What is the probability of all three companies going bankrupt? So I had a question on a test which I can't stop thinking about.
The question is: say that there are three companies $(A,B,C)$ that are independent of each other. The probability of company $B$ going bankrupt is $0.20$. The probability of both $A$ and $C$ going bankrupt is $0.07$, and the probability of $B$ and $C$ going bankrupt is $0.08$. If we know that at least two of the companies have gone bankrupt, what is the probability of all three of them having gone bankrupt?
$$P(B) = 0.2$$
$$P(A\cap C) = 0.07$$
$$P(B\cap C) = 0.08$$
Isn't this just $P(\textrm{all three bankrupt}\; |\; \textrm{at least two of the companies having gone bankrupt})$?
And if we use the rule $P(A\;|\;B) = P(A)$ for independent events (if we can?), shouldn't the answer to this question just be $P(\textrm{All three bankrupt})$ Which would just be $P(A) \cdot P(B) \cdot P(C)$?
Is there something I've done wrong? This was a question on an exam for a statistics basic course at UNI.
Just confused if this is really the correct answer or if its supposed to be way harder?


$P(\textrm{all three bankrupt}\; |\; \textrm{at least two bankrupt})$.......if we use the rule $P(A\;|\;B)= P(A)$ for independent events (if we can?), shouldn't the answer to this question just be $P(\textrm{all three bankrupt})?$

But the event that all three have gone bankrupt is not independent of the event that at least two have gone bankrupt.

$$P(\text{all three bankrupt}|\text{at least two bankrupt})\\=
\frac{P(\text{all three bankrupt})}{P(\text{at least two bankrupt})}\\=
\frac{P(A\cap B\cap C)}{P(A\cap B)+P(B\cap C)+P(C\cap A)-2P(A\cap B\cap C)}\\=
\frac{0.014}{0.035+0.08+0.07-2(0.014)}\\=8.92\%$$
