Since the game is a home game, let's suppose that 80% of the fans attending the game are Penn State fans, while 20% are Notre Dame fans. That is, \(P(P)=0.8\) and \(P(N)=0.2\). Then, by independence:

\(P(X=0)=P(NNN)=0.2\times0.2\times0.2=0.008\)

And, by independence and mutual exclusivity of \(NNP, NPN\), and \(PNN\):

\(P(X=1)=P(NNP)+P(NPN)+P(PNN)=3\times0.2\times0.2\times0.8=0.096\)

Likewise, by independence and mutual exclusivity of \(PPN, PNP\), and \(NPP\):

\(P(X=2)=P(PPN)+P(PNP)+P(NPP)=3\times0.8\times0.8\times0.2=0.384\)

Finally, by independence:

\(P(X = 3) = P(PPP) = 0.8\times0.8\times0.8 = 0.512\)

There are a few things to note here:

• The results make sense! Given that 80% of the fans in the stands are Penn State fans, it shouldn't seem surprising that we would be most likely to select 2 or 3 Penn State fans.
• The probabilities behave well in that (1) the probabilities are all greater than 0, that is, \(P(X=x)>0\) and (2) the probability of the sample space is 1, that is,\(P(\mathbf{S}) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1\).
• Because the values that it takes on are random, the variable \(X\) has a special name. It is called a random variable! Ta-daaaa!

Answer

The key to finding \(c\) is to use item #2 in the definition of a p.m.f.

Answer

Again, the key to finding \(c\) is to use item #2 in the definition of a p.m.f.

Solution

This problem is very similar to the example on the previous page in which we were interested in finding the p.m.f. of \(X\), the number of defective bulbs selected in a sample of 4 bulbs. Here, we are interested in finding \(X\), the number of tagged fish selected in a sample of 15 fish. That is, \(X\) is a hypergeometric random variable with \(m = 80\), \(N = 600\), and \(n = 15\). Therefore, the p.m.f. of \(X\) is:

for the support \(x=0, 1, 2, \ldots, 15\).

Solution

The random variable \(X\) here also follows the hypergeometric distribution. Here, there are \(N=52\) total cards, \(n=5\) cards sampled, and \(m=4\) aces. Therefore, the p.m.f. of \(X\) is:

\(f(x)=\dfrac{\dbinom{4}{x} \dbinom{48}{5-x}}{\dbinom{52}{5}}\)

for the support \(x=0, 1, 2, 3, 4\).

Solution

If the random variable \(X\) is the top face of a tossed, fair, six-sided die, then the p.m.f. of \(X\) is:

\(f(x)=\dfrac{1}{6}\)

for \(x=1, 2, 3, 4, 5, \text{and } 6\). Therefore, the average toss, that is, the expected value of \(X\), is:

\(E(X)=1\left(\dfrac{1}{6}\right)+2\left(\dfrac{1}{6}\right)+3\left(\dfrac{1}{6}\right)+4\left(\dfrac{1}{6}\right)+5\left(\dfrac{1}{6}\right)+6\left(\dfrac{1}{6}\right)=3.5\)

Hmm... if we toss a fair, six-sided die once, should we expect the toss to be 3.5? No, of course not! All the expected value tells us is what we would expect the average of a large number of tosses to be in the long run. If we toss a fair, six-sided die a thousand times, say, and calculate the average of the tosses, will the average of the 1000 tosses be exactly 3.5? No, probably not! But, we can certainly expect it to be close to 3.5. It is important to keep in mind that the expected value of \(X\) is a theoretical average, not an actual, realized one!

