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?
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?
Hello, solace!
They don't expect you to list all the possible outcomes and count them up.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 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