Corrected exercises: probability and distributions (201-337)

$P(X=2)=\binom{6}{2}(0.3)^{2}(0.7)^{4}=15\times 0.09\times 0.2401=0.324135\approx 0.3241$. Pitfall: forgetting $\binom{6}{2}=15$, which gives the probability of ONE single sequence, $0.0216$.

$P(X\leq 1)=P(X=0)+P(X=1)=(0.8)^{5}+5(0.2)(0.8)^{4}=0.32768+0.4096=0.73728\approx 0.7373$. Pitfall: computing only $P(X=1)$.

$P(X\geq 1)=1-P(X=0)=1-(0.85)^{8}=1-0.27249=0.72751\approx 0.7275$. Adding the terms from 1 to 8 would give the same result with eight times as much work.

$E(X)=np=40\times 0.35=14$ and $\sigma=\sqrt{np(1-p)}=\sqrt{40\times 0.35\times 0.65}=\sqrt{9.1}\approx 3.02$. Pitfall: giving the variance $9.1$ as the standard deviation.

$P(X=3)=\frac{e^{-2.5}(2.5)^{3}}{3!}=\frac{0.082085\times 15.625}{6}\approx 0.2138$.

$P(X\leq 2)=e^{-1.2}\left(1+1.2+\frac{1.2^{2}}{2}\right)=0.301194\times(1+1.2+0.72)=0.301194\times 2.92\approx 0.8795$. Pitfall: forgetting the term $k=0$, which equals $e^{-1.2}$ and carries a lot of weight.

The table gives $P(Z<1.28)=0.8997$ directly. By the symmetry of the bell curve, $P(Z>-0.64)=P(Z<0.64)=0.7389$. Pitfall: writing $1-0.7389=0.2611$, the area on the wrong side.

The two bounds give $z_{1}=\frac{42-50}{8}=-1.00$ and $z_{2}=\frac{62-50}{8}=1.50$. So $P(42<X<62)=0.9332-0.1587=0.7745$. Pitfall: adding the two areas instead of subtracting them.

$P(X>x)=0.10$ leaves $0.90$ to the LEFT: look for $0.90$ inside the table, which gives $z\approx 1.28$. So $x=\mu+z\sigma=200+1.28\times 25=232.0$. Pitfall: looking for $0.10$ in the table, which gives $z=-1.28$ and $x=168$, the cutoff of the lowest 10%.

$\mu=np=75\times 0.2=15$ and $\sigma=\sqrt{75\times 0.2\times 0.8}=\sqrt{12}\approx 3.464$; the conditions $np=15\geq 5$ and $n(1-p)=60\geq 5$ are met. “At least 20” starts at $19.5$: $z=\frac{19.5-15}{3.464}\approx 1.30$, so $P(X\geq 20)\approx 1-0.9032=0.0968$. Pitfall: keeping $20$, or taking $20.5$, which removes the value 20 from the event.

Let $M$ be taking the mathematics course and $P$ taking the physics course. Then $P(M)=\dfrac{22}{40}=0.550$, $P(P)=\dfrac{18}{40}=0.450$ and $P(M\cap P)=\dfrac{9}{40}=0.225$. So $P(M\cup P)=0.550+0.450-0.225=0.775$. The intersection must be subtracted, otherwise the nine students taking both would be counted twice.

It is the complementary event: $P\left(\overline{M\cup P}\right)=1-0.775=0.225$, that is 9 students out of 40. Check by direct counting: $22+18-9=31$ students are taking at least one of the courses, so $40-31=9$ remain, which confirms it.

They are NOT mutually exclusive, since $P(M\cap P)=0.225\neq 0$: nine students do take both. For independence, compare $P(M\cap P)$ with the product $P(M)\times P(P)=0.550\times 0.450=0.2475$. The intersection is 0.225, a DIFFERENT value: the two events are therefore not independent. The gap is small but real, and points to a slight tendency not to combine the two courses.

