Conditional Probability and Independence, Worked Solutions

Ques­tion 1: Con­di­tional prob­a­bil­ity from a for­mula#

The prob­lem#

$P(E) = 0.6$, $P(F) = 0.3$, $P(E \cap F) = 0.2$. Find $P(E \mid F)$ and $P(F \mid E)$.

Under­stand­ing the prob­lem#

You are given the prob­a­bil­i­ties of two events and of both hap­pen­ing together. $P(E \mid F)$ is the prob­a­bil­ity of $E$ once you know $F$ has hap­pened; $P(F \mid E)$ is the other way round. They are usu­ally dif­fer­ent.

The idea#

Use the def­i­n­i­tion $\displaystyle P(E \mid F) = \frac{P(E \cap F)}{P(F)}$: divide the prob­a­bil­ity of "both" by the prob­a­bil­ity of the event you are told has hap­pened.

Step-by-step solu­tion#

Step 1. For $P(E \mid F)$, the given event is $F$, so divide by $P(F)$.

$$\displaystyle P(E \mid F) = \frac{P(E \cap F)}{P(F)} = \frac{0.2}{0.3} = \frac{2}{3}$$

Step 2. For $P(F \mid E)$, the given event is $E$, so divide by $P(E)$.

$$\displaystyle P(F \mid E) = \frac{P(E \cap F)}{P(E)} = \frac{0.2}{0.6} = \frac{1}{3}$$

Check­ing the answer#

Both lie between $0$ and $1$. Also, the mul­ti­pli­ca­tion rule should give the same $P(E \cap F)$ both ways: $\displaystyle 0.3 \times \tfrac{2}{3} = 0.2$ and $\displaystyle 0.6 \times \tfrac{1}{3} = 0.2$ ✓.

Answer#

$\displaystyle P(E \mid F) = \tfrac{2}{3}$; $\displaystyle P(F \mid E) = \tfrac{1}{3}$.

Com­mon mis­take to avoid#

Divid­ing by the wrong event. The event after the bar "$\mid$" is always the denom­i­na­tor.

Ques­tion 2: Coin tosses with a con­di­tion#

The prob­lem#

A coin is tossed three times. Find $P(\text{at least two heads} \mid \text{first toss is a head})$.

Under­stand­ing the prob­lem#

Let $E$ = "at least two heads" and $F$ = "first toss is a head". You are told $F$ hap­pened, so only out­comes start­ing with H count.

The idea#

With equally likely out­comes, $\displaystyle P(E \mid F) = \frac{n(E \cap F)}{n(F)}$: the sam­ple space shrinks to $F$.

Step-by-step solu­tion#

Step 1. List the full sam­ple space: $2^3 = 8$ equally likely out­comes.

$$\text{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}$$

Step 2. Keep only the out­comes in $F$ (first toss H).

$$F = \{\text{HHH, HHT, HTH, HTT}\}, \qquad n(F) = 4$$

Step 3. Among these, find those with at least two heads.

$$E \cap F = \{\text{HHH, HHT, HTH}\}, \qquad n(E \cap F) = 3$$

Step 4. Divide.

$$\displaystyle P(E \mid F) = \frac{3}{4}$$

Check­ing the answer#

Another view: once the first toss is a head, you need at least one more head in the remain­ing two tosses. The only fail­ure is TT, with prob­a­bil­ity $\displaystyle \tfrac{1}{4}$, so the answer is $\displaystyle 1 - \tfrac{1}{4} = \tfrac{3}{4}$ ✓.

Answer#

$\displaystyle \tfrac{3}{4}$.

Ques­tion 3: Dice with a con­di­tion#

The prob­lem#

Two dice are thrown. Find $P(\text{sum} > 9 \mid \text{one die shows } 5)$.

Under­stand­ing the prob­lem#

Let $E$ = "sum is more than $9$" (sum $10$, $11$ or $12$) and $F$ = "at least one die shows $5$". The con­di­tion "one die shows $5$" is read as "a $5$ appears on at least one die".