Solution

Assuming that the ball has an equally likely chance of landing on each number, the p.m.f of \(X\) is:

\(f(x)=\dfrac{1}{38}\)

for \(x=0, 00, 1, 2, 3, \ldots, 36\). Therefore, the expected value of \(u(X)\) is:

\(E(u(X))=\$5\left(\dfrac{1}{38}\right)+\$10\left(\dfrac{1}{38}\right)+\left[\$1\left(\dfrac{1}{38}\right)\times 18 \right]+\left[\$2\left(\dfrac{1}{38}\right)\times 18 \right]=\$1.82\)

Note that the 18 that is multiplied by the \$1 and \$2 is because there are 18 odd and 18 even numbers on the wheel. Our calculation tells us that, in the long run, Hannah's House of Gambling would expect to have to pay out \$1.82 for each spin of the roulette wheel. Therefore, in order to ensure that the House made money, the House would have to charge at least \$1.82 per play.

Solution

If you play and lose, you are guaranteed to lose \$1! An expected loss of 20 cents means that if you played the game over and over and over and over .... again, the average of your \$3 winnings and your \$1 losses would be a 20 cent loss. "In the long run" means that you can't draw conclusions about one or two plays, but rather thousands and thousands of plays.

Solution

The expected value is calculated as follows:

\(E(X)=\sum\limits_{x=1}^\infty xf(x)=\sum\limits_{x=1}^\infty x\left(\dfrac{c}{x^2}\right)=c\sum\limits_{x=1}^\infty \dfrac{1}{x}\)

The first equal sign arises from the definition of the expected value. The second equal sign just involves replacing the generic p.m.f. notation \(f(x)\) with the given p.m.f. And, the third equal sign is because the constant \(c\) can be pulled through the summation sign, because it does not depend on the value of \(x\).

Now, to finalize our calculation, all we need to do is determine what the summation:

\(\sum\limits_{x=1}^\infty \dfrac{1}{x}\)

equals. Oops! You might recognize this quantity from your calculus studies as the divergent harmonic series, whose sum is infinity. Therefore, as the above definition of expectation suggests, we say in this case that the expected value of \(X\) doesn't exist.

Using part (b) of the first theorem, we can determine that:

\(E(4X^2)=4E(X^2)=4(4.4)=17.6\)

And using the second theorem, we can determine that:

\(E(3X+2X^2)=3E(X)+2E(X^2)=3(1.8)+2(4.4)=14.2\)

Answer The mean can be calculated as:

\(\mu_Y=E(Y)=\sum\limits_{y=3}^{13} yf(y)=3(0.03)+4(0.05)+\cdots+12(0.03)+13(0.01)=7.9\)

That is, the average patent life for a new drug is 7.9 years.

Solution

Recalling the p.m.f. of a hypergeometric distribution and using the definition of the expected value of \(X\), we have:

\(E(X)=\sum\limits_{x\in S} x \dfrac{\dbinom{m}{x} \dbinom{N-m}{n-x}}{\dbinom{N}{n}}\)

You should be getting the idea already that this is going to be messy! So, we're going to work on it in parts. First, note that the first term of the summation equals 0 when \(x=0\). And, note that some of the terms can be written differently:

That is:

\(\dbinom{m}{x}=\dfrac{m!}{x!(m-x)!}\)

and:

\(\dbinom{N}{n}=\dfrac{N!}{n!(N-n)!}=\dfrac{N(N-1)!}{n \cdot (n-1)!(N-n)!}=\dfrac{N}{n} \cdot \dfrac{(N-1)!}{(n-1)!(N-1-(n-1))!}=\dfrac{N}{n} \cdot \dbinom{N-1}{n-1}\)

Therefore, replacing these quantities in our formula for \(E(X)\), we have:

My voice gets caught off at the end there, but we still managed to finish the proof in the nick of time! We've shown that, in general, the mean of a hypergeometric random variable \(X\), in which \(n\) objects are selected from \(N\) objects with \(m\) of the objects being one type, is:

\(E(X)=\dfrac{mn}{N}\)

Solution

The variance of \(X\) is calculated as:

\(\sigma^2_X=E[(X-\mu)^2]=(3-4)^2(0.3)+(4-4)^2(0.4)+(5-4)^2(0.3)=0.6\)

And, therefore, the standard deviation of \(X\) is:

\(\sigma_X=\sqrt{0.6}=0.77\)

Now, the variance of \(Y\) is calculated as:

\(\sigma_Y^2=E[(Y-\mu)^2]=(1-4)^2(0.4)+(2-4)^2(0.1)+(6-4)^2(0.3)+(8-4)^2(0.2)=8.4\)

And, therefore, the standard deviation of \(Y\) is:

\(\sigma_Y=\sqrt{8.4}=2.9\)

As you can see, the expected variation in the random variable \(Y\), as quantified by its variance and standard deviation, is much larger than the expected variation in the random variable \(X\). Given the p.m.f.s of the two random variables, this result should not be surprising.

