However, we can be certain that P(A or B) cannot be less than P(A), and P(A and B) cannot be less than P(B). How do we know this?

Well, if P(A) = 1/2, then it cannot be the case that P(A OR B) is less than 1/2.

In other words, P(A or B) cannot be less than 1/2, and P(A or B) cannot be less than 1/3

Since 1/3 < 1/2, we can conclude that P(A or B) cannot be less than 1/2

So, P(A or B) must be less than or equal to 5/6 AND greater than or equal to 1/2

This allows us to ELIMINATE answer choice A, and we are left with answer choices B and C

ASIDE: Here's an experiment in which P(A or B) = 1/2

Select ONE number from the set: {1, 2, 3, 10, 15, 20}

Event A: The selected number is a multiple of 5

Event B: The selected number is a multiple of 10

Here, P(A) = 3/6 = 1/2 and P(B) = 2/6 = 1/3

Also, P(A and B) = 2/6 = 1/3

So, P(A or B) = P(A) + P(B) - P(A and B)

= 1/2 + 1/3 - 1/3

= 1/2

Does that help?

## " Well, if P(A) = 1/2, then

In other words, P(A or B) cannot be less than 1/2, and P(A or B) cannot be less than 1/3

Since 1/3 < 1/2, we can conclude that P(A or B) cannot be less than 1/2

So, P(A or B) must be less than or equal to 5/6 AND greater than or equal to 1/2 "

above what you explained was crucial to solve this, but i am quite not sure how P(A or B) can not be less than P(A) 1/2 and P(B) 1/3.

## If I add another possible

If I add another possible event to the probability, then the probability of one event OR another event cannot be less than the probability of one even alone. Here's an example:

Let's say the probability is 0.5 that it rains tomorrow, and let's also say the probability is 0.3 that my dog eats a bee tomorrow.

What is the probability that it rains tomorrow OR my dog eats a bee tomorrow OR both? That is P(rains OR dog eats bee tomorrow)

Well, P(rains tomorrow) = 0.5

No matter what, there's a 50% likelihood that it rains tomorrow. So, adding a second possible event (e.g., dog eating a bee) must be AT LEAST AS likely as it raining tomorrow.

That is, P(rains OR dog eats bee tomorrow) must be greater than or equal to P(rain tomorrow).

Put another way: Let's say I'll give you one million dollars if a certain event happens. Furthermore, I'm going to give you three cases to choose from:

case a: It rains tomorrow

case b: My dog eats a bee tomorrow

case c: It rains tomorrow OR my dog eats a bee tomorrow

Which of these cases is most likely to occur? Well, P(rain tomorrow) = 0.5. So, P(rain tomorrow OR dog eats a bee tomorrow cannot be less than 0.5

Does that help?

## I always get confused with P

## In the OR probability formula

In the OR probability formula, we subtract P(A and B) in order to account for any duplication that occurs with P(A) + P(B).

Let's examine an example:

One number is randomly selected from the set {3, 5, 6, 15}

This means P(prime) = 2/4 = 1/2

And P(odd) = 3/4

Now that's say we want to find P(prime OR odd)

As you can imagine, adding 1/4 and 3/4 is not the way to calculate P(prime OR odd), since the resulting probability is greater than 1.

We get a probability greater than the one because 3 and 5 satisfy the condition that the number is prime, AND the same values satisfy the condition that the number is odd.

In other words, 3 and 5 have been considered twice

So, in order to find P(A or B), we need to subtract the probability that both events happen.

In other words, P(prime OR odd) = P(prime) + P(odd) - P(prime AND odd)

At this point we need to find P(prime AND odd)

There are two ways in which we can calculate P(prime AND odd)

Option i: Recognize that 2 of the 4 numbers are both prime and odd.

So, P(prime AND odd) = 2/4 = 1/2

Option ii: Use one of the two formulas

Since the two events are NOT independent we'll use the conditional probability.

So, P(prime AND odd) = P(prime) x P(odd|prime)

= 2/4 x 1

= 1/2

So, if P(prime OR odd) = P(prime) + P(odd) - P(prime AND odd)

We can write: P(prime OR odd) = 1/2 + 3/4 - 1/2 = 3/4

ASIDE: P(odd|prime) = 1, because if it's given that the number is prime, we know that the number must be either 3 or 5.

If we know the number is either 3 or 5, and the probability is 1 (guaranteed) that the number is odd.

Does that help?

## Oh wow!, I have understood

## Why have you said that answer

## Hi Deepak,

Hi Deepak,

That's not quite what I say in the video. I ask "Do you feel that the probability is greater than or less than 0.5?"

If you feel that the probability is GREATER THAN 0.5, then we can ELIMINATE answer choices A and B since those values are less than 0.5.

Cheers,

Brent

## I want to be member

## Great - here's where to start

Great - here's where to start: https://www.greenlighttestprep.com/prices

Cheers,

Brent

## I solved this question using

My question is how can we tackle this question with other approaches?

I used the "Or Probability Approach"

Can they occur together? No, Exclusive

With the "Or Approach", I got or tried ... :

P(A (1 tails) or B (2 tails) or C (3 tails) ) = P(A) + P(B) + P(C) - P(A and B and C)

What must occur?

P(A) => Tails, Heads, Heads => (0.3)(0.7)(0.7) = 0.147

P(B) => Tails, Tails, Heads => (0.3)(0.3)(0.7) = 0.063

P(C) => Tails, Tails, Tails => (0.3)(0.3)(0.3) = 0.027

P(A (1 tails) or B (2 tails) or C (3 tails) ) = 0.147 + 0.063 + 0.027 - 0

P(A (1 tails) or B (2 tails) or C (3 tails) ) = 0.237 ???

Which I am sure I am wrong

## For P(A) and P(B), you are

For P(A) and P(B), you are missing some possible cases.

For example, P(A) = P(1 tails)

There are 3 different ways to get exactly 1 tails:

#1) Tails, Heads, Heads

#2) Heads, Tails, Heads

#3) Heads, Heads, Tails

In your solution, you looked at #1 only.

P(#1) = 0.147 (as per your calculations)

Also, P(#2) = 0.147

And P(#3) = 0.147

So, P(A) = 0.147 + 0.147 + 0.147 = 0.441

------------------------------------

The same applies to P(B) = P(2 tails)

You calculated P(Tails, Tails, Heads) but you didn't calculate the following:

P(Tails, Heads, Tails) = 0.063

P(Heads, Tails, Tails) = 0.063

So, P(B) = 0.063 + 0.063 + 0.063 = 0.189

-----------------------------

Your calculation of P(C) is fine, since there's only one way to get P(all tails)

----------------------------

So, P(A or B or C) = 0.441 + 0.189 + 0.027

= 0.657

Does that help?

Cheers,

Brent

## Can you solve this using

## You bet!

You bet!

P(at least 1 tails) = 1 - P(no tails)

= 1 - P(all 3 tosses are heads)

---------------------------

P(all 3 tosses are heads) = P(1st toss is heads AND 2nd toss is heads AND 3rd toss is heads)

= P(1st toss is heads) x P(2nd toss is heads) x P(3rd toss is heads)

= 0.7 x 0.7 x 0.7

= 0.343

---------------------------

So, P(at least 1 tails) = 1 - 0.343 = 0.657

Cheers,

Brent

## This is really an amazing way