The idea#

Count within the reduced sam­ple space $F$: $\displaystyle P(E \mid F) = \frac{n(E \cap F)}{n(F)}$.

Step-by-step solu­tion#

Step 1. Count $F$. Out­comes with a $5$ on the first die: $(5, 1), \ldots, (5, 6)$, that is $6$. With a $5$ on the sec­ond die: $(1, 5), \ldots, (6, 5)$, another $6$. The out­come $(5, 5)$ has been counted twice, so

$$n(F) = 6 + 6 - 1 = 11$$

Step 2. Among these $11$, find sums greater than $9$. With a $5$ show­ing, the other die must show at least $5$:

$$E \cap F = \{(5, 5), (5, 6), (6, 5)\}, \qquad n(E \cap F) = 3$$

Step 3. Divide.

$$\displaystyle P(E \mid F) = \frac{3}{11}$$

Check­ing the answer#

Using the for­mula: $\displaystyle P(E \cap F) = \tfrac{3}{36}$ and $\displaystyle P(F) = \tfrac{11}{36}$, so $\displaystyle \tfrac{3/36}{11/36} = \tfrac{3}{11}$ ✓.

Answer#

$\displaystyle \tfrac{3}{11}$.

Com­mon mis­take to avoid#

Count­ing $n(F) = 12$ by for­get­ting that $(5, 5)$ appears in both lists.

Ques­tion 4: Draw­ing bulbs with­out replace­ment#

The prob­lem#

A box has $4$ good and $2$ faulty bulbs. Two are drawn with­out replace­ment. Find (a) $P(\text{both good})$ and (b) $P(\text{one good, one faulty})$.

Under­stand­ing the prob­lem#

There are $6$ bulbs. "With­out replace­ment" means the first bulb is not put back, so the sec­ond draw is from $5$ bulbs and its prob­a­bil­i­ties depend on the first draw. In (b), "one good, one faulty" can hap­pen in two orders.

The idea#

Use the mul­ti­pli­ca­tion rule $P(E \cap F) = P(E)\,P(F \mid E)$, as in Exam­ple 4. For (b), add the prob­a­bil­i­ties of the two orders, since they can­not both hap­pen.

Step-by-step solu­tion#

Part (a)

Step 1. First bulb good: $4$ good out of $6$.

$$\displaystyle P(G_1) = \frac{4}{6}$$

Step 2. Given the first was good, $3$ good remain out of $5$.

$$\displaystyle P(G_2 \mid G_1) = \frac{3}{5}$$

Step 3. Mul­ti­ply.

$$\displaystyle P(\text{both good}) = \frac{4}{6}\cdot\frac{3}{5} = \frac{12}{30} = \frac{2}{5}$$

Part (b)

Step 1. Good then faulty: after a good bulb, $2$ faulty remain out of $5$.

$$\displaystyle \frac{4}{6}\cdot\frac{2}{5} = \frac{8}{30}$$

Step 2. Faulty then good: after a faulty bulb, $4$ good remain out of $5$.

$$\displaystyle \frac{2}{6}\cdot\frac{4}{5} = \frac{8}{30}$$

Step 3. Add the two orders.

$$\displaystyle P(\text{one good, one faulty}) = \frac{8}{30} + \frac{8}{30} = \frac{16}{30} = \frac{8}{15}$$

Check­ing the answer#

The third pos­si­bil­ity, both faulty, has prob­a­bil­ity $\displaystyle \tfrac{2}{6}\cdot\tfrac{1}{5} = \tfrac{1}{15}$. The three cases must add to $1$: $\displaystyle \tfrac{6}{15} + \tfrac{8}{15} + \tfrac{1}{15} = 1$ ✓.

Answer#

(a) $\displaystyle \tfrac{2}{5}$; (b) $\displaystyle \tfrac{8}{15}$.

Ques­tion 5: Hearts and kings#

The prob­lem#

Are $E$ = "a card is a heart" and $F$ = "a card is a king" inde­pen­dent?

Under­stand­ing the prob­lem#

One card is drawn from a well-shuf­fled pack of $52$. You must decide whether know­ing it is a king changes the chance it is a heart.

