Probability: Mixed Practice, Worked Solutions

Ques­tion 1: Inter­sec­tion, con­di­tional prob­a­bil­ity and a test for inde­pen­dence#

The prob­lem#

$P(A) = 0.7$, $P(B) = 0.5$ and $P(A \cup B) = 0.85$. Find $P(A \cap B)$ and $P(A \mid B)$, and decide whether $A$ and $B$ are inde­pen­dent.

Under­stand­ing the prob­lem#

You know the prob­a­bil­ity of each event and of "$A$ or $B$". You must find the prob­a­bil­ity of "both $A$ and $B$", then the prob­a­bil­ity of $A$ once you know $B$ has hap­pened, and finally check the inde­pen­dence con­di­tion $P(A \cap B) = P(A)\,P(B)$.

The idea#

Use the addi­tion rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ to find the inter­sec­tion, then the def­i­n­i­tion of con­di­tional prob­a­bil­ity.

Step-by-step solu­tion#

Step 1. Rearrange the addi­tion rule for $P(A \cap B)$.

$$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.7 + 0.5 - 0.85 = 0.35$$

Step 2. Con­di­tional prob­a­bil­ity of $A$ given $B$.

$$\displaystyle P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.35}{0.5} = 0.7$$

Step 3. Test inde­pen­dence: com­pare $P(A \cap B)$ with $P(A)P(B)$.

$$P(A)\,P(B) = 0.7 \times 0.5 = 0.35 = P(A \cap B)$$

They are equal, so $A$ and $B$ are inde­pen­dent.

Check­ing the answer#

Another way to see it: $P(A \mid B) = 0.7 = P(A)$. Know­ing $B$ hap­pened does not change the chance of $A$, which is exactly what inde­pen­dence means.

Answer#

$P(A \cap B) = 0.35$; $P(A \mid B) = 0.7$; $A$ and $B$ are inde­pen­dent, since $0.35 = 0.7 \times 0.5$.

Ques­tion 2: Draw­ing two cards with­out replace­ment#

The prob­lem#

Two cards are drawn one after the other with­out replace­ment from a pack of $52$. Find (a) the prob­a­bil­ity that both are aces, and (b) the prob­a­bil­ity that the sec­ond is an ace given that the first was not.

Under­stand­ing the prob­lem#

"With­out replace­ment" means the first card is not put back, so the pack for the sec­ond draw has $51$ cards, and what it con­tains depends on the first card. There are $4$ aces in a pack.

The idea#

Use the mul­ti­pli­ca­tion rule $P(E_1 \cap E_2) = P(E_1)\,P(E_2 \mid E_1)$, work­ing out the sec­ond prob­a­bil­ity from the pack that is left.

Step-by-step solu­tion#

Part (a)

Step 1. First card an ace: $4$ aces out of $52$ cards.

$$\displaystyle P(\text{first ace}) = \frac{4}{52}$$

Step 2. Given the first was an ace, $3$ aces remain among $51$ cards.

$$\displaystyle P(\text{second ace} \mid \text{first ace}) = \frac{3}{51}$$

Step 3. Mul­ti­ply and sim­plify.

$$\displaystyle P(\text{both aces}) = \frac{4}{52}\cdot\frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$$

Part (b)

Step 1. If the first card was not an ace, all $4$ aces are still in the $51$ cards that remain.

$$\displaystyle P(\text{second ace} \mid \text{first not ace}) = \frac{4}{51}$$

Check­ing the answer#

Count­ing check for (a): the num­ber of ways to choose $2$ aces is $^4C_2 = 6$, and to choose any $2$ cards is $^{52}C_2 = 1326$; $\displaystyle \tfrac{6}{1326} = \tfrac{1}{221}$.

Answer#

(a) $\displaystyle \frac{1}{221}$; (b) $\displaystyle \frac{4}{51}$.

Ques­tion 3: Wait­ing for the first six#

The prob­lem#

A die is thrown until a six appears. Find the prob­a­bil­ity that it takes exactly three throws.

