Conditional probability problem - help need

  • Thread starter ORACLE
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  • #1
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hi

I got a stats problem infornt of me. I figured out that it is abaut conditional probability. But I am stuck :confused: .

# hurricanes 0 1 2 3 4 5 6
probability .25 .33 .24 .11 .04 .02 .01

prob >6 is 0
questions are independent.
a.) what is prob 2< X <5?
b.) 2 hurr are already observed. what is the prob for 4 or more?
c.) 1 hurr is already observed. what is the prob for at least 1 or more?
thanks
 

Answers and Replies

  • #2
arildno
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Let us take b) as an example:

Now, we must have P(4)=P(2)*P(4|2)
P(4|2) is the one you are looking after; that is:

P(4|2)=P(4)/P(2)=0.04/0.24=4/24=1/6.
 
  • #3
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thanks arildno

but one doubt.
as the question asks for prob of 4 or more, do i have to find for 5 and 6 the same way and add the probs?
 
  • #4
arildno
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Dear oh dear, I made a grievous mistake here!

In the conditional probability, I should have used the "probability of getting a first hurricane", P(>=1), rather than "the prob of getting exactly 1 hurricane", P(1).
Sorry about that!

Now, P(>=1) equals the sum of the probabilities of non-zero events, i.e P(>=1)=0.75.
In addition, I misreadread in b) the probability to be that of EXACTLY 4, rather than 4 or more.

Thus, in b) P(>=4|>=1)=P(>=4)/P(>=1), P(>=4)=0.07

And in c)
P(>=2)|>=1)=P(>=2)/P(>=1), P(>=2)=0.42.
 
  • #5
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thanks a lot arildno
 

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