Solution

First, we need to calculate the expected value of \(X^2\):

\(E(X^2)=3^2(0.3)+4^2(0.4)+5^2(0.3)=16.6\)

Earlier, we determined that \(\mu\), the mean of \(X\), is 4. Therefore, using the shortcut formula for the variance, we verify that indeed the variance of \(X\) is 0.6:

\(\sigma^2_X=E(X^2)-\mu^2=16.6-4^2=0.6\)

Solution

On the previous page, we determined that the mean of the discrete uniform random variable \(X\) is:

\(\mu=E(X)=\dfrac{m+1}{2}\)

If we can calculate \(E(X^2)\), we can use the shortcut formula to calculate the variance of \(X\). Let's do that:

Solution

First, recall that the conversion from Fahrenheit (F) to Celsius (C) is:

\(C=\dfrac{5}{9}(F-32)\)

Therefore, the mean temperature in degrees Celsius is calculated as:

\(\mu_C=E(C)=E\left[\dfrac{5}{9}F-\dfrac{160}{9}\right]= \dfrac{5}{9}E(F)-\dfrac{160}{9}=\dfrac{5}{9}(50)-\dfrac{160}{9}=\dfrac{250-160}{9}=\dfrac{90}{9}=10\)

And, the standard deviation in degrees Celsius is calculated as:

\(\sigma_C=|\dfrac{5}{9}|\sigma_F=\dfrac{5}{9}(8)=\dfrac{40}{9}=4.44\)

Solution

The sample mean is:

\(\bar{x}=\dfrac{7+6+8+4+2+7+6+7+6+5}{10}=5.8\)

Solution

The sample variance is:

\(s^2=\dfrac{1}{9}\left[(7-5.8)^2+(6-5.8)^2+\cdots+(5-5.8)^2\right]=\dfrac{1}{9}(27.6)=3.067\)

Therefore, the sample standard deviation is:

\(s=\sqrt{3.067}=1.75\)

Solution

The sample variance is:

\(s^2=\dfrac{1}{9}\left[(7^2+6^2+\cdots+6^2+5^2)-10(5.8)^2\right]=3.067\)

Therefore, the sample standard deviation is:

\(s=\sqrt{3.067}=1.75\)

Solution

Keeping in mind that we need to take the first derivative of \(M(t)\) with respect to \(t\), we get:

\(M'(t)=n[1-p+pe^t]^{n-1} (pe^t)\)

And, setting \(t=0\), we get the binomial mean \(\mu=np\):

To find the variance, we first need to take the second derivative of \(M(t)\) with respect to \(t\). Doing so, we get:

\(M''(t)=n[1-p+pe^t]^{n-1} (pe^t)+(pe^t) n(n-1)[1-p+pe^t]^{n-2} (pe^t)\)

And, setting \(t=0\), and using the formula for the variance, we get the binomial variance \(\sigma^2=np(1-p)\):

Solution

We previously determined that the moment generating function of a binomial random variable is:

\(M(t)=[(1-p)+p e^t]^n\)

for \(-\infty<t<\infty\). Comparing the given moment generating function with that of a binomial random variable, we see that \(X\) must be a binomial random variable with \(n = 20\) and \(p=\frac{1}{4}\). Therefore, the p.m.f. of \(X\) is:

\(f(x)=\dbinom{20}{x} \left(\dfrac{1}{4}\right)^x \left(\dfrac{3}{4}\right)^ {20-x}\)

for \(x=0, 1, \ldots, 20\).

