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I am stuck on the following question and am not sure where to go from what I have currently done.

Donald Trump has gone moose hunting. When he fires his gun at a moose he scores a hit with a probability of 0.8. On a typical hunt, he will fire his gun 30 times. Let X denote the number of shots that actually hit a moose.

I know i need to find out n, p and q. So n = 30, p = 0.8 and q = 0.2

i) What is the probability that at least 3 of the shots are misses?

For this do I need to find the probability that x=3, x=4, x=5 ... x=29, x=30?
that is all that I can think to do.

ii) What is the probability that more than 20 of the shots are hits?

for this one do I need to find the probability that x=20, x=21, ... x=30?

I know I have to put it into an equation, but I am not sure how to do that equation. any help will be great. thank you

Daniel Fischer
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guala
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  • "At least 20" and "More than 20" are different phrases. I would have interpreted the second part as $21,22,23,\dots$ but specifically not $20$. As for the first part, sure you can do that but it will be a lot of arithmetic. You could instead find the opposite probability being "fewer than 3 are misses" to only have to calculate the numbers for $x\in{0,1,2}$. Once you have that you can subtract from $1$. Also you should notice that the one question asks for misses while the other part asks for hits... so you might have to reverse your values of $p$ and $q$ on one of the parts – JMoravitz May 12 '17 at 20:52
  • Do not deface your questions. – Daniel Fischer May 12 '17 at 21:44

1 Answers1

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  1. For this do I need to find the probability that x=3, x=4, x=5 ... x=29, x=30? that is all that I can think to do.

    Yes, specifically you want to find $P(X \geq 3) = \sum_{x = 3}^{30}P(X = x)$ where we can add because the events are disjoint. However, there are two problems. First, that's a lot of work. You instead might want to consider the complement and do $$P(X\geq 3) = 1- P(X<3).$$

    Second, notice how $X$ was defined. It counts the number of hits (which are considered "successes"), but in part i., we want to count the number of misses. We consider the misses to be the "successes". So for clarity, define a new random variable $Y$ that follows the binomial distribution with $n = 30$ and $p = 0.2$, where $p$ is the probability of missing ("success") and compute $$P(Y\geq 3) = 1-P(Y<3).$$

  2. "More than 20" means strictly greater than 20. It does not include 20. Notice that here we are in fact counting the number of hits ("successes"). The number of hits is accurately described by $X$ as given in the problem, and $n = 30$ and $p = 0.8$ like you provided. Like you described, you want to find $$P(X> 20) = \sum_{k = 21}^{30}P(X = k).$$

Finally, recall that for a random variable $X$ that follows the binomial distribution with $n$, $p$, the probability that $X = k$ for $0\leq k\leq n$ is $$P(X= k) = \binom{n}{k}p^k(1-p)^{n-k}$$

where $\binom{n}{k}$ is the binomial coefficient and $p+q = 1$.

Em.
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