Stat: Binomial Distribution Formula

[Guest]

I had to miss a class and though I have the class notes, I am having trouble understanding how to use this formula. My notes say:

P(x successes from n trials)
= P(x successes and (n-x) failures from n trials)
= P(X = x)
= (n)p^x(1-p)^n-x
.. (x)
x= 0, 1, 2, ..., n

The notes give examples to try, but don't explain how to plug in the information to find the desired outcome.

A fair coin is tossed 10 times and labeled Head as success.
a) Find probablity that all 10 tosses are tails
b) Find probability that half the tosses are heads
c) Find probability that number of heads is at most 2.

Would anyone be willing to take one of the examples (a, b, or c) and show me how I'm supposed to use this formula? I'm happy to do the rest on my own, I just don't even know where to start! Thank you in advance.


-- Alena

 

[pka]

b) Find probability that half the tosses are heads.
In ten tosses half is five. So you will have 5 H’s and 5 T’s.
Thus there are 5 places in the string of 10 to put the H’s.
That is combination of 10 taking 5: \(\displaystyle \L
\left( \begin{array}{c}
10 \\
5 \\
\end{array} \right) = \frac{{10!}}{{(5!)(5!)}}\) ways to do this.

Now multiply by the probabilities: \(\displaystyle \L
\frac{{10!}}{{(5!)(5!)}}\left( {\frac{1}{2}} \right)^5 \left( {\frac{1}{2}} \right)^5\)

 

[Guest]

Thanks very much for answering, and I believe I understand what you said. I worked it out and got a logical answer.. Unfortunately, I apparently am missing the concept because I don't seem to be successful at applying this to the other examples, so I am likely going to have to arrange a meeting with the prof to get clarification.


-- Alena

 

[pka]

Toss a die 15 times, what is the probability of getting 10 sixes?
Probability of success is 1/6, that is p. (1−1/6)=5/6 that is q.
In the string of 15 there are ten places for successes and (15−10)=5 failures, q’s.
The probability of 10 sixes is: \(\displaystyle \L
\left( \begin{array}{l}
15 \\
10 \\
\end{array} \right)\left( {\frac{1}{6}} \right)^{10} \left( {\frac{5}{6}} \right)^5\).

I we shoot at a target 20 times with 75% hit rate, the probability of exactly 12 hits is:
\(\displaystyle \L
\left( \begin{array}{l}
20 \\
12 \\
\end{array} \right)\left( {0.75} \right)^{12} \left( {0.25} \right)^8\).
The probability of at least 12 hits is:
\(\displaystyle \L
\sum\limits_{k = 12}^{20} {\left( \begin{array}{c}
20 \\
k \\
\end{array} \right)\left( {0.75} \right)^k \left( {0.25} \right)^{20 - k} }\)