Expert: Abe Mantell - 7/22/2009

Question
QUESTION: when tossing a unbiased ,find the probaility that
1)the first head occurs on the third trial
2)at most three trials are neccesary to observe a head .
(use geometric distribution)
solution i tried plz check 2nd part can u check .
geometric distribution i,e
g(x,p)=p.(1-p)^x-1 where x=0,12...n
1).g(3,.5)=(.5)*(.5)^2
      =.125
2).in second we find g(3,.5)+g(2,.5)+g(1,.5)
         .125+.25+.5=.875

ANSWER: 1. P(TTH)=(1/2)(1/2)(1/2)=1/8=0.125
2. P(H)+P(TH)+P(TTH)=sum((1/2)^k (1/2), k=0..2)=(1/2)+(1/2)^2+(1/2)^3
.  =1/2+1/4+1/8=7/8=0.875

Looks fine to me!

Abe


---------- FOLLOW-UP ----------

QUESTION: a telephone exchange receives,on average ,five calls per minute.
find the probability that 1).in a one-minute period ,no calls are received.
         2).in a two -mins period fewer than four calls are received .
3).out of separate one-minute periods,there are exactly  four in which two or more calls are received.


Answer
1. P(x=0)=(e^-5)(5^0)/0!=e^-5 or about 0.006738
2. Let x=# of calls for the first minute, and y=# of calls for the second minute.
.  Here are the ways x+y<4, (x,y):
.  (0,0),(1,0),(0,1),(1,1),(2,0),(0,2),(2,1),(1,2),(3,0) or (0,3)
.  Now just calculate each one, then add them to get the total probability.
.  P(x=0)P(y=0)+P(x=1)P(y=0)+...+P(x=0)P(y=3)...which turns out to be (683/3)*(exp(-10))
.  or about 0.01034
3. How many separate periods?

Abe