Question: 2 heads

Comment on 2 heads

You can also solve this simply by multiplying .9*.9*.1*.1*.1 to get 0.0081 since order doesn't matter.
greenlight-admin's picture

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

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 tossed 5 times. What is the probability that exactly 3 of the 5 tosses result in heads?

greenlight-admin's picture

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 binomial formula:

(n choose k)Xp^kX(1-p)^n-k
(5 choose 2)X0.9^2X0.1^3
greenlight-admin's picture

That works too!

Thanks for the wonderful lessons of probability..they were very helpful and cleared many small doubts.
greenlight-admin's picture

Thanks Deepak!

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?
greenlight-admin's picture

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 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?

Looking forward,
Best Regards,
greenlight-admin's picture

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

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 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).
greenlight-admin's picture

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 understand why do we need to add all 10 different orders, since at which place does the head turn up doesn't matter? why is the answer not 0.9*0.9*0.1^3 = 0.00081?
greenlight-admin's picture

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

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