Probability problem: $60$ workers in an industry

Question

Let's say there is an industry that employs $60$ workers, among these $60$ workers, $57$ are honest and the other $3$ are spies of another industry. Let's say that the industry starts a new project, so $4$ workers from the $60$ are randomly chosen (without replacement). We guess that the project will leak if among those $4$ workers selected, there is at least $1$ spy.

i) What's the probability of choosing all $3$ spies?

ii) What's the probability of project not to leak?

iii) Taking as a fact that the project leaked, what's the probability of that all 3 spies are chosen?

What I have achieved so far:

i) There are $\binom{60}{4}$ ways to chose $4$ workers from $60$. Chosing all the spies mean that the team created has $3$ spies and $1$ honest worker. So there are $\binom {3}{3}$ ways to chose $3$ spies and $\binom {57}{1}$ to choose $1$ honest worker. Let $A$ be the probability of chosing all spies, the answer is :

$$P(A) = \frac{\binom {3}{3}\cdot\binom{57}{1}}{\binom {60}{4}}.$$

ii) If we want the project not to leak, we need to create a team with $4$ honest workers and $0$ spies. So, there are $\binom {57}{4}$ ways to chose $4$ workers from $57$ honest ones. Let $B$ be the probability of project not to leak, the answer is:

$$P(B) = \frac{\binom {3}{0}\cdot\binom{57}{4}}{\binom {60}{4}}.$$

iii) Here we have conditional probability and we need to find $$P(A \mid B') = \dfrac{P(A\cap B')}{B'}.$$ Now I am not really sure what to do exactly. Any hint would be valueable.

Answer 1

Whereas in the previous cases you divided by the "total number of possibilities", now you have to divide by the "number of possibilities such that it is leaked". So this can be computed by: $${3\choose 1} \cdot {57\choose 3} + {3\choose 2} \cdot {57\choose 2} + {3\choose 3} \cdot {57\choose 1}$$

The number of events in which all 3 spies are chosen is: $${3\choose 3} \cdot {57\choose 1}$$

Thus the probability would be:

$$\dfrac{{3\choose 3} \cdot {57\choose 1}}{{3\choose 1} \cdot {57\choose 3} + {3\choose 2} \cdot {57\choose 2} + {3\choose 3} \cdot {57\choose 1}}$$

Answer 2

Turning things around by choosing $3$ spies (instead of $4$ workers) we find:

i)

$$P(\text{3 spies among the 4 workers})=\frac{\binom43\binom{56}0}{\binom{60}3}$$

ii)$$P(\text{not leaked})=\frac{\binom40\binom{56}3}{\binom{60}3}$$

iii)$$P(\text{3 spies among the 4 workers}\mid\text{leaked})P(\text{leaked})=$$$$P(\text{3 spies among the 4 workers and leaked})=P(\text{3 spies among the 4 workers})$$so that:

$$P(\text{3 spies among the 4 workers}\mid\text{leaked})\left(1-\frac{\binom40\binom{56}3}{\binom{60}3}\right)=\frac{\binom43\binom{56}0}{\binom{60}3}$$This equality enables you to find an expression for the mentioned conditional probability.