Under­stand­ing the prob­lem#

"Exactly three throws" means the first throw is not a six, the sec­ond is not a six, and the third is a six. The throws are inde­pen­dent of each other.

The idea#

Write the event as a sequence of inde­pen­dent out­comes and mul­ti­ply their prob­a­bil­i­ties.

Step-by-step solu­tion#

Step 1. Prob­a­bil­ity of not a six on one throw: $\displaystyle \frac{5}{6}$. Prob­a­bil­ity of a six: $\displaystyle \frac{1}{6}$.

Step 2. The event is "not six, not six, six". Mul­ti­ply, since the throws are inde­pen­dent.

$$\displaystyle P = \frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6} = \frac{25}{216}$$

Check­ing the answer#

$\displaystyle \tfrac{25}{216} \approx 0.116$, which is less than the chance of a six on the first throw ($\approx 0.167$), as it should be: wait­ing longer is less likely.

Answer#

$\displaystyle \left(\frac{5}{6}\right)^2\frac{1}{6} = \frac{25}{216}$.

Com­mon mis­take to avoid#

Answer­ing $\displaystyle \left(\tfrac{1}{6}\right)^3$ or for­get­ting that the first two throws must fail.

Ques­tion 4: Exactly one of three peo­ple solves a puz­zle#

The prob­lem#

Three peo­ple inde­pen­dently try a puz­zle, with suc­cess prob­a­bil­i­ties $\displaystyle \tfrac{1}{2}$, $\displaystyle \tfrac{1}{3}$ and $\displaystyle \tfrac{1}{5}$. Find the prob­a­bil­ity that exactly one of them solves it.

Under­stand­ing the prob­lem#

"Exactly one" means one per­son suc­ceeds and the other two fail. There are three ways this can hap­pen, depend­ing on who the suc­cess­ful per­son is. These three ways can­not hap­pen together, so their prob­a­bil­i­ties add.

The idea#

For each per­son, mul­ti­ply their suc­cess prob­a­bil­ity by the fail­ure prob­a­bil­i­ties of the other two (inde­pen­dence). Then add the three cases.

Step-by-step solu­tion#

Step 1. Fail­ure prob­a­bil­i­ties: $\displaystyle 1 - \tfrac{1}{2} = \tfrac{1}{2}$, $\displaystyle 1 - \tfrac{1}{3} = \tfrac{2}{3}$, $\displaystyle 1 - \tfrac{1}{5} = \tfrac{4}{5}$.

Step 2. Only the first suc­ceeds.

$$\displaystyle \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5} = \frac{8}{30}$$

Step 3. Only the sec­ond suc­ceeds.

$$\displaystyle \frac{1}{2}\cdot\frac{1}{3}\cdot\frac{4}{5} = \frac{4}{30}$$

Step 4. Only the third suc­ceeds.

$$\displaystyle \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{5} = \frac{2}{30}$$

Step 5. Add the three mutu­ally exclu­sive cases.

$$\displaystyle \frac{8 + 4 + 2}{30} = \frac{14}{30} = \frac{7}{15}$$

Check­ing the answer#

The prob­a­bil­ity that nobody solves it is $\displaystyle \tfrac{1}{2}\cdot\tfrac{2}{3}\cdot\tfrac{4}{5} = \tfrac{8}{30}$. The prob­a­bil­ity that two or more solve it works out to $\displaystyle \tfrac{8}{30}$ as well ($\displaystyle \tfrac{4 + 1 + 2 + 1}{30}$, count­ing "first and sec­ond only", "first and third only", "sec­ond and third only" and "all three"). Then $\displaystyle \tfrac{8}{30} + \tfrac{14}{30} + \tfrac{8}{30} = 1$.

Answer#

$\displaystyle \frac{7}{15}$.

