Hi

how do i do the following

Binomial random variable
number of trails = 10
Pr success 0.3

what is the probability of

P(X=3)
P(X<=3)
P(X>3)
P(0<=X<5)

please show me the steps
makbool

Hi makbool!!

First of all read this document for reference and formulas:
"BERNOULLI AND BINOMIAL RANDOM VARIABLES":
http://www.math.nus.edu.sg/~matlimch/CS1232-EIGHT.pdf

After read this you must know that for the random variable X -->
Bin(n, p) the probability that X is a certain value k is given by the
formula:

p(X = k) = C(n, k).p^k.(1 − p)^(n−k) , (k = 0, 1, ..., n).

where C(n, k) = n! / k!.(n - k)!

For this problem we have:  X --> Bin(10, 0.3)
Note: visit this page to calculate Combinations:
"Permutations and Combinations Calculator":
http://fclass.vaniercollege.qc.ca/web/mathematics/real/Calculators/PermsCombs_Intro.htm

a) P(X=3) = C(10, 3) . 0.3^3 . 0.7^7 =
          = 120 . 0.027 . 0,0823543 =
          = 0.266828

b) P(X<=3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)

P(X=0) = C(10,0) . 0.3^0 . 0.7^10 = 
       = 1 . 1 . 0,0282475249 =
       = 0.0282475249

P(X=1) = C(10, 1) . 0.3^1 . 0.7^9 =
       = 10 . 0.3 . 0.0282475249 =
       =  0.0847425747

P(X=2) = C(10, 2) . 0.3^2 . 0.7^8 =
       = 45 . 0,09 . 0,05764801 = 
       = 0.2334744405

P(X=3) = 0.266828

then
P(X<=3) = 0.0282475249 + 0.0847425747 + 0.2334744405 + 0.266828 =
        = 0.6132925401


c) P(X>3) = 1 - P(X=3) = 1 - 0.6132925401 = 0,3867074599


d) P(0<=X<5) = P(X<=3) + P(X=4) =

P(X<=3) = 0.6132925401
          
P(X=4) = C(10, 4) . 0.3^4 . 0.7^6 =
       = 210 . 0.0081 . 0.117649 =
       = 0.200120949 

P(0<=X<5) = 0.6132925401 +  0.200120949  = 0.8134134891


I hope this helps you. Also I hope that the numbers are right, but the
step by step solution (your requested answer) is showed.
If you need a clarification feel free to post a request of an answer
clarification.

Best regards.
livioflores-ga

---

Clarification of Answer by livioflores-ga on 21 Aug 2003 02:02 PDT

Oops!!

I donīt know what happens but this is the formula for the probability
that X is a certain value k for the random variable X --> Bin(n, p):
 
p(X = k) = C(n, k).p^k .(1 - p)^(n - k) , (k = 0, 1, ..., n).

Hope this clarify the answer.
livioflores-ga

---

Clarification of Answer by livioflores-ga on 21 Aug 2003 06:40 PDT

Hi makbool!!
Please excuse me for the following mistakes:

The problem b) has errors in the numbers and I will right it again
here:
b) P(X<=3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) 
 
P(X=0) = C(10,0) . 0.3^0 . 0.7^10 =  
       = 1 . 1 . 0,0282475249 = 
       = 0.0282475249 
 
P(X=1) = C(10, 1) . 0.3^1 . 0.7^9 = 
       = 10 . 0.3 . 0.040353607 =  (here I put 0,0282475249)
       =  0.121060821      (the result change)  

Note: this copy and paste fault lead to several bad results.        
 
P(X=2) = C(10, 2) . 0.3^2 . 0.7^8 = 
       = 45 . 0,09 . 0,05764801 =  
       = 0.2334744405 
 
P(X=3) = 0.266828 
 
then 
P(X<=3) = 0.0282475249 + 0.121060821 + 0.2334744405 + 0.266828 = 
        = 0.6496107864  (here is a new result) 


There is a typo in the problem c):
I wrote P(X>3) = 1 - P(X=3) instead of P(X>3) = 1 - P(X<=3), despite
this typo the answer is right, I only forgot to write the symbol '<' .

Also I used the bad number for this problem:
I wrote P(X>3) = 1 - P(X<=3) = 1 - 0.6132925401 = 0,3867074599 , and
the correct result is:
P(X>3) = 1 - P(X<=3) = 1 - 0.6496107864  = 0.3503892136 .

This also lead me to commit a mistake in the exercise d) where I used
the wrong value of P(X<=3), the correct result is:
P(0<=X<5) = 0.6496107864 + 0.200120949  = 0.8497317354

Please excuse for these errors, they was originated in a copy and
paste fault.
Thank you for your understanding.
livioflores-ga

Thanks for yours clarification