Solution

Since the game is a home game, let's again suppose that 80% of the fans attending the game are Penn State fans, while 20% are Notre Dame fans. That is, \(P(P) = 0.8\) and \(P(N) = 0.2\). Then, by independence:

\(P(X = 0) = P(NNN) = 0.2 \times 0.2 \times 0.2 = 1 \times (0.8)^0\times (0.2)^3\)

And, by independence and mutual exclusivity of \(NNP\), \(NPN\), and \(PNN\):

\(P(X = 1) = P(NNP) + P(NPN) + P(PNN) = 3 \times 0.8\times 0.2\times 0.2 = 3\times (0.8)^1\times (0.2)^2\)

Likewise, by independence and mutual exclusivity of \(PPN\), \(PNP\), and \(NPP\):

\(P(X = 2) = P(PPN) + P(PNP) + P(NPP) = 3\times 0.8 \times 0.8 \times 0.2 = 3\times (0.8)^2\times (0.2)^1\)

Finally, by independence:

\(P(X = 3) = P(PPP) = 0.8\times 0.8\times 0.8 = 1\times (0.8)^3\times (0.2)^0\)

Do you see a pattern in our calculations? It seems that, in each case, we multiply the number of ways of obtaining \(x\) Penn State fans first by the probability of \(x\) Penn State fans \((0.8)^x\) and then by the probability of \(3-x\) Nebraska fans \((0.2)^{3-x}\).

Answer

Yes, \(X\) is a binomial random variable, because:

1. The coin is tossed in exactly the same way 100 times.
2. Each toss results in either a head (success) or a tail (failure).
3. One toss doesn't affect the outcome of another toss. The trials are independent.
4. The probability of getting a head is 0.70 for each toss of the coin.
5. \(X\) equals the number of heads (successes).

Answer

No, \(X\) is not a binomial random variable, because the number of trials \(n\)was not fixed in advance, and \(X\) does not equal the number of volunteers in the sample.

Answer

No, \(X\) is not a binomial random variable, because \(p\), the probability that a randomly selected skein has acceptable color changes from trial to trial. For example, suppose, unknown to the QCI, that 9 of the 15 skeins of yarn in the lot are acceptable. For the first trial, \(p\)equals \(\frac{9}{15}\). However, for the second trial, \(p\)equals either \(\frac{9}{14}\) or \(\frac{8}{14}\)depending on whether an acceptable or unacceptable skein was selected in the first trial. Rather than being a binomial random variable, \(X\) is a hypergeometric random variable. If we continue to assume that 9 of the 15 skeins of yarn in the lot are acceptable, then \(X\) has the following probability mass function:

\(f(x)=P(X=x)=\dfrac{\dbinom{9}{x} \dbinom{6}{5-x}}{\dbinom{15}{5}}\) for \(x=0, 1, \ldots, 5\)

Answer

No, \(X\) is technically a hypergeometric random variable, not a binomial random variable, because, just as in the previous example, sampling takes place without replacement. Therefore, \(p\), the probability of selecting an SUV owner, has the potential to change from trial to trial. To make this point concrete, suppose that Americans own a total of \(N=270,000,000\) cars. Suppose too that half (135,000,000) of the cars are SUVs, while the other half (135,000,000) are not. Then, on the first trial, \(p\)equals \(\frac{1}{2}\) (from 135,000,000 divided by 270,000,000). Suppose an SUV owner was selected on the first trial. Then, on the second trial, \(p\) equals 134,999,999 divided by 269,999,999, which equals.... punching into a calculator... 0.499999... Hmmmmm! Isn't that 0.499999... close enough to \(\frac{1}{2}\) to just call it \(\frac{1}{2}\)?Yes...that's what we do!

In general, when the sample size \(n\)is small in relation to the population size \(N\), we assume a random variable \(X\), whose value is determined by sampling without replacement, follows (approximately) a binomial distribution. On the other hand, if the sample size \(n\)is close to the population size \(N\), then we assume the random variable \(X\) follows a hypergeometric distribution.

