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What am I given?

In an AI model of a tennis simulation video game, it is assumed that the number of shots until someone wins a point, X, has a geometric distribution with probability mass function:

$$Pr(X = x) = 0.8 \cdot 0.2^{x-1} \space,\space x\geq 1$$

(a) What is the probability of at least 2 shots until a point is won?

(b) In 10 randomly selected points, find the probability that there are at least 4 points that required at least 2 shots to win the point.

(c) Find the probability that there are exactly 10 shots until 5 points are won.

What have I attempted?

I've had a bit of difficulty with this one. So normally I would like to list a fair bit of stuff I've tried but this is all I have...

(a) $$Pr(X > 2) = Pr(X \geq 3)$$ $$= \text{What next?}$$

(b) Genuinely just don't know where to start.

(c)Something to do with binomials? maybe ${10 \choose 5}$...Something?

$${10 \choose 5}p^5(1-p)^{10} * p$$ $$\text{I just don't know}$$

$$\text{THANKS IN ADVANCE!}$$

Rubicon
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    "At least two shots" corresponds to greater than or equal to two usually. As such, you are asked to calculate $Pr(X\geq 2)$, not $Pr(X>2)$ (there is a difference for discrete random variables). Now... how to calculate it, the easiest way is to look at the opposite event. You should know that $Pr(A)+Pr(A^c)=1$. What is the opposite of $Pr(X\geq 2)$? – JMoravitz Apr 25 '16 at 03:42
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    For (b), approach via a binomial distribution. You should have calculated the probability that a randomly selected point requires at least two shots in part (a). You should know how to calculate the probability that in ten flips of an unfair coin with probability of heads, $p$, that there are at least four heads. What if $p$ is the answer you got in part (a)? – JMoravitz Apr 25 '16 at 03:45
  • Thanks for the help mate, but I don't know how to take it any further in part (a). I appreciate the detailed comments, but I'm new to CDFs, PDFs and PMFs. So I'm still pretty confused. How do I actually get a value for (a)? – Rubicon Apr 25 '16 at 04:03
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    Since this random variable only takes positive integer values, the following simplifcations are allowed: $Pr(X\geq 2) = 1- Pr(X<2) = 1 - Pr(X\leq 1) = 1 - Pr(X=1)$. Now, you should know $Pr(X=1)$ (it was given to you in the problem statement, but was written for arbitrary $x$) – JMoravitz Apr 25 '16 at 04:05

3 Answers3

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a) I would probably model the answer as:

Pr(x≥2)= 1 - Pr(x=1) as x≥1

  • So then how did you find the answer after that? Have you done B and C? Thanks :) – Rubicon Apr 25 '16 at 03:59
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    You can find Pr(x=1) by subbing 1 in to the original equation given in the question. As for B, I have tried a number of different methods... As it is Geometric as outlined in the question, I tried doing X~Geometric(Pr(x≥2)) where Pr(X≥4)=1-Pr(NOT(X≥4)). But I am unsure about this method. For C, I have tried Pr(4 out of 9 shots result in a point)xPr(10th shot results in a point), but I am also unsure about this method. Sorry I can't be that much of a help! –  Apr 25 '16 at 04:09
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Given information: $Pr(X=x) = .8\cdot .2^{x-1}$ for each $x\geq 1$

Side note: it is worth checking on your own that this is indeed a probability distribution. I.e. that $\left(\sum\limits_{n=1}^\infty Pr(X=n)\right) = 1$. It is and it does.

  • (a) Find the probability that a point requires at least two shots. I.e. find $Pr(X\geq 2)$.

By inclusion-exclusion, we know that $Pr(A) + Pr(A^c) = Pr(A\cup A^c) = 1$ or worded a different way, $Pr(A) = 1 - Pr(A^c)$. For this specific problem, $Pr(X\geq 2) = 1 - Pr(X<2)$ since the opposite event of a number being greater than or equal to two is being strictly less than two.

Further, since the only values that our random variable can take are positive integers, we can simplify further: $\dots=1-Pr(X<2) = 1-Pr(X=1)$

We were told what $Pr(X=1)$ is in the problem statement.

$Pr(X\geq 2) = 1 - Pr(X=1) = 1 - 0.8\cdot 2^{1-1} = 1-0.8 = 0.2$

  • (b) In 10 randomly selected points, find the probability that there are at least 4 points that required at least 2 shots to win the point.

Now, let us approach this as a binomial distribution. We have ten trials, and in each trial we consider a "success" if it takes at least two shots to win the point. Let $Z$ be the random variable defined as the number of successes in these ten trials.

The problem asks us to find $Pr(Z\geq 4)$

We know, since it is a binomial distribution, that $Pr(Z=z) = \binom{10}{z}p^z(1-p)^{10-z}$, where $p$ is the answer we found for part (a).

We have then:

$Pr(Z\geq 4) = \sum\limits_{k=4}^{10} Pr(Z=k) = \sum\limits_{k=4}^{10}\binom{10}{k}0.2^k\cdot 0.8^{10-k}$. Recommend use of calculator if you want an exact value, it doesn't simplify horribly well.

  • (c) Find the probability that there are exactly 10 shots until 5 points are won.

Consider a sequence of hits, labelled $H$, and misses, labelled $m$. If exactly ten shots occur to get the fifth hit, that implies that the sequence of hits and misses are something like $HmmHmHHmmH$ with specifically: $5$ $m$'s, $5$ $H$'s, and one of the $H$'s must be at the end.

How many ways can you arrange $5$ $m$'s, $5$ $H$'s, with one $H$ at the end?

In the first nine spaces in the sequence, you must have five $m$'s and four $H$'s. Pick which four of the nine spaces are used by $H$'s. The rest must be $m$'s. There are then $\binom{9}{4}$ possible sequences of hits and misses.

For each of those sequences, what is the probability that a sequence of ten random shots happens to follow that specific sequence?

Remember, that hits have a probability of $0.8$ to occur and misses have a probability of $0.2$ to occur. (implied from the pdf)

$0.8^5\cdot 0.2^5$

Thus, the total probability is:

$0.8^5\cdot 0.2^5\cdot \binom{9}{4}$

JMoravitz
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a.) $Pr(X > 2) = 1 - Pr(X \leq 2)= 1 - F(2)$ where $F(X)$ is the CDF of your PMF.

Gage
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  • $Pr(X>2) = 1- Pr(X\leq 2)$. You probably mean to use $\geq 2$ instead. – JMoravitz Apr 25 '16 at 03:43
  • @JMoravitz You're right. Thanks. – Gage Apr 25 '16 at 03:45
  • You should notice my original comment on the question. It is highly unlikely that $Pr(X>2)$ is useful here. The OP was likely confused and wrote $Pr(X>2)$ when he in fact is supposed to calculate $Pr(X\geq 2)$. This also does not provide any insight as to the remaining parts of the problem. – JMoravitz Apr 25 '16 at 03:47
  • Yes, I see, because I can see how to do part (a), but I do not see how it helps with (b) or (c) – Gage Apr 25 '16 at 15:59
  • then read my answer above. – JMoravitz Apr 25 '16 at 16:08