Binomial-type distribution with different probabilities

Question

I fear this may be too simple for many of you, but I'm stuck as to how to start solving this problem. I have five agents each performing a single task which has a yes/no answer. Four of them have a probability of 0.75 of getting a yes answer. The fifth has a 0.9 probability. What is the probability that three or more of them will answer yes? A (Philosophy!) paper I was reading suggested that it was around 0.93, but I would quite like to be able to prove this for myself. If the agents had the same probability I (think I) could use binomial distributions. However, I am stuck when one of the agents has a different probability. How should I proceed?

(I am not a hard-line mathematician, so the simpler the better…)

Kind Regards, Peter

Answer 1

For $i=1,2,3,4$ let $X_i=1$ if agent $i$ gets answer "yes" and let $X_i=0$ otherwise.

Then $X:=X_1+X_2+X_3+X_4$ has binomial distribution with parameters $n=4$ and $p=0.75$.

Let $Y=1$ if agent $5$ gets answer "yes" and let $Y=0$ otherwise.

Then $Y$ has Bernoulli distribution with parameter $0.9$.

Then to be found is $P\left(X+Y\geq3\right)$ or equivalently:$$P\left(X+Y=3\right)+P\left(X+Y=4\right)+P\left(X+Y=5\right)$$where $X$ and $Y$ are independent.

Each of the terms can be split up like this:$$P\left(X+Y=k\right)=P\left(X+Y=k\mid Y=0\right)P(Y=0)+P\left(X+Y=k\mid Y=1\right)P(Y=1)=$$$$P\left(X=k\mid Y=0\right)0.1+P\left(X=k-1\mid Y=1\right)0.9=$$$$P\left(X=k\right)0.1+P\left(X=k-1\right)0.9$$where the last equation rests on independence of $X$ and $Y$.

To finish this properly you only need familiarity with binomial distribution.

Can you take it from here?

Answer 2

As you said, since the probabilities are different, the binomial distribution won't work as it is. My first inclination is to split this into cases:

1) The fifth agent answers yes, with probability .9. In this case, at least 2 of the 4 remaining agents have to say yes. You can model this sub-problem using a binomial distribution, since all agents have the same probability.

In short, the resulting probability of this case would be .9 * (Probability of 2 of the 4 remaining agents say yes) since this case occurs with .9 probability.

2) The fifth agent answers no, with probability .1. Now we need at least 3 of the 4 remaining agents to say yes. This, again, can be modeled using binomial distribution.

Once each case is evaluated, you can add them to get the overall probability of at least 3 people answering yes.

Hope this helps! :)