# Can independent probabilities add up to more than 1?

I need help with the following question:

The Hsp70 chaperone has a unique binding site with different affinity for three individual binding sites on a protein, denoted as sites A, B and C. The independent probabilities that Hsp70 binds to the A, B and C sites are PA=0.35, PB=0.24 PC=0.61, respectively. More than one chaperone molecule may independently bind to a single protein. What is the probability that Hsp70 binds somewhere, i.e., to at least one binding site?

My idea was to add the three probabilities PA + PB + PC, but their sum > 1

• Think about the probability of flipping heads at least once on three tosses of a coin that is 60% in favor of heads. How would you solve that?
– Dave
Jan 31, 2020 at 2:03
• Yes. Example, Assume $P(A \cup B) = 1$ and $A$ and $B$ are independent events with non-zero probability. Then we have $P(A) + P(B) - P(A)P(B) = 1$, the last term comes from independence. Then $P(A)+P(B) = 1+P(A)P(B)$, which is greater than 1. Jan 31, 2020 at 2:10
• I stand by my earlier comment, but to answer the question in the title, consider three independent flips of a fair coin. Each time, there probability of flipping heads is $0.5$, so $0.5+0.5+0.5>1$. Anyway, the probability of flipping heads at least once in those three flips isn’t $1.5$; it’s $0.875$.
– Dave
Jan 31, 2020 at 2:18

Yes, you are confusing independent events with disjoint events. Independence means that the probability of $$A$$ and $$B$$ occuring is $$P(A\cap B) = P(A)P(B)$$. Disjointedness means that the probability of $$A$$ and $$B$$ occuring is $$P(A\cap B) = 0$$.

For this question, we can compute the answer complementarily since the complements of independent events are also independent; it is easy to compute the probability that none of the events happen. $$P(A\cup B\cup C) = 1-(1-0.35)(1-0.24)(1-0.61)=0.80734.$$

• Ok thanks. This is similar logic to survival analysis? Feb 1, 2020 at 8:44

Expanding what @Anon said (+1), the key phrase in the quote is:

[...] More than one chaperone molecule may independently bind to a single protein [...]

The probability of either of the events happening, given that they are mutually exclusive is their sum

$$P(A \cup B) = P(A \text{ or } B) = P(A) + P(B)$$

In your case, they are obviously not mutually exclusive, as it can happen then "more then one" molecule binds. That is why you need to use the formula in the other answer, that shows the probability of either of them binding is the complement of probability of neither of them binding.