If $A$ and $B$ are mutually exclusive, then $P(A\cap B)=0$ by definition. If they were also independent, we would have $P(A\cap B)=P(A)\times P(B)$, so $P(A)\times P(B)=0$, which forces at least one of the two probabilities to be zero. As soon as both are strictly positive, the contradiction is immediate. Intuitively, two mutually exclusive events are in fact strongly DEPENDENT: knowing that one has occurred guarantees that the other has not, which is the most extreme form of dependence.

Out of 100,000 people, $0.50$% are ill, that is 500 ill people and 99,500 healthy people. Among the 500 ill people, the test detects 99%, that is 495 TRUE POSITIVES, and misses the other 5, who are FALSE NEGATIVES. Among the 99,500 healthy people, 95% are correctly declared negative, that is 94,525 TRUE NEGATIVES, and the remaining 5%, that is 4,975 people, are FALSE POSITIVES.

The total number of positive tests is $495+4975=5470$. Of these, only 495 correspond to people who are really ill, that is $\dfrac{495}{5470}=0.0905$, or 9.05%. In other words, more than 90% of the people with a positive test are NOT ill.

$P\left(M\mid +\right)=\dfrac{P\left(+\mid M\right)P(M)}{P(+)}$ with $P(+)=P\left(+\mid M\right)P(M)+P\left(+\mid\overline{M}\right)P\left(\overline{M}\right)$. Numerically, $P(+)=0.99\times 0.005+0.05\times 0.995=0.00495+0.04975=0.0547$. Then $P\left(M\mid +\right)=\dfrac{0.00495}{0.0547}=0.0905$, or 9.05%: identical to the count in b, as it should be.

The gap comes from the size of the two groups. Ill people are RARE, 500 of them, so even detecting 99% of them yields only 495 true positives. Healthy people, on the other hand, are very NUMEROUS, 99,500 of them, so the 5% false positive rate, which looks small, produces 4,975 positives on its own, ten times as many. A small percentage of a very large number swamps a large percentage of a very small number. It is the PREVALENCE, missing from the talk of “99% reliability”, that drives the result.

That it should be avoided, or paired with a second test. Mass screening for a rare disease produces an overwhelming majority of false positives, with needless anxiety, costly and sometimes invasive confirmation tests, and a loss of trust in screening as a result. The rational strategy is to screen only HIGH-prevalence populations, that is, people with risk factors, which raises $P(M)$ and therefore the positive predictive value; or else to confirm every positive with a second, independent and more specific test. That is exactly how real screening protocols are designed.

Three possible outcomes. The 6 pays 12 dollars for 3 bet, so $X=+9$ with probability $\frac{1}{6}$. The 5 pays 6 dollars, so $X=+3$ with probability $\frac{1}{6}$. The four other faces pay nothing, so $X=-3$ with probability $\frac{4}{6}$. Check: $\frac{1}{6}+\frac{1}{6}+\frac{4}{6}=1$.

$E(X)=(9)\frac{1}{6}+(3)\frac{1}{6}+(-3)\frac{4}{6}=\frac{9+3-12}{6}=0$. The game is therefore perfectly FAIR: in the long run, neither the player nor the operator wins anything. Be careful with the interpretation: the expected value is an AVERAGE gain over a large number of games, not a prediction for one game. Here, the player can never win exactly 0 dollars: the expected value is a value the variable never takes.

$E(X^{2})=(81)\frac{1}{6}+(9)\frac{1}{6}+(9)\frac{4}{6}=\frac{81+9+36}{6}=21$. So $V(X)=E(X^{2})-[E(X)]^{2}=21-0=21$ and $\sigma=\sqrt{21}\approx 4.58$ dollars. The expected value says the game is neutral; the standard deviation says it is RISKY: in each game, the result typically deviates from zero by more than four and a half dollars. Two games with the same expected value can carry radically different risks, and it is the variance that tells them apart. A game where you bet one dollar to win one has the same zero expected value as a million-dollar lottery: nobody would confuse the two.

