Conditional Binomial Distribution

[mrmt]

Hi guys,

I can't get my head around this, if anyone could help that would be great.

"A robotic assembly line contains 20 stations. Suppose that the probability
that each individual station will fail is 0.3 and that the stations fail indepen-
dently of each other. Given that at least one of the stations has failed, what is
the probability that at least 2 of the stations have failed?"

Binomial distribution as:
fixed no. of identical trials
only 2 outcomes (success/failure)
the trials are independent

Slightly confusing wording, but if we call a "success" failure of a station, what is:

P(X≥2 l X≥1) ?

First I played around with just a reduced sample space i.e. for X≥2 out of the original 20 stations give that X≥1 already, would mean that X≥1 out of the remaining 19? i.e. 1-P(X=0)

(For the above I assumed that although the wording says "at least one station" has already failed, that would have to mean "only one station has failed" as if any more that one station had failed already the probability that X≥2 would be 1 (a certainty).)

I've seen two other similar examples floating around the web, one with

P(X≥2)/P(X≥1)

and another I still can't understand: (figures changed to suit this problem, wording unchanged)

"failure has occurred it is no longer just a possibility

the binomial from 0 to n C(20,s).7^(20-s).3^s no longer applies s the number of failures

since failure is known to have happened we exclude .7^20

we are in that bleak place where success is impossible probability (1 - .7^20) which is parcelled out among the various possibilities of failure to sum to one. We scale these by dividing through by (1 - .7^20)

so the probability of two or more failures is (certain failure - (C(20,1).1^1*.9^9)/(1-.9^10)) = (certain failure - p(only one failure))

(1 - (C(20,1).3^1*.7^19)/(1-.7^20)"

Any help would be great

Cheers

 

[Stephen Tashi]

To find $P(X \ge 2 | X \ge 1)$ by "first principles" it is correct to do what you are attempting, namely to visualize the probability space defined by the restriction $X \ge 1$ as "the space of all possible outcomes". However, I think it is simpler to apply Bayes formula to this problem (and thus do all the calculations in the original probability space).

$P(X \ge 2 | X \ge 1)$ = $P( (X \ge 2)$ and $(X \ge 1))/ P(X \ge 1)$ = $P(X \ge 2)/ P(X \ge 1)$.

If you want to visualize the probability space defined by the restriction $X \ge 1$, visualize the binomial PDF. The restriction is going to "throw out" the possibility of $X = 0$. If you look at the values that are left, the sum of their probabilities will add up to less than 1. They add up to $1 - P(X = 0)$. How can you fix that? You can divide them all by $1 - P(X= 0)$ = $P(X \ge 1)$. This gives the new PDF for the events in the probability space defined by $X \ge 1$. The probability for the event $X \ge 2$ in the revised PDF is found by summing the probabilities corresponding to that event in the orginal probability space, except that each probability is "inflated" by dividing it by $P(X \ge 1)$.

 

[mrmt]

Stephen,

That's still slowly sinking in - I think I need more coffee

To be completely honest my brain is stuck on how we go from here:

= P((X≥2) and (X≥1))/P(X≥1)

to here:

= P(X≥2)/P(X≥1).

I know that's 101 somewhere but my brain just won't work - it's been something that's been bugging me for days!

Thank you

 

[Stephen Tashi]

how we go from here:

= P((X≥2) and (X≥1))/P(X≥1)

to here:

= P(X≥2)/P(X≥1).

The event $X \ge 2$ is the same event as the event $X \ge 2$ and $X \ge 1$. If you know $X \ge 2$, it is superflous to say $X \ge 1$.

Abstractly, if $A \subset B$ then $(A \cap B) = A$ so $P(A \cap B) = P(A)$.