Ques­tion 5: Three chil­dren, given at least one boy#

The prob­lem#

A fam­ily has three chil­dren. Given that at least one is a boy, find the prob­a­bil­ity that all three are boys.

Under­stand­ing the prob­lem#

Assume each child is equally likely to be a boy or a girl, inde­pen­dently. The sam­ple space has $2^3 = 8$ equally likely out­comes (BBB, BBG, …, GGG). You are told that at least one child is a boy, which removes only GGG.

The idea#

Use $\displaystyle P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ with $A$ = "all boys" and $B$ = "at least one boy". Since "all boys" already includes "at least one boy", $A \cap B = A$.

Step-by-step solu­tion#

Step 1. $\displaystyle P(A) = P(BBB) = \frac{1}{8}$.

Step 2. $\displaystyle P(B) = 1 - P(GGG) = 1 - \frac{1}{8} = \frac{7}{8}$.

Step 3. $A \cap B = A$, so

$$\displaystyle P(A \mid B) = \frac{1/8}{7/8} = \frac{1}{7}$$

Check­ing the answer#

Count directly: of the $7$ out­comes with at least one boy, only BBB has three boys. So the answer is $\displaystyle \tfrac{1}{7}$.

Answer#

$\displaystyle \frac{1}{7}$.

Ques­tion 6: Bayes' the­o­rem with glasses#

The prob­lem#

In a class, $45\%$ of stu­dents are girls. $20\%$ of girls and $30\%$ of boys wear glasses. A stu­dent cho­sen at ran­dom wears glasses. Find the prob­a­bil­ity that the stu­dent is a girl.

Under­stand­ing the prob­lem#

You are given the chance of glasses for girls and for boys, but you are asked the reverse: the chance of a girl given glasses. Revers­ing a con­di­tion is the job of Bayes' the­o­rem. Boys are $100\% - 45\% = 55\%$ of the class.

The idea#

Let $G$ = girl, $B$ = boy, $E$ = wears glasses. Bayes' the­o­rem:

$$\displaystyle P(G \mid E) = \frac{P(G)\,P(E \mid G)}{P(G)\,P(E \mid G) + P(B)\,P(E \mid B)}$$

Step-by-step solu­tion#

Step 1. Write down the given val­ues.

$$P(G) = 0.45,\quad P(B) = 0.55,\quad P(E \mid G) = 0.2,\quad P(E \mid B) = 0.3$$

Step 2. Prob­a­bil­ity of a girl with glasses.

$$P(G)\,P(E \mid G) = 0.45 \times 0.2 = 0.09$$

Step 3. Prob­a­bil­ity of a boy with glasses.

$$P(B)\,P(E \mid B) = 0.55 \times 0.3 = 0.165$$

Step 4. Total prob­a­bil­ity of glasses.

$$P(E) = 0.09 + 0.165 = 0.255$$

Step 5. Apply Bayes' the­o­rem and sim­plify ($\displaystyle 0.09 = \tfrac{90}{1000}$, $\displaystyle 0.255 = \tfrac{255}{1000}$, and $\displaystyle \tfrac{90}{255} = \tfrac{6}{17}$).

$$\displaystyle P(G \mid E) = \frac{0.09}{0.255} = \frac{90}{255} = \frac{6}{17}$$

Check­ing the answer#

Imag­ine $1000$ stu­dents: $450$ girls, of whom $90$ wear glasses; $550$ boys, of whom $165$ wear glasses. Of the $255$ with glasses, $90$ are girls: $\displaystyle \tfrac{90}{255} = \tfrac{6}{17} \approx 0.35$.

Answer#

$\displaystyle P(\text{girl} \mid \text{glasses}) = \frac{6}{17}$.

Ques­tion 7: Which bag did the red ball come from?#

The prob­lem#

Bag A has $4$ red and $6$ black balls; bag B has $7$ red and $3$ black. A die is thrown: on $1$ or $2$ a ball is drawn from A, oth­er­wise from B. The ball drawn is red. Find the prob­a­bil­ity that it came from bag A.