The idea#

Test the con­di­tion $P(E \cap F) = P(E)\,P(F)$. If it holds, the events are inde­pen­dent.

Step-by-step solu­tion#

Step 1. $P(E)$: there are $13$ hearts.

$$\displaystyle P(E) = \frac{13}{52} = \frac{1}{4}$$

Step 2. $P(F)$: there are $4$ kings.

$$\displaystyle P(F) = \frac{4}{52} = \frac{1}{13}$$

Step 3. $E \cap F$ is "the king of hearts", only $1$ card.

$$\displaystyle P(E \cap F) = \frac{1}{52}$$

Step 4. Com­pare.

$$\displaystyle P(E)\,P(F) = \frac{1}{4}\cdot\frac{1}{13} = \frac{1}{52} = P(E \cap F)$$

Check­ing the answer#

Directly: among the $4$ kings, exactly $1$ is a heart, so $\displaystyle P(E \mid F) = \tfrac{1}{4} = P(E)$. Know­ing it is a king does not change the chance of a heart ✓.

Answer#

Yes, $E$ and $F$ are inde­pen­dent, because $\displaystyle P(E \cap F) = \tfrac{1}{52} = P(E)\,P(F)$.

Ques­tion 6: Work­ing with inde­pen­dent events#

The prob­lem#

$P(A) = 0.3$, $P(B) = 0.4$, $A$ and $B$ inde­pen­dent. Find (a) $P(A \cup B)$, (b) $P(A' \cap B')$, (c) $P(A \mid B')$.

Under­stand­ing the prob­lem#

Inde­pen­dence gives $P(A \cap B) = P(A)\,P(B)$. $A'$ and $B'$ are the com­ple­ments ("not $A$", "not $B$").

The idea#

Use $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. For com­ple­ments, use the lesson's fact that if $A$ and $B$ are inde­pen­dent, so are $A'$ and $B'$, and so are $A$ and $B'$.

Step-by-step solu­tion#

Part (a)

Step 1. Find $P(A \cap B)$ using inde­pen­dence.

$$P(A \cap B) = 0.3 \times 0.4 = 0.12$$

Step 2. Apply the addi­tion rule.

$$P(A \cup B) = 0.3 + 0.4 - 0.12 = 0.58$$

Part (b)

Step 1. $A' \cap B'$ means "nei­ther hap­pens", which is the com­ple­ment of "at least one hap­pens".

$$P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.58 = 0.42$$

Step 2. Cross-check with inde­pen­dence of $A'$ and $B'$: $0.7 \times 0.6 = 0.42$ ✓.

Part (c)

Step 1. Since $A$ and $B'$ are inde­pen­dent, know­ing $B'$ hap­pened does not change $P(A)$.

$$P(A \mid B') = P(A) = 0.3$$

Step 2. Con­firm with the def­i­n­i­tion. $P(A \cap B') = P(A)\,P(B') = 0.3 \times 0.6 = 0.18$, and $P(B') = 0.6$.

$$\displaystyle P(A \mid B') = \frac{0.18}{0.6} = 0.3$$

Check­ing the answer#

All three val­ues lie between $0$ and $1$, and (a) and (b) add to $1$, as "at least one" and "nei­ther" must.

Answer#

(a) $0.58$; (b) $0.42$; (c) $0.3$.

Ques­tion 7: Three shoot­ers#

The prob­lem#

The prob­a­bil­i­ties that three shoot­ers hit a tar­get are $\displaystyle \tfrac{1}{2}$, $\displaystyle \tfrac{1}{3}$, $\displaystyle \tfrac{1}{4}$. Find the prob­a­bil­ity that the tar­get is hit at least once.

Under­stand­ing the prob­lem#

The shoot­ers fire inde­pen­dently. "Hit at least once" means one, two or all three hit, which has many cases. The oppo­site, "nobody hits", is a sin­gle case.