Solution

Since \(n=15\) is small relative to the population of \(N\) = 300,000,000 Americans, and all of the other criteria pass muster (two possible outcomes, independent trials, ....), the random variable \(X\) can be assumed to follow a binomial distribution with \(n=15\) and \(p=0.2\). Using the probability mass function for a binomial random variable, the calculation is then relatively straightforward:

\(P(X=3)=\dbinom{15}{3}(0.20)^3 (0.80)^{12}=0.25\)

That is, there is a 25% chance, in sampling 15 random Americans, that we would find exactly 3 that had no health insurance.

Solution

"At most one" means either 0 or 1 of those sampled have no health insurance. That is, we need to find:

\(P(X\le 1)=P(X=0)+P(X=1)\)

Using the probability mass function for a binomial random variable with \(n=15\) and \(p=0.2\), we have:

\(P(X \leq 1)=\dbinom{15}{0}(0.2)^0 (0.8)^{15}+ \dbinom{15}{1}(0.2)^1(0.8)^{14}=0.0352+0.1319=0.167\)

That is, we have a 16.7% chance, in sampling 15 random Americans, that we would find at most one that had no health insurance.

Solution

Yikes! "More than seven" in the sample means 8, 9, 10, 11, 12, 13, 14, 15. As the following picture illustrates, there are two ways that we can calculate \(P(X>7)\):

We could calculate \(P(X>7)\) by adding up \(P(X=8), P(X=9)\), up to \(P(X=15)\). Alternatively, we could calculate \(P(X>7)\) by finding \(P(X\le 7)\) and subtracting it from 1. But to find \(P(X\le 7)\), we'd still have to add up \(P(X=0), P(X=1)\) up to \(P(X=7)\). Either way, it becomes readily apparent that answering this question is going to involve more work than the previous two questions. It would clearly be helpful if we had an alternative to using the binomial p.m.f. to calculate binomial probabilities. The alternative typically used involves cumulative binomial probabilities.

Solution

The probability that at most 1 has no health insurance can be written as \(P(X\le 1)\). To find \(P(X\le 1)\) using the binomial table, we:

1. Find \(n=15\) in the first column on the left.
2. Find the column containing \(p=0.20\).
3. Find the 1 in the second column on the left, since we want to find \(F(1)=P(X\le 1)\).

Now, all we need to do is read the probability value where the \(p=0.20\) column and the (\(n=15, x=1\)) row intersect. What do you get?

We've used the cumulative binomial probability table to determine that the probability that at most 1 of the 15 sampled has no health insurance is 0.1671. For kicks, since it wouldn't take a lot of work in this case, you might want to verify that you'd get the same answer using the binomial p.m.f.

Solution

As we determined previously, we can calculate \(P(X>7)\) by finding \(P(X\le 7)\) and subtracting it from 1:

The good news is that the cumulative binomial probability table makes it easy to determine \(P(X\le 7)\) To find \(P(X\le 7)\) using the binomial table, we:

1. Find \(n=15\) in the first column on the left.
2. Find the column containing \(p=0.20\).
3. Find the 7 in the second column on the left, since we want to find \(F(7)=P(X\le 7)\).

Now, all we need to do is read the probability value where the \(p=0.20\) column and the (\(n = 15, x = 7\)) row intersect. What do you get?

The cumulative binomial probability table tells us that \(P(X\le 7)=0.9958\). Therefore:

\(P(X>7) = 1 − 0.9958 = 0.0042\)

That is, the probability that more than 7 in a random sample of 15 would have no health insurance is 0.0042.

Solution

We can calculate \(P(X=3)\) by finding \(P(X\le 2)\) and subtracting it from \(P(X\le 3)\), as illustrated here:

To find \(P(X\le 2)\) and \(P(X\le 3)\) using the binomial table, we:

1. Find \(n=15\) in the first column on the left.
2. Find the column containing \(p=0.20\).
3. Find the 3 in the second column on the left, since we want to find \(F(3)=P(X\le 3)\). And, find the 2 in the second column on the left, since we want to find \(F(2)=P(X\le 2)\).