Under­stand­ing the prob­lem#

The die decides the bag: bag A with prob­a­bil­ity $\displaystyle \tfrac{2}{6} = \tfrac{1}{3}$, bag B with prob­a­bil­ity $\displaystyle \tfrac{4}{6} = \tfrac{2}{3}$. You know the result (red) and want the prob­a­bil­ity of the cause (bag A). That is Bayes' the­o­rem.

The idea#

Let $E_1$ = bag A cho­sen, $E_2$ = bag B cho­sen, $R$ = red ball. Com­pute $P(E_1)P(R \mid E_1)$ and $P(E_2)P(R \mid E_2)$, then divide the first by their sum.

Step-by-step solu­tion#

Step 1. Prob­a­bil­i­ties of choos­ing each bag.

$$\displaystyle P(E_1) = \frac{1}{3}, \qquad P(E_2) = \frac{2}{3}$$

Step 2. Prob­a­bil­ity of red from each bag.

$$\displaystyle P(R \mid E_1) = \frac{4}{10} = \frac{2}{5}, \qquad P(R \mid E_2) = \frac{7}{10}$$

Step 3. Mul­ti­ply for each path.

$$\displaystyle P(E_1)P(R \mid E_1) = \frac{1}{3}\cdot\frac{2}{5} = \frac{2}{15}, \qquad P(E_2)P(R \mid E_2) = \frac{2}{3}\cdot\frac{7}{10} = \frac{7}{15}$$

Step 4. Bayes' the­o­rem.

$$\displaystyle P(E_1 \mid R) = \frac{2/15}{2/15 + 7/15} = \frac{2}{9}$$

Check­ing the answer#

The total prob­a­bil­ity of red is $\displaystyle \tfrac{9}{15} = \tfrac{3}{5}$. Bag A is cho­sen less often and has fewer red balls, so a small answer like $\displaystyle \tfrac{2}{9} \approx 0.22$ makes sense.

Answer#

$\displaystyle P(\text{bag A} \mid \text{red}) = \frac{2}{9}$.

Ques­tion 8: Inde­pen­dent events and their com­ple­ments#

The prob­lem#

$A$ and $B$ are inde­pen­dent, $P(A) = 0.4$ and $P(A \cap B) = 0.16$. Find $P(B)$ and $P(A' \cap B')$.

Under­stand­ing the prob­lem#

Inde­pen­dence gives $P(A \cap B) = P(A)\,P(B)$, which lets you find $P(B)$. $A' \cap B'$ means "nei­ther $A$ nor $B$". If $A$ and $B$ are inde­pen­dent, so are $A'$ and $B'$.

The idea#

Solve $0.16 = 0.4 \times P(B)$, then mul­ti­ply the com­ple­ment prob­a­bil­i­ties.

Step-by-step solu­tion#

Step 1. Use inde­pen­dence.

$$\displaystyle P(A)\,P(B) = P(A \cap B) \quad\Longrightarrow\quad P(B) = \frac{0.16}{0.4} = 0.4$$

Step 2. Com­ple­ments: $P(A') = 1 - 0.4 = 0.6$ and $P(B') = 1 - 0.4 = 0.6$.

Step 3. Since $A'$ and $B'$ are inde­pen­dent,

$$P(A' \cap B') = P(A')\,P(B') = 0.6 \times 0.6 = 0.36$$

Check­ing the answer#

With­out using inde­pen­dence of com­ple­ments: $P(A \cup B) = 0.4 + 0.4 - 0.16 = 0.64$, so $P(A' \cap B') = 1 - P(A \cup B) = 0.36$. Same answer.

Answer#

$P(B) = 0.4$; $P(A' \cap B') = 0.36$.