With a 4-dollar bet, the net gains become $+8$, $+2$ and $-4$: $E(X)=\frac{8+2-16}{6}=-\frac{6}{6}=-1.00$ dollar per game for the player, so $+1.00$ dollar for the operator. Over 1000 games, the expected profit is 1000 dollars. The operator can be almost certain of it thanks to the LAW OF LARGE NUMBERS: the average gain over $n$ games converges to the expected value, and its standard deviation is only $\frac{\sigma}{\sqrt{n}}$, here about $\frac{4.6}{31.6}\approx 0.15$ dollar. In other words, the randomness that dominates one game becomes negligible over a thousand. That is exactly the business model of every casino: a slightly favourable expected value, repeated a very large number of times.

For the insurer: $E=300-(0.012)(20{,}000)=300-240=+60$ dollars per policy per year. For the policyholder, it is the opposite: $-60$ dollars. In terms of expected value, insurance is therefore a bad investment, and it always will be, otherwise no insurer would survive. People buy it all the same because the expected value is not the only decision criterion: a 20,000-dollar loss, however unlikely, can be RUINOUS, whereas a 300-dollar premium can be absorbed. Paying 60 dollars a year amounts to buying the removal of the variance, which is perfectly rational for anyone who cannot afford the worst case. The insurer, for its part, pools tens of thousands of policies and finds itself in the position of the operator in d): for the insurer, the variance has practically disappeared.

The three conditions are met. First, the number of trials is FIXED: $n=10$ questions. Second, each trial has only TWO outcomes, a right or a wrong answer, with a CONSTANT probability of success $p=\dfrac{1}{4}=0.25$, since there is one correct option out of four on every question. Third, the trials are INDEPENDENT, as the student answers at random and the answers do not influence one another. So $X\sim B(10, 0.25)$.

$\binom{10}{3}=\dfrac{10!}{3!\,7!}=\dfrac{10\times 9\times 8}{3\times 2\times 1}=120$. This number counts the WAYS of choosing which 3 of the 10 questions are answered correctly, regardless of order: the sequence of successes on questions 1, 2 and 3 is counted once, just like the one on questions 4, 7 and 9.

$P(X=3)=\binom{10}{3}\left(0.25\right)^{3}\left(0.75\right)^{7}=120\times 0.015625\times 0.13348=0.250$. About one chance in four. Each specific sequence has probability $0.25^{3}\times 0.75^{7}$, and there are 120 of them: the binomial coefficient really is just a counter.

$\mu=np=10\times 0.25=2.5$ correct answers on average, consistent with the intuition of a quarter of 10. The standard deviation is $\sigma=\sqrt{np(1-p)}=\sqrt{10\times 0.25\times 0.75}=\sqrt{1.875}=1.37$.

“At most 1” means 0 or 1, two mutually exclusive cases whose probabilities are added. $P(X=0)=\binom{10}{0}\left(0.25\right)^{0}\left(0.75\right)^{10}=0.0563$ and $P(X=1)=\binom{10}{1}\left(0.25\right)^{1}\left(0.75\right)^{9}=10\times 0.25\times 0.0751=0.1877$. The sum is $0.244$. A student answering at random therefore has almost one chance in four of getting no more than one correct answer, which is why this kind of quiz is often marked with a penalty for wrong answers. On the bar chart, each bar equals $\binom{n}{k}p^{k}(1-p)^{n-k}$ and the sum of all the bars, from $k=0$ to $k=n$, is exactly 1.

Convert: $z=\dfrac{130-100}{15}=2.00$. The table gives the area to the left of $z=2.00$, which is 0.9772. The proportion we want is therefore $1-0.9772=0.0228$, about 2.3% of the population.

The two bounds correspond to $z=\dfrac{85-100}{15}=-1.00$ and $z=\dfrac{115-100}{15}=+1.00$. The area between these two z-scores is $0.8413-0.1587=0.6827$, about 68.3%.

The empirical rule states that about 68% of the data lie within one standard deviation of the mean. Now 85 and 115 are exactly $\mu-\sigma$ and $\mu+\sigma$, since $\sigma=15$. The result of 68.3% therefore matches perfectly: the empirical rule is nothing more than the table read at $z=\pm 1$, $\pm 2$ and $\pm 3$, rounded.

