Probability- Help!!

L

luvugodm

New member
Joined
May 17, 2007
Messages
11
One more question and maybe I will be ready for my exam.


List all the possible outcomes of the four flips I have started it and then I get confused.
hhhh
hhht
hhth
hthh
httt
htth tttt thhh tthh ttth
Then I get confused,
What is the total possible outcomes?
Then how do I figure the probability of two of the flips come up heads
thank you whoever can show me how to do this. Probability has been the hardest thing in algebra for me
 
The number of outcomes is \(\displaystyle 2^{n}\), where n is the number of coins. In your case n=4. I'll leave the listing up to you.

You can use the binomial probability to find whatever you need.

For the probability of 2 heads from the 4 flips: \(\displaystyle \L\\C(4,k)(0.5)^{k}(0.5)^{4-k}\), for k=2.

Once you calculate the probability and list all the outcomes, they should match. You can count the number of 2 tail outcomes from the list. Divide the number of them by the total number of outcomes and it should match the result from your binomial calculation. See?. It ain't hard. :wink:
 
Re: probability- Help!!

luvugodm said:
I am still lost. I am not understanding this at all
Click to expand...
You've been given the exact formula, and the exact value to plug in. How can you have no understanding of any portion of this? This makes it sound as though you're waiting for that one guy to give you the complete worked solution again, which surely isn't what you mean.

Please reply with specifics. Thank you.

Eliz.
 
Would this be right
hhhh
hthh
htth
httt
tttt
thhh
tthh
ttth
thth
hhtt
hhht
hhth
htht
thht
ttht
thtt
Total of 16 outcomes
Is this correct
 
then to find out if exactly two of the flips come up heads would that be 6/16= 3/8
 
Did you notice the relationship?. How many out of the 16 had 2 t's?. 6 of them. Right?. 6/16=3/8.

Now, what did the binomial give you?. 3/8. See?.
 
Hello, luvugodm!

There is a simple routine for this type of Listing . . .

List all the possible outcomes of the four flips.
Click to expand...

First, each of the four flips has two possible outcomes (H or T).
. . Hence, there are: \(\displaystyle \,2^4\:=\:16\) outcomes.

Now make a four-coumn chart.
In the first column, make the first half H's and last half T's: \(\displaystyle \;\begin{array}{c}H\\H\\H\\H\\H\\H\\H\\H\\T\\T\\T\\T\\T\\T\\T\\T\end{array}\)
In the second column, change every four terms: \(\displaystyle \;\;\;\;\begin{array}{c}H\\H\\H\\H\\T\\T\\T\\T\\H\\H\\H\\H\\T\\T\\T\\T\end{array}\)
In the third column, change every two terms: \(\displaystyle \;\;\;\begin{array}{c}H\\H\\T\\T\\H\\H\\T\\T\\H\\H\\T\\T\\H\\H\\T\\T\end{array}\)
In the fourth column, alternate H's and T's: \(\displaystyle \;\begin{array}{c}H\\T\\H\\T\\H\\T\\H\\T\\H\\T\\H\\T\end{array}\)


Of course, we write them in one table:

. . \(\displaystyle \begin{array}{cccc}H & H & H & H \\H & H & H & T \\ H & H & T & H \\ H & H & T & T \\ H & T & H & H \\ H & T & H & T \\ H & T & T & H \\ H & T & T & T \\
T & H & H & H \\ T & H & H & T \\ T & H & T & H \\ T & H & T & T \\ T & T & H & H \\ T & T & H & T \\ T & T & T & H \\ T & T & T & T\end{array}\)

And we have listed the 16 possible outcomes.

 
You must log in or register to reply here.
Top