# probability distribution for 6 fair coins

• May 18th 2007, 05:01 PM
solace
probability distribution for 6 fair coins
In a certain experiment, 6 fair coins are flipped. Let the random variable x be the number of heads observed. Construct the probability distribution for the experiment.

The only way i can think to do this may take FOREVER.

Here is my method

1-1-1-1-1-1, 1-2-1-1-1-1, 1-1-2-1-1-1, 1-1-1-2-1-1, 1-1-1-1-2-1, 1-1-1-1-1-2, 1-2-2-1-1-1, 1-1-2-2-1-1, ... 6-6-6-6-6-6

Etc!!!

Is there a more efficient method to create this probability distribuition than listing all (who knows how many) possiblities out?
• May 18th 2007, 06:40 PM
solace
Oops sorry i was thinking for some reason that i was dealing with 6 dice.

Listing all possible things for 6 coins consisting of only heads and tails shouldnt be too bad.

My mistake.

Here is my new question: How many possible outcomes are there?
• May 19th 2007, 04:19 AM
Plato
Quote:

Originally Posted by solace
Listing all possible things for 6 coins consisting of only heads and tails shouldnt be too bad.
Here is my new question: How many possible outcomes are there?

2^6=64
• May 19th 2007, 11:04 AM
Soroban
Hello, solace!

Quote:

In a certain experiment, 6 fair coins are flipped.
Let the random variable x be the number of heads observed.
Construct the probability distribution for the experiment.

They don't expect you to list all the possible outcomes and count them up.
. . They probably think that anyone can do that ... given enough time.
Try some Thinking.

You're expected to know that are: 2^6 = 64 possible outcomes.

How many ways are there to get 0 heads?
. . There is one way: TTTTTT

How many ways are there to get 1 head (and five tails)?
. . There are six ways . . . why?
Because the one head can appear in any of six positions.

How many ways are there to get 2 heads (and four tails)?
. . There are C(6,2) = 15 ways. .(Do you see why?)

How many ways are there to get 3 heads (and three tails)?
. . There are C(6,3) = 20 ways.

By now, these values may (or may not) seem familiar to you.
They are the cofficients for the expansion of (a + b)^6.
Pascal's Triangle gives us: .1, 6, 15, 20, 15, 6, 1

Therefore, if x = number of heads, we have:
. . . P(0)
= 1/64
. . . P(1)
= 6/64
. . . P(2)
= 15/64
. . . P(3)
= 20/64
. . . P(4)
= 15/64
. . . P(5)
= 6/64
. . . P(6)
= 1/64