Being in the top 2.5% means leaving 97.5% of the area to the left. The table gives $z=1.96$ for that area. Back on the original scale: $x=\mu+z\sigma=100+1.96\times 15=100+29.4=129.4$. You must therefore exceed about 129, which is consistent with a, where the cutoff 130 left 2.3%.

Because of the CENTRAL LIMIT THEOREM: whenever a quantity is the sum of a large number of small, independent contributions of similar size, its distribution tends to a normal distribution, whatever the individual distributions. That is the case for most natural traits: a person's height results from many genes and nutritional factors, a measurement error results from many tiny causes, and so on. This theorem explains why the bell curve is everywhere, and it also underpins all of the statistical inference in the next chapter.

With replacement, the urn has the same composition at every draw: the probability of red is $\frac{5}{12}$ each time, and the draws are INDEPENDENT. All the conditions of the binomial are therefore met: a fixed number of trials, two outcomes, a constant probability and independence. $P(3\text{ red})=\left(\frac{5}{12}\right)^{3}=\frac{125}{1728}\approx 0.0723$.

Without replacement, each draw changes the urn. With the tree: $P=\frac{5}{12}\times\frac{4}{11}\times\frac{3}{10}=\frac{60}{1320}=\frac{1}{22}\approx 0.0455$. With combinations: choose 3 red balls out of 5, over all the ways of choosing 3 balls out of 12, which gives $\frac{\binom{5}{3}}{\binom{12}{3}}=\frac{10}{220}=\frac{1}{22}$. The two methods give the same result, which is reassuring: the first follows the draws in order, the second directly counts the possible hands, ignoring order. Note that the probability has dropped from 0.0723 to 0.0455: removing the red balls as you go makes a triple clearly harder to get.

Exactly one red means one red out of the 5 and two blue out of the 7: $P=\frac{\binom{5}{1}\binom{7}{2}}{\binom{12}{3}}=\frac{(5)(21)}{220}=\frac{105}{220}\approx 0.477$. This is the HYPERGEOMETRIC distribution, the without-replacement version of the binomial.

The shortest route is the complement: “at least one red” is the opposite of “no red”, that is, of three blue. $P(\text{no red})=\frac{\binom{7}{3}}{\binom{12}{3}}=\frac{35}{220}$, so $P(\text{at least one})=1-\frac{35}{220}=\frac{185}{220}\approx 0.841$. The other route would require adding the cases “exactly 1”, “exactly 2” and “exactly 3”, three calculations instead of one. Whenever a question says “at least one”, the reflex must be the complement.

$P=\frac{5000}{12000}\times\frac{4999}{11999}\times\frac{4998}{11998}\approx 0.0723$, that is, the value in a) to within a tenth of a percent. The reason is that removing three balls from an urn of twelve thousand hardly changes the proportions, whereas removing them from an urn of twelve upsets them completely. Rule of thumb: a draw without replacement can be treated as a binomial as long as the sample is no more than about 5% of the population. That is the case for almost every survey, where a thousand people are polled in a country of several million, and it is what justifies using the binomial, and then its normal approximation, throughout the rest of the course.

Four-digit code: order matters and repetition is allowed, so it is a permutation WITH repetition: $10^{4}=10{,}000$. Podium with 8 runners: order matters (gold is not silver) but there is no repetition, so it is a permutation WITHOUT repetition: $\frac{8!}{5!}=(8)(7)(6)=336$. Lottery: order does not matter and there is no repetition, so it is a COMBINATION: $\binom{49}{6}=13{,}983{,}816$. This last number is why a single ticket has about one chance in fourteen million, and why playing every week for a whole lifetime makes almost no difference.

ANAGRAMME has 9 letters, including three A's and two M's, the other four, N, G, R and E, appearing once each. If there were no repeated letters, there would be $9!=362{,}880$ arrangements. But the three A's are indistinguishable: each visible anagram corresponds to $3!=6$ ways of permuting the A's among themselves, all of which give the same word. Likewise for the two M's, with $2!=2$. So we divide: $\frac{9!}{3!\,2!}=\frac{362{,}880}{12}=30{,}240$ distinct anagrams. The principle is general: count as if everything were distinguishable, then divide by the number of invisible permutations.