Ques­tion 9: Bayes' the­o­rem for an insur­ance claim#

The prob­lem#

An insur­ance com­pany insures $3000$ scooter rid­ers, $5000$ car dri­vers and $2000$ truck dri­vers. The prob­a­bil­i­ties of an acci­dent are $0.02$, $0.04$ and $0.10$ respec­tively. One of the insured peo­ple has an acci­dent. Find the prob­a­bil­ity that the per­son is a scooter rider.

Under­stand­ing the prob­lem#

There are $3000 + 5000 + 2000 = 10000$ insured peo­ple. You know the acci­dent chance for each group, and you want the reverse: given an acci­dent, the chance the per­son is from the scooter group. That is Bayes' the­o­rem.

The idea#

Let $E_1, E_2, E_3$ be "scooter rider", "car dri­ver", "truck dri­ver", and $A$ = acci­dent. Then

$$\displaystyle P(E_1 \mid A) = \frac{P(E_1)P(A \mid E_1)}{\sum_i P(E_i)P(A \mid E_i)}$$

Work­ing with the actual num­bers of expected acci­dents is sim­pler and gives the same ratio.

Step-by-step solu­tion#

Step 1. Prob­a­bil­i­ties of each group.

$$\displaystyle P(E_1) = \frac{3000}{10000} = 0.3,\quad P(E_2) = 0.5,\quad P(E_3) = 0.2$$

Step 2. Prod­ucts for each group.

$$0.3 \times 0.02 = 0.006,\qquad 0.5 \times 0.04 = 0.02,\qquad 0.2 \times 0.10 = 0.02$$

Step 3. Total prob­a­bil­ity of an acci­dent.

$$P(A) = 0.006 + 0.02 + 0.02 = 0.046$$

Step 4. Bayes' the­o­rem.

$$\displaystyle P(E_1 \mid A) = \frac{0.006}{0.046} = \frac{6}{46} = \frac{3}{23}$$

Check­ing the answer#

In peo­ple: expected acci­dents are $3000 \times 0.02 = 60$ scooter rid­ers, $5000 \times 0.04 = 200$ car dri­vers and $2000 \times 0.10 = 200$ truck dri­vers, $460$ in all. The share from scoot­ers is $\displaystyle \tfrac{60}{460} = \tfrac{3}{23} \approx 0.13$.

Answer#

$\displaystyle \frac{3}{23}$.

Ques­tion 10: A head fol­lowed by a six#

The prob­lem#

A coin is tossed. If it shows heads, a die is thrown and the num­ber noted; if tails, a card is drawn from a pack. Find the prob­a­bil­ity of a head fol­lowed by a six.

Under­stand­ing the prob­lem#

The event needs two things in order: the coin shows heads, and then the die (which is thrown only after heads) shows six. The card branch is irrel­e­vant to this event.

The idea#

Mul­ti­pli­ca­tion rule: $P(H \cap 6) = P(H)\,P(6 \mid H)$.

Step-by-step solu­tion#

Step 1. $\displaystyle P(H) = \frac{1}{2}$.

Step 2. Given heads, the die is fair: $\displaystyle P(6 \mid H) = \frac{1}{6}$.

Step 3. Mul­ti­ply.

$$\displaystyle P(H \cap 6) = \frac{1}{2}\cdot\frac{1}{6} = \frac{1}{12}$$

Check­ing the answer#

On the heads branch there are $6$ equally likely out­comes $H1, \ldots, H6$, shar­ing prob­a­bil­ity $\displaystyle \tfrac{1}{2}$, so each has $\displaystyle \tfrac{1}{12}$.

Answer#

$\displaystyle \frac{1}{12}$.

Ques­tion 11: Mutu­ally exclu­sive events can­not be inde­pen­dent#

The prob­lem#

Prove that mutu­ally exclu­sive events $A$ and $B$ with $P(A) > 0$ and $P(B) > 0$ can­not be inde­pen­dent.

Under­stand­ing the prob­lem#

You must show that two con­di­tions can­not both hold. Mutu­ally exclu­sive means $A \cap B = \varnothing$. Inde­pen­dent means $P(A \cap B) = P(A)\,P(B)$. Both prob­a­bil­i­ties are given to be pos­i­tive.

