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I have been doing the following example, and there is something I do not fully understand in the solutions. Namely, the problem goes like this:

In biathlon, contestants ski around a track stopping at each of four shooting stations. At each shooting station the contestant attempts to hit five targets. The contestant may use up to eight shots, however the three last shots must be loaded one at a time. If despite eight shots spent, the contestant has not hit all five targets, the contestant must ski penalty laps, one for each missed target. Given that a certain contestant hits a target with a probability of 0.7, calculate the probability that no more than two out of the contestant’s four shootings result in penalty laps. Hits are considered independent of one another.

Solution: Firstly, it is obvious that $X: Bin(8,0.7)$. Then we calculate $P(X\ge5)$. After we calculate that, we have a new distribution $Z$ with p we got from calculating the previous one, so $Z: Bin(4,0.25413)$. Finally, we get the solution by calculating $P(Z\ge2)$.

The thing that I do not understand: When we calculate $P(X\ge5)$, why do we calculate that, and not just $P(X=5)$, since, after all, according to the question there are 5 targets, not 5 or more?

Any help would be appreciated.

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2 Answers 2

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Well, the Binomial Random variable we use is the count of hits among eight attempts.   So when the shooter actually stops after just five hits, it goes off model.   We compensate by accounting for those unused attempts, where the shooter may have hit more that five times should they have had more targets available.

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The event under consideration is the athlete hits all five targets, and the complement of this is that they missed at least one target.
The probability of the complement is $P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)$.
Since you want the event under consideration and its complement to have probabilities which sum to 1, then the probability that all 5 targets are hit must be $P(X\geq5)$.

So if $A$ is the event that all 5 targets are hit and $A^c$ is its complement, then $P(A^c)=P(X<5)$.
Since $P(A)+P(A^c)=1$, then $P(A)=1-P(A^c)=P(X\geq5)$.

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