Now, all we need to do is (1) read the probability value where the \(p = 0.20\) column and the (\(n = 15, x = 3\)) row intersect, and (2) read the probability value where the \(p = 0.20\) column and the (\(n = 15, x = 2\)) row intersect. What do you get?

The cumulative binomial probability table tells us that finding \(P(X\le 3)=0.6482\) and \(P(X\le 2)=0.3980\). Therefore:

\(P(X = 3) = P(X \le 3) - P(X\le 2) = 0.6482-0.3980 = 0.2502\)

That is, there is about a 25% chance that exactly 3 people in a random sample of 15 would have no health insurance. Again, for kicks, since it wouldn't take a lot of work in this case, you might want to verify that you'd get the same answer using the binomial p.m.f.

Solution

We can calculate \(P(X\ge 1)\) by finding \(P(X\le 0)\) and subtracting it from 1, as illustrated here:

To find \(P(X\le 0)\) using the binomial table, we:

1. Find \(n=15\) in the first column on the left.
2. Find the column containing \(p=0.20\).
3. Find the 0 in the second column on the left, since we want to find \(F(0)=P(X\le 0)\).

Now, all we need to do is read the probability value where the \(p = 0.20\) column and the (\(n = 15, x = 0\)) row intersect. What do you get?

The cumulative binomial probability table tells us that \(P(X\le 0)=0.0352\). Therefore:

\(P(X\le 1) = 1-0.0352 = 0.9648\)

That is, the probability that at least one person in a random sample of 15 would have no health insurance is 0.9648.

Solution

"Fewer than 5" means 0, 1, 2, 3, or 4. That is, \(P(X<5)=P(X\le 4)\), and \(P(X\le 4)\) can be readily found using the cumulative binomial table. To find \(P(X\le 4)\), we:

1. Find \(n=15\) in the first column on the left.
2. Find the column containing \(p=0.20\).
3. Find the 4 in the second column on the left, since we want to find \(F(4)=P(X\le 4)\).

Now, all we need to do is read the probability value where the \(p = 0.20\) column and the (\(n = 15, x = 4\)) row intersect. What do you get?

The cumulative binomial probability table tells us that \(P(X\le 4)= 0.8358\). That is, the probability that fewer than 5 people in a random sample of 15 would have no health insurance is 0.8358.

Solution

If we let \(X\) denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with \(n=10\) and \(p=0.70\). And, if we let \(Y\) denote the number of subscribers who don't qualify for favorable rates, then \(Y\), which equals \(10-X\), is a binomial random variable with \(n=10\) and \(q=1-p=0.30\). We are interested in finding \(P(X\ge 4)\). We can't use the cumulative binomial tables, because they only go up to \(p=0.50\). The good news is that we can rewrite \(P(X\ge 4)\)as a probability statement in terms of \(Y\):

\(P(X\ge 4) = P(-X\le -4) = P(10 -X\le 10 - 4) = P(Y\le 6)\)

Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find \(P(Y\le 6)\), we:

1. Find \(n=10\) in the first column on the left.
2. Find the column containing \(p=0.30\).
3. Find the 6 in the second column on the left, since we want to find \(F(6)=P(Y\le 6)\).

Now, all we need to do is read the probability value where the \(p = 0.30\) column and the (\(n = 10, y = 6\)) row intersect. What do you get?

The cumulative binomial probability table tells us that \(P(Y\le 6)=P(X\ge 4)=0.9894\). That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.

Solution

Because \(X\) is a binomial random variable, the mean of \(X\) is \(np\). Therefore, the gardener could expect, on average, \(9\times 0.80=0.72\) seeds to germinate.

What does it mean that the average is 7.2 seeds? Obviously, a seed either germinates or not. You can't have two-tenths of a seed germinating. Recall that the mean is a long-run (population) average. What the 7.2 means is... if the gardener conducted this experiment... that is, planting nine radish seeds and observing the number that germinated... over and over and over again, the average number of seeds that would germinate would be 7.2. The number observed for any particular experiment would be an integer (that is, whole seeds), but when you take the average of all of the integers from the repeated experiments, you need not obtain an integer, as is the case here. In general, the average of a discrete random variable need not be an integer.

