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## Comment on

2 heads## Why not A=0.00081

0.9 x 0.9 x 0.1 x 0.1 x 0.1

## That is the probability ONE

That is the probability ONE particular outcome: 1st toss is heads, 2nd toss is heads, 3rd toss is tails, 4th toss is tails, and 5th toss is tails. However, you need to consider ALL outcomes that yield 2 heads and 3 tails, like H-T-T-H-T

The video explains how to handle ALL outcomes.

## This is quite confusing. I

Does HHTTT different with HTHTT? Does order matter here?

## yes, HHTTT and HTHTT are

yes, HHTTT and HTHTT are different outcomes.

## Hello,

Can we consider another way to evaluate the number of possible outcomes? (#possible arrangements of "H" and "T"). I started with 5C2 = 10 and avoided the missisipi rule. it's not really saving time, but was wondering if this could be a valid alternative approach.

Thx,

Ben

## Yes, that's a valid approach

Yes, that's a valid approach if you are only arranging TWO types of objects (in this case, H's and T's).

You are essentially saying "We have 5 spaces, and we want to place H's in 2 of those spaces. We can select 2 spaces (to place the H's) from 5 spaces in 5C2 ways."

ASIDE: You'll find that your calculations look identical to the calculations used in the MISSISSIPPI approach.

## Hi Brent,

Denominator will be 5c2 but will be the numerator by this method? Can you explain how will you solve using missippi approach?

## We can use combinations (5C2)

We can use combinations (5C2) to determine the number of different ways to get exactly 2 heads. However, we can't continue using counting techniques to determine the denominator, because the outcomes are NOT equally likely.

At 1:20 of the Intro to Probability video (https://www.gmatprepnow.com/module/gmat-probability/video/742), I note that we can use the general probability formula AS LONG AS the outcomes are equally likely.

In the video question above, each outcome is not equally likely.

For example, getting 5 heads on 5 flips is MUCH MORE LIKELY than getting 5 tails on 5 flips.

Does that help?

Cheers,

Brent

## WHen we use combinations, we

## When we use counting methods

When we use counting methods to solve Probability questions, we must be sure that the outcomes are equally likely.

Let's take this concept to the extreme.

A bag contains 998 blue balls and 2 red balls.

If we randomly choose 2 balls (without replacement), what is the probability that both balls are red?

If we say that "order does not matter," we MIGHT conclude that there are only 3 possible outcomes:

1) both balls are red

2) both balls are blue

3) 1 ball is red and 1 ball is blue

So, P(both red) = 1/3

The problem here is that each of the 3 outcomes are NOT equally likely.

If I were to generalize, I'd suggest two approached:

1) If you have the option of using counting methods or probability rules to solve a question, the better/faster/more accurate approach is usually to use probability rules

2) If you are going to use counting methods to calculate the numerator and denominator, it's often better (if possible) to do so as though order matters.

By the way, on the topic of applying probability rules, here's the solution to the above question:

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

= P(1st ball is red) x P(2nd ball is red)

= 2/1000 x 1/999

ASIDE: When we're using counting methods to answer Probability questions, we can sometimes calculate the numerator and denominator two ways:

1) as though order matters

2) as though order does not matter

What matters is that we use the same counting approach for both the numerator and the denominator

Cheers,

Brent

## Question asks probability of

In every case probability will be the same. i.e 0.00081.

We never consider sequence in ball's question such as selection of exactly 2 G balls from set of 3G and 4R.

## Yes, you must consider every

Yes, you must consider every possible way in which we can get exactly 2 heads.

Since there are 10 ways to get exactly 10 heads, we must add 0.00081 ten times to find the final probability.

## ohh...i marked A

## Why does ORDER matter in coin

## If you try to treat order as

If you try to treat order as NOT mattering, it becomes very hard to count outcomes.

For example, if we toss a coin two times, there are 4 possible outcomes:

1) heads on 1st toss and heads on 2nd toss

2) heads on 1st toss and tails on 2nd toss

3) tails on 1st toss and heads on 2nd toss

4) tails on 1st toss and tails on 2nd toss

IMPORTANT: Each of the above outcomes is equally likely.

e.g., P(heads on 1st toss and heads on 2nd toss) = P(tails on 1st toss and tails on 2nd toss)

What happens if we say order does not matter?

We get 3 outcomes:

1) 2 heads

2) 2 tails

3) 1 head and 1 tail

This, however, does NOT mean that P(1 head and 1 tail) = 1/3, because the 3 outcomes are NOT EQUALLY LIKELY.

Does that help?

Cheers,

Brent

## Brent, wont 5C2 be a better

## It turns out that the 5C2

It turns out that the 5C2 calculation is IDENTICAL to the MISSISSIPPI rule calculation (5!/2!3!)

Here's why:

We need to get 2 Heads and 3 Tails.

So, some favorable outcomes include THTTH, TTHHT, HTTTH, etc

One way to determine the number of favorable outcomes is to apply the MISSISSIPPI rule.

We want to arrange 5 letters, of which 2 are identical H's and 3 are identical T's

We get: 5!/2!3! = 10

Another way to look at it is to recognize that we have 5 possible positions (1st toss, 2nd toss, . . . 5th toss), and we want to select 2 of those slots to hold T's (tails). Whatever slots are left not selected will be later filled with H's.

So, we can select 2 of the 5 slots in 5C2 ways.

5C2 = 5!/2!3! = 10

Cheers,

Brent

## Can we also say that we want

5C3 = 5!/3!2! = 10

are they interchangable? It is hard to know whether you picked the correct one..

## Yes, they're interchangable.

Yes, they're interchangable.

Here's an analogous question:

I have 5 cars and I'm going to paint 2 of them green and 3 of them red.

In how many ways can I do this?

Well, I can select 2 cars to paint green (5C2 = 10), and then paint the remaining cars red.

Or I can select 3 cars to paint red (5C3 = 10), and then paint the remaining cars green.

In both cases, the outcomes are the same.

Cheers,

Brent

## Hi Brent,

I saw a probability formula on GMAT Club for problems including a combination of independent and mutually exclusive events. The formula goes as follows:

P = nCr*p^r(1-p)^(n-r)

In this question,

r = 2 (heads or tail)

n = number of time the coin tossed

p = 0.9

5C2 = 10

0.9^2 = 0.81

(1-0.9)^(5-2) = 0.001

Hence P=10*0.81*0.001

Which is,

P = 0.0081

Please let me know where I'm wrong.

PS: There is another variation of the formula which says we do not need to use nCr when sequence doesn't matter and it goes like this:

P = p^r(1-p)^(n-r)

This formula gives us the correct answer which is 0.00081. But I'm failed to understand the usage as sequence matters in above-mentioned question and hence we have to count combination that is nCr.

## P = nCr*p^r(1-p)^(n-r) is

P = nCr*p^r(1-p)^(n-r) is known as the binomial probability formula

You have correctly used it to get 0.0081, which is the correct answer (the correct answer is B, not A).

Cheers,

Brent

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