Let's say there is a tournament where a competitor can shoot a ball $10$ times. A competitor is fit with probability $0.6$, so making a succesful shot has $0.7$ probability, and he is not fit with probability $0.4$, so making a succesful shot has $0.2$ probability. Shots are independent.
(a)Taking as a fact that the competitor is fit, what is the probability that he gets exactly $6$ succesful shots?(b) And what is the probability when he is not fit?
(c)The competitor gets exactly $6$ succesful shots, what is the probability of him to be fit?
What i have thought: Let $X$ dentote the number of succesful shots. We are gonna use binomial distribution.
(a) $$P(X=6)=\binom{10}{6}(0.7)^6(0.3)^4$$ (b)$$P(X=6)=\binom{10}{6}(0.2)^6(0.8)^4$$ (c)$$P(X=6)=\binom{10}{6}(0.6)^6(0.4)^4$$ (I am almost sure that i have made a mistake but still can't figure out how to correct it)
(a) and (b) are correctly solved.
Let $F$ denote the event that the competitor is fit.
Let $F^{\complement}$ denote the event that the competitor is not fit.
Then in (c) you are asked to find $$P(F\mid X=6)\tag1$$
For this note that: $$P(F\mid X=6)P(X=6)=P(F\text{ and } X=6)=P(X=6\mid F)P(F)$$
Concerning the RHS you already know $P(F)$ and you already calculated $P(X=6\mid F)$.
So to find an expression for $(1)$ is is enough now to find and expression for $P(X=6)$.
For this you can use:$$P(X=6)=P(X=6\mid F)P(F)+P(X=6\mid F^{\complement})P(F^{\complement})$$
All probabilities on RHS are known or are calculated in (a) or (b).