Solution

The variance of \(X\) is:

\(np(1-p)=9\times 0.80\times 0.2=1.44\)

Therefore, the standard deviation of \(X\) is the square root of 1.44, or 1.20.

Solution

Solution

To find the desired probability, we need to find \(P(X=4\), which can be determined readily using the p.m.f. of a geometric random variable with \(p=0.20\), \(1-p=0.80\), and \(x=4\):

\(P(X=4)=0.80^3 \times 0.20=0.1024\)

There is about a 10% chance that the marketing representative would have to select 4 people before he would find one who attended the last home football game.

Solution

To find the desired probability, we need to find \(P(X>6)=1-P(X\le6)\), which can be determined readily using the c.d.f. of a geometric random variable with \(1-p=0.80\), and \(x=6\):

\(P(X >6)=1-P(X \leq 6)=1-[1-0.8^6]=0.8^6=0.262\)

There is about a 26% chance that the marketing representative would have to select more than 6 people before he would find one who attended the last home football game.

Solution

The average number is:

\(\mu=E(X)=\dfrac{1}{p}=\dfrac{1}{0.20}=5\)

That is, we should expect the marketing representative to have to select 5 people before he finds one who attended the last football game. Of course, on any given try, it may take 1 person or it may take 10, but 5 is the average number. The variance is 20, as determined by:

\(\sigma^2=Var(X)=\dfrac{1-p}{p^2}=\dfrac{0.80}{0.20^2}=20\)

Solution

Solution

To find the requested probability, we need to find \(P(X=3\). Note that \(X\)is technically a geometric random variable, since we are only looking for one success. Since a geometric random variable is just a special case of a negative binomial random variable, we'll try finding the probability using the negative binomial p.m.f. In this case, \(p=0.20, 1-p=0.80, r=1, x=3\), and here's what the calculation looks like:

\(P(X=3)=\dbinom{3-1}{1-1}(1-p)^{3-1}p^1=(1-p)^2 p=0.80^2\times 0.20=0.128\)

It is at the second equal sign that you can see how the general negative binomial problem reduces to a geometric random variable problem. In any case, there is about a 13% chance thathe first strike comes on the third well drilled.

Solution

To find the requested probability, we need to find \(P(X=7\), which can be readily found using the p.m.f. of a negative binomial random variable with \(p=0.20, 1-p=0.80, x=7, r=3\):

\(P(X=7)=\dbinom{7-1}{3-1}(1-p)^{7-3}p^3=\dbinom{6}{2}0.80^4\times 0.20^3=0.049\)

That is, there is about a 5% chance that the third strike comes on the seventh well drilled.

Solution

The mean number of wells is:

\(\mu=E(X)=\dfrac{r}{p}=\dfrac{3}{0.20}=15\)

with a variance of:

\(\sigma^2=Var(x)=\dfrac{r(1-p)}{p^2}=\dfrac{3(0.80)}{0.20^2}=60\)

We can find the requested probability directly from the p.m.f. The probability that \(X\) is at least one is:

\(P(X\ge 1)=1-P(X=0)\)

Therefore, using the p.m.f. to find \(P(X=0)\), we get:

\(P(X \geq 1)=1-\dfrac{e^{-3}3^0}{0!}=1-e^{-3}=1-0.0498=0.9502\)

That is, there is just over a 95% chance of finding at least one typo on a randomly selected page when the average number of typos per page is 3.

The probability that \(X\) is at most one is:

\(P(X\le 1)=P(X=0)+P(X=1)\)

Therefore, using the p.m.f., we get:

\(P(X \leq 1)=\dfrac{e^{-3}3^0}{0!}+\dfrac{e^{-3}3^1}{1!}=e^{-3}+3e^{-3}=4e^{-3}=4(0.0498)=0.1992\)

That is, there is just under a 20% chance of finding at most one typo on a randomly selected page when the average number of typos per page is 3.