Possible committees: $\binom{14}{5}=2002$. Exactly 3 women: choose 3 women out of 8 AND 2 men out of 6, so $\binom{8}{3}\binom{6}{2}=(56)(15)=840$. At least one man: the quickest route is the complement, “no man” meaning five women, which gives $\binom{8}{5}=56$ committees; so $2002-56=1946$. Notice the “AND” in the second calculation, which translates into a multiplication, whereas the “at least” in the third calls for a subtraction.

With distinct roles, order matters: $(14)(13)(12)=2184$ ways. Simply choosing 3 people without assigning roles gives $\binom{14}{3}=364$. The ratio is $\frac{2184}{364}=6=3!$: each group of three people can share out the three roles in $3!$ ways. This is exactly the relation between permutations and combinations, $\binom{n}{k}=\frac{P(n,k)}{k!}$, where $P(n,k)=\frac{n!}{(n-k)!}$ counts the ordered selections: a combination is a permutation whose order has been forgotten.

On a bench, the seats are distinct: $5!=120$ ways. Around a round table, only the RELATIVE arrangement matters: moving everyone one seat over changes nothing about who sits next to whom. Each circular arrangement therefore corresponds to 5 linear arrangements, which leaves $\frac{5!}{5}=4!=24$ ways. The rule for circular permutations, $(n-1)!$, illustrates the same principle as in b): as soon as a transformation leaves the configuration unchanged, you must divide by the number of such transformations.

Three conditions: the events must be RARE on an instantaneous scale, meaning that two calls cannot arrive at exactly the same instant; they must be INDEPENDENT, the arrival of one call making the next neither more nor less likely; and the average RATE must be constant over the period studied. Here, callers do not coordinate with one another and the switchboard is assumed to be equally busy from one minute to the next: the conditions are reasonable. They would stop holding when offices open, where the rate shoots up, or if a group call triggered several callbacks. Note a fundamental difference with the binomial: here $k$ is not bounded by any number of trials, the tail of the distribution never stops and any whole number keeps a probability, however small.

$P(X=0)=e^{-3}\approx 0.0498$, so about 5% of minutes have no call at all. $P(X=2)=\frac{e^{-3}(3)^{2}}{2!}=\frac{9e^{-3}}{2}\approx 0.224$. $P(X\geq 1)=1-P(X=0)\approx 0.950$: in 95% of minutes, the switchboard rings at least once.

For a Poisson distribution, $E(X)=\lambda=3$ calls and $V(X)=\lambda=3$, so $\sigma=\sqrt{3}\approx 1.73$ calls. The remarkable property is that the mean and the VARIANCE are equal, which no other common distribution does. It provides an immediate diagnostic test: if the counts you record have a variance clearly larger than their mean, the Poisson model must be rejected; this is called overdispersion, a typical sign that the events are not independent or that the rate varies. That is exactly what is observed for road accidents by time of day, or for sales by season.

The parameter is proportional to the length of the observation window: over 5 minutes, $\lambda=(3)(5)=15$ calls. $P(X=0)=e^{-15}\approx 3.1\times 10^{-7}$, about one chance in three million. The contrast with the 5% of silent minutes is striking, and it shows clearly that it is not the probability per minute that gets multiplied, but the parameter.

Exact distribution: binomial with parameters $n=1000$ and $p=0.0030$. $P(X=2)=\binom{1000}{2}(0.0030)^{2}(0.9970)^{998}\approx 0.2242$. Poisson approximation with $\lambda=np=(1000)(0.0030)=3$: $P(X=2)=\frac{9e^{-3}}{2}\approx 0.2240$. The difference is under a tenth of a percent, for a far lighter calculation: the binomial requires a 998th power and a six-digit coefficient, the Poisson fits on one line. The approximation is legitimate when $n$ is LARGE and $p$ SMALL, in practice $n\geq 100$ and $np\leq 10$; the Poisson distribution is in fact the mathematical limit of the binomial as $n\to\infty$ and $p\to 0$ with the product $np$ held constant. That explains why it is everywhere: radioactive decays, insurance claims, typos per page, goals scored in a game, any count of rare events among a very large number of opportunities.