### You can also solve this

You can also solve this simply by multiplying .9*.9*.1*.1*.1 to get 0.0081 since order doesn't matter. ### 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:
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

Q.N.: A fair coin is to be tossed 5 times. What is the probability that exactly 3 of the 5 tosses result in heads?

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

Or we can also choose the binomial formula:

(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!

### Thanks for the wonderful

Thanks for the wonderful lessons of probability..they were very helpful and cleared many small doubts. Thanks Deepak!

### Does solving this through

Does solving this through combinations work? 5c2= 5*4/1*2= 10. Then 2/10= 0.2. The using compliment 1-0.2= 0.8? ### 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,

Hello Brent,

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

Looking forward,
Thanks
Best Regards,
Revathi ### hey brent! why can't we use

hey brent! why can't we use combinations to find out the number of ways 2 Hs and 3Ts can be arranged? ### 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

But in this case, order does matter right? Like having H1 in the first spot vs having it in the second spot are different variations.. is that accounted for in this combination?(Since order does not matter in a combination). ### 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?