Solution

The probability that a randomly selected page has four typos on it can be written as \(P(X=4)\). We can calculate \(P(X=4)\) by subtracting \(P(X\le 3)\) from \(P(X\le 4)\). To find \(P(X\le 3)\) and \(P(X\le 4)\) using the Poisson table, we:

1. Find the column headed by \(\lambda=3\).
2. Find the 3 in the first column on the left, since we want to find \(F(3)=P(X\le 3)\). And, find the 4 in the first column on the left, since we want to find \(F(4)=P(X\le 4)\).

The cumulative Poisson probability table tells us that finding \(P(X\le 4)=0.815\) and \(P(X\le 3)=0.647\). Therefore:

\(P(X=4)=P(X\le 4)-P(X\le 3)=0.815-0.647=0.168\)

That is, there is about a 17% chance that a randomly selected page would have four typos on it. Since it wouldn't take a lot of work in this case, you might want to verify that you'd get the same answer using the Poisson p.m.f.

Solution

Solving this problem involves taking one additional step. Recall that \(X\) denotes the number of typos on one printed page. Then, let's define a new random variable \(Y\) that equals the number of typos on three printed pages. If the mean of \(X\) is 3 typos per page, then the mean of \(Y\) is:

\(\lambda_Y=3 \text{ typos per one page }\times 3\text{ pages }=9 \text{ typos per three pages}\)

Finding the desired probability then involves finding:

\(P(Y>8)=1-P(Y\le 8)\)

The cumulative Poisson probability table tells us that finding \(P(X\le 8)=0.456\). Therefore:

\(P(Y>8)=1-P(Y\le 8)=1-0.456=0.544\)

That is, there is a 54.4% chance that three randomly selected pages would have more than eight typos on it.

Solution

Can you convince yourself that \(X\) is a binomial random variable? Hmmm.... let's see... there are two possible outcomes (defective or not), the 100 trials of selecting the bulbs from the assembly line can be assumed to be performed in an identical and independent manner, and the probability of getting a defective bulb can be assumed to be constant from trial to trial. So, \(X\) is indeed a binomial random variable. Well, calculating the probability is easy enough then... we just need to use the cumulative binomial table with \(n=100\) and \(p=0.05\).... Oops! The table won't help us here, will it? Even many standard calculators would have trouble calculating the probability using the p.m.f.:

\(P(X\leq 3)=\dbinom{100}{0}(0.05)^0 (0.95)^{100}+\cdots+\dbinom{100}{3}(0.05)^3 (0.95)^{97}\)

Using a statistical software package (Minitab), I was able to use the binomial p.m.f. to determine that:

\(P(X\le 3)=0.0059205+0.0311607+0.0811818+0.1395757=0.25784\)

But, if you recall the way that we derived the Poisson distribution,... we started with the binomial distribution and took the limit as n approached infinity. So, it seems reasonable then that the Poisson p.m.f. would serve as a reasonable approximation to the binomial p.m.f. when your \(n\) is large (and therefore, \(p\) is small). Let's calculate \(P(X\le 3)\) using the Poisson distribution and see how close we get. Well, the probability of success was defined to be:

\(p=\dfrac{\lambda}{n}\)

Therefore, the mean \(\lambda\) is:

\(\lambda=np\)

The cumulative Poisson probability table tells us that finding \(P(X\le 3)=0.265\). That is, if there is a 5% defective rate, then there is a 26.5% chance that the a randomly selected batch of 100 bulbs will contain at most 3 defective bulbs. More importantly, since we have been talking here about using the Poisson distribution to approximate the binomial distribution, we should probably compare our results. When we used the binomial distribution, we deemed \(P(X\le 3)=0.258\), and when we used the Poisson distribution, we deemed \(P(X\le 3)=0.265\). Not too bad of an approximation, eh?

It is important to keep in mind that the Poisson approximation to the binomial distribution works well only when \(n\) is large and \(p\) is small. In general, the approximation works well if \(n\ge 20\) and \(p\le 0.05\), or if \(n\ge 100\) and \(p\le 0.10\).