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## Comment on

2 heads## You can also solve this

## That's not quite correct.

That's not quite correct.

Your equation only accounts for one of many possible cases. That is, your equation is for calculating the probability of Heads - Heads - Tails - Tails - Tails

You also need to consider other cases like:

- Tails - Heads - Heads - Tails - Tails

- Tails - Tails - Heads - Tails - Heads

- Tails - Tails - Tails - Heads - Heads

etc.

If you evaluate your equation (.9*.9*.1*.1*.1), you get 0.00081 (not 0.0081)

## Q.N.: A fair coin is to be

7⁄32

1⁄4

5⁄16

3⁄8

7⁄16

## Let's examine ONE case in

Let's examine ONE case in which we get exactly 3 heads: HHHTT

P(HHHTT) = (1/2)(1/2)(1/2)(1/2)(1/2) = 1/32

This, of course, is just ONE possible way to get exactly 3 heads.

Another possible outcome is HHTTH

Here, P(HHTTH) = (1/2)(1/2)(1/2)(1/2)(1/2) = 1/32

As you might guess, each possible outcome will have the same probability (1/32). So, the question becomes "In how many different ways can we get exactly 3 heads and 2 tails?"

In other words, in how many different ways can we arrange the letters HHHTT?

Well, we can apply the MISSISSIPPI rule (from the counting module) to see that the number of arrangements = 5!/(3!)(2!) = 10

So P(exactly 3 heads) = (1/32)(10) = 10/32 = 5/16

## Or we can also choose the

(n choose k)Xp^kX(1-p)^n-k

(5 choose 2)X0.9^2X0.1^3

10X0.81X0.001

0.0081

## That works too!

That works too!

## Thanks for the wonderful

## Thanks Deepak!

Thanks Deepak!

## Does solving this through

## I'm not sure what role the 2

I'm not sure what role the 2 (as in 2/10) plays in your solution.

Can you show me all of the steps in your solution?

## Hello Bernt,

Can you please best explain the following problem.

Eight points are equally spaced on a circle. If 4 of the 8 points are to be chosen at random, what is the probability that a quadrilateral having the 4 points chosen as vertices will be a square?

https://gre.myprepclub.com/forum/eight-points-are-equally-spaced-on-a-circle-if-4-of-the-8-p-1918.html

Looking forward,

Thanks

Best Regards,

Revathi

## Here's my full solution:

Here's my full solution: https://gre.myprepclub.com/forum/eight-points-are-equally-spaced-on-a-ci...

## hey brent! why can't we use

## Great question!!!

Great question!!!

We most certainly can use combinations here to determine the number of different ways to arrange 2 H's and 3 T's. Think of it this way:

We have 5 spaces in which we'll place 2 H's and 3 T's

Let's select the 2 places where the H's will go.

We can do this in 5C2 ways (10 ways)

Once the H's are placed, the T's will go in the remaining three spaces, and we're done!

You may also notice that, if we use the Mississippi rule, number of ways to arrange 2 H's and 3 T's = 5!/(2!)(3!), and if we use combinations, the number of Arrangements = 5C2, which also equals 5!/(2!)(3!)

## But in this case, order does

## Another great question!

Another great question!

Let's call the 5 spaces spaceA, spaceB, spaceC, spaceD, and spaceE

We want to select 2 of those 5 spaces (where we will place the two H's)

Does the order in which we SELECT the two spaces does not matter, we can use combinations.

In other words, selecting spaceB and then spaceD is the same as selecting spaceD and then spaceB

Does that help?

## Hi Brent! i don't quite

## HHTTT represents heads on 1st

HHTTT represents heads on 1st toss, heads on 2nd toss, tails on 3rd toss, tails on 4th toss, and tails on 5th toss.

This outcome is different from HHHTT, and different from HTHTH, etc