The idea#

Use the com­ple­ment, as in Exam­ple 7: $P(\text{at least one hit}) = 1 - P(\text{all miss})$. Because the shoot­ers are inde­pen­dent, the prob­a­bil­ity all miss is the prod­uct of the indi­vid­ual miss prob­a­bil­i­ties.

Step-by-step solu­tion#

Step 1. Find each miss prob­a­bil­ity.

$$\displaystyle 1 - \frac{1}{2} = \frac{1}{2}, \qquad 1 - \frac{1}{3} = \frac{2}{3}, \qquad 1 - \frac{1}{4} = \frac{3}{4}$$

Step 2. Mul­ti­ply (inde­pen­dence).

$$\displaystyle P(\text{all miss}) = \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4} = \frac{6}{24} = \frac{1}{4}$$

Step 3. Take the com­ple­ment.

$$\displaystyle P(\text{hit at least once}) = 1 - \frac{1}{4} = \frac{3}{4}$$

Check­ing the answer#

The answer is larger than the best sin­gle shooter's $\displaystyle \tfrac{1}{2}$, as it should be when three peo­ple try.

Answer#

$\displaystyle \tfrac{3}{4}$.

Com­mon mis­take to avoid#

Adding $\displaystyle \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{4}$. That gives more than $1$, because the events over­lap.

Ques­tion 8: Com­ple­ments of inde­pen­dent events#

The prob­lem#

Show that if $E$ and $F$ are inde­pen­dent, so are $E'$ and $F'$.

Under­stand­ing the prob­lem#

Given: $P(E \cap F) = P(E)\,P(F)$. To show: $P(E' \cap F') = P(E')\,P(F')$.

The idea#

$E' \cap F'$ ("nei­ther") is the com­ple­ment of $E \cup F$ ("at least one"), by De Mor­gan's law. Write $P(E \cup F)$ with the addi­tion rule, use inde­pen­dence, and fac­torise.

Step-by-step solu­tion#

Step 1. Use De Mor­gan's law: $E' \cap F' = (E \cup F)'$. So

$$P(E' \cap F') = 1 - P(E \cup F)$$

Step 2. Apply the addi­tion rule.

$$P(E' \cap F') = 1 - \left[P(E) + P(F) - P(E \cap F)\right]$$

Step 3. Use inde­pen­dence: replace $P(E \cap F)$ by $P(E)\,P(F)$.

$$P(E' \cap F') = 1 - P(E) - P(F) + P(E)\,P(F)$$

Step 4. Fac­torise (group the first two and last two terms).

$$1 - P(E) - P(F)\left(1 - P(E)\right) = \left(1 - P(E)\right)\left(1 - P(F)\right)$$

Step 5. Recog­nise the com­ple­ments: $1 - P(E) = P(E')$ and $1 - P(F) = P(F')$.

$$P(E' \cap F') = P(E')\,P(F')$$

So $E'$ and $F'$ are inde­pen­dent.

Check­ing the answer#

Test with Ques­tion 6: $P(A) = 0.3$, $P(B) = 0.4$. We found $P(A' \cap B') = 0.42 = 0.7 \times 0.6$ ✓.

Answer#

$P(E' \cap F') = (1 - P(E))(1 - P(F)) = P(E')\,P(F')$, so $E'$ and $F'$ are inde­pen­dent.

Ques­tion 9: Test­ing inde­pen­dence with two dice throws#

The prob­lem#

A die is thrown twice. $A$ = first throw even, $B$ = sec­ond throw odd, $C$ = sum odd. Are $A$ and $B$ inde­pen­dent? $A$ and $C$?

Under­stand­ing the prob­lem#

There are $36$ equally likely ordered out­comes $(a, b)$. You must test two pairs of events for inde­pen­dence.

The idea#

For each pair, com­pute both sides of $P(X \cap Y) = P(X)\,P(Y)$ by count­ing out­comes.

Step-by-step solu­tion#

Step 1. Find $P(A)$, $P(B)$, $P(C)$. Half the faces are even, so $\displaystyle P(A) = \tfrac{1}{2}$ and $\displaystyle P(B) = \tfrac{1}{2}$. The sum is odd exactly when one throw is even and the other odd: $3 \times 3 + 3 \times 3 = 18$ out­comes, so $\displaystyle P(C) = \tfrac{18}{36} = \tfrac{1}{2}$.

