# Thread: Conditional probability with Binomial distribution

1. ## Conditional probability with Binomial distribution

Hello all

I am trying to make a calculation here, and the result doesn't make sense to me, I don't know what I am doing wrong...

I have a random group of 23 people. Each of them has independently a probability of 6% to have some event. I want to know, what is the probability of having at least 2 events, knowing that there is at least one.

I did:

X~Bin(23,0.06)

P(X>1|X>0) = P(X>1 AND X>0)/(P(X>0)=P(X>1)/P(X>0)=1-P(X<=1)/1-P(X=0)=

=1-[P(X=0)+P(X=1)]/1-P(X=0)=1-[0.24+0.3537]/1-0.24=

=1-0.5937/1-0.24 = 0.4063/0.76 = 0.534 = 53.4%

I don't get it. The probability of an event is 0.06. I already know there was 1, so does it makes sense that I have 53% of having more than one ? I though the number will be much smaller.

Are my calculations makes sense ? Is there a way, maybe using Bayesian Statistics to do it better ?

Thanks !

Edit: I didn't give you all details. I know that there is at least one event, because after 17 people there was one event. There are 6 to go, and I want to know the probability of having more events

2. ## Re: Conditional probability with Binomial distribution

hi,
I believe this could make sense. The way I would explain it is thatbthe probability is the number of "successes" divided by the number of possible outcomes, right? If you already had one event, then the number of possible outcome is proportional to that smaller area of the binomial distribution that lies to the left of the value one and the number of successes is proportional to the area of the distribution that lies to the left of the value 2. So, this proportion can easily be a lot larger then 0,06 .

BTW, I just had the exact same question yesterday from a colleague - how probable is that?

regards
rogojel

3. ## Re: Conditional probability with Binomial distribution

Originally Posted by NN_STAT
I don't get it. The probability of an event is 0.06. I already know there was 1, so does it makes sense that I have 53% of having more than one ? I though the number will be much smaller.
The probabilities for 2 and 3 are still quite large so it's not surprising to me that this occurs.
Are my calculations makes sense ? Is there a way, maybe using Bayesian Statistics to do it better ?
You aren't doing statistics at the moment - just probability. So I guess the answer is no?

Edit: I didn't give you all details. I know that there is at least one event, because after 17 people there was one event. There are 6 to go, and I want to know the probability of having more events
Well that changes things and you aren't calculating the correct probability for the question of interest anymore. Really you just want to know if there will be any success in the next six trials. You're just looking for P(X > 0) where X ~ Bin(6, .06).

4. ## Re: Conditional probability with Binomial distribution

well Dason, what confuses me is that I know the Geometric random variable has the "lack of memory" characteristic, I didn't know the Binomial does too. That's why I didn't follow the intuition of X~Bin(6,0.06)

5. ## Re: Conditional probability with Binomial distribution

It doesn't have that property. But that's not what you're talking about. You've split your trial up into two different independent sections - the first 17 observations and the last 6 observations.

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