The idea#

Com­pute $P(A \cap B)$ in two ways, from each con­di­tion, and show the results clash.

Step-by-step solu­tion#

Step 1. Since $A$ and $B$ are mutu­ally exclu­sive, $A \cap B = \varnothing$, so

$$P(A \cap B) = P(\varnothing) = 0$$

Step 2. Since $P(A) > 0$ and $P(B) > 0$, their prod­uct is pos­i­tive.

$$P(A)\,P(B) > 0$$

Step 3. So $P(A \cap B) = 0 \ne P(A)\,P(B)$. The con­di­tion for inde­pen­dence fails, and $A$ and $B$ can­not be inde­pen­dent.

Check­ing the answer#

Exam­ple: on a die, $A = \{1\}$ and $B = \{2\}$. They are mutu­ally exclu­sive; $\displaystyle P(A)P(B) = \tfrac{1}{36}$ but $P(A \cap B) = 0$. Know­ing $A$ hap­pened tells you for cer­tain that $B$ did not, which is the oppo­site of inde­pen­dence.

Answer#

$P(A \cap B) = 0$, whereas $P(A)P(B) > 0$; since these dif­fer, $A$ and $B$ are not inde­pen­dent.

Ques­tion 12: Bat­ter­ies tested with­out replace­ment#

The prob­lem#

A shelf has $3$ faulty and $7$ good bat­ter­ies. Two are tested one after the other with­out replace­ment. Given that the sec­ond is good, find the prob­a­bil­ity that the first was faulty.

Under­stand­ing the prob­lem#

Let $F_1$ = first bat­tery faulty, $G_1$ = first good, $G_2$ = sec­ond good. You want $P(F_1 \mid G_2)$: a con­di­tion on a later event, ask­ing about an ear­lier one. That is a rever­sal, so use the con­di­tional prob­a­bil­ity for­mula with the total prob­a­bil­ity of $G_2$.

The idea#

$$\displaystyle P(F_1 \mid G_2) = \frac{P(F_1 \cap G_2)}{P(G_2)},$$

where $P(G_2) = P(F_1)P(G_2 \mid F_1) + P(G_1)P(G_2 \mid G_1)$ by total prob­a­bil­ity.

Step-by-step solu­tion#

Step 1. Numer­a­tor: first faulty ($\displaystyle \tfrac{3}{10}$), then, with $9$ left of which $7$ are good, sec­ond good ($\displaystyle \tfrac{7}{9}$).

$$\displaystyle P(F_1 \cap G_2) = \frac{3}{10}\cdot\frac{7}{9} = \frac{21}{90} = \frac{7}{30}$$

Step 2. The other way to get a good sec­ond bat­tery: first good ($\displaystyle \tfrac{7}{10}$), then $6$ good among $9$ ($\displaystyle \tfrac{6}{9}$).

$$\displaystyle P(G_1 \cap G_2) = \frac{7}{10}\cdot\frac{6}{9} = \frac{42}{90}$$

Step 3. Total prob­a­bil­ity of $G_2$.

$$\displaystyle P(G_2) = \frac{21}{90} + \frac{42}{90} = \frac{63}{90} = \frac{7}{10}$$

(This matches the sym­me­try argu­ment: before any­thing is known, the sec­ond bat­tery is as likely to be good as the first, $\displaystyle \tfrac{7}{10}$.)

Step 4. Divide.

$$\displaystyle P(F_1 \mid G_2) = \frac{7/30}{7/10} = \frac{10}{30} = \frac{1}{3}$$

Check­ing the answer#

Think of it this way: given the sec­ond bat­tery is good, the first is one of the other $9$ bat­ter­ies, of which $3$ are faulty and $6$ good. So the chance it is faulty is $\displaystyle \tfrac{3}{9} = \tfrac{1}{3}$.

Answer#

$\displaystyle \frac{1}{3}$.