Step 2. $A \cap B$: first even ($3$ choices) and sec­ond odd ($3$ choices).

$$\displaystyle n(A \cap B) = 3 \times 3 = 9, \qquad P(A \cap B) = \frac{9}{36} = \frac{1}{4}$$

Step 3. Com­pare: $\displaystyle P(A)\,P(B) = \tfrac{1}{2}\cdot\tfrac{1}{2} = \tfrac{1}{4}$. Equal, so $A$ and $B$ are inde­pen­dent.

Step 4. $A \cap C$: first throw even and sum odd. If the first is even, the sum is odd exactly when the sec­ond is odd. So $A \cap C$ is the same set as $A \cap B$.

$$\displaystyle P(A \cap C) = \frac{9}{36} = \frac{1}{4}$$

Step 5. Com­pare: $\displaystyle P(A)\,P(C) = \tfrac{1}{2}\cdot\tfrac{1}{2} = \tfrac{1}{4}$. Equal, so $A$ and $C$ are inde­pen­dent too.

Check­ing the answer#

Intu­itively, the two throws do not affect each other, so $A$ and $B$ being inde­pen­dent is expected. For $A$ and $C$: whether the first is even or odd, the sum is odd with prob­a­bil­ity $\displaystyle \tfrac{1}{2}$ (it depends only on the sec­ond throw), so $\displaystyle P(C \mid A) = \tfrac{1}{2} = P(C)$ ✓.

Answer#

Yes, $A$ and $B$ are inde­pen­dent; yes, $A$ and $C$ are inde­pen­dent (in each case $\displaystyle P(\cap) = \tfrac{1}{4} = \tfrac{1}{2}\cdot\tfrac{1}{2}$).

Ques­tion 10: An urn that changes after the first draw#

The prob­lem#

An urn has $3$ white and $5$ black balls. One ball is drawn, its colour noted, and it is returned with $2$ more of the same colour. A sec­ond ball is drawn. Find $P(\text{second is white})$.

Under­stand­ing the prob­lem#

The con­tents of the urn for the sec­ond draw depend on the first ball. After the first draw, the urn has $8 + 2 = 10$ balls. If the first was white, there are then $5$ white; if black, $3$ white and $7$ black.

The idea#

Split by the colour of the first ball. For each case use the mul­ti­pli­ca­tion rule $P(\text{first}) \times P(\text{second white} \mid \text{first})$, then add the two cases, since exactly one of them hap­pens.

Step-by-step solu­tion#

Step 1. Case 1: first ball white, prob­a­bil­ity $\displaystyle \tfrac{3}{8}$. The urn becomes $3 + 2 = 5$ white and $5$ black ($10$ balls).

$$\displaystyle P(W_1 \cap W_2) = \frac{3}{8}\cdot\frac{5}{10} = \frac{15}{80}$$

Step 2. Case 2: first ball black, prob­a­bil­ity $\displaystyle \tfrac{5}{8}$. The urn becomes $3$ white and $5 + 2 = 7$ black ($10$ balls).

$$\displaystyle P(B_1 \cap W_2) = \frac{5}{8}\cdot\frac{3}{10} = \frac{15}{80}$$

Step 3. Add the cases.

$$\displaystyle P(W_2) = \frac{15}{80} + \frac{15}{80} = \frac{30}{80} = \frac{3}{8}$$

Check­ing the answer#

The answer equals the orig­i­nal chance of white, $\displaystyle \tfrac{3}{8}$. That is a known fea­ture of this "add more of the same colour" scheme: before you see the first ball, the sec­ond draw is no more likely to be white than the first.

Answer#

$\displaystyle P(\text{second is white}) = \tfrac{3}{8}$.

Com­mon mis­take to avoid#

Using $8$ as the denom­i­na­tor for the sec­ond draw. After return­ing the ball and adding $2$, there are $10$ balls.