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)

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    $\begingroup$ I assume you mean "exactly $6$", yes? If so then the first two parts are correct, but not the third. For that one, you need Bayes' Theorem or the equivalent. From the first two parts you have the total probability that exactly $6$ shots are made. What portion of that is explained by the scenario in which the competitor is fit? $\endgroup$ – lulu Apr 3 '18 at 10:40

(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).

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