Rules of Probability and Mixed Practice, Worked Solutions

Ques­tion 1: Two dice#

The prob­lem#

Two dice are rolled. Find $P(\text{sum} = 9)$, $P(\text{doublet})$, $P(\text{sum} \ge 10)$.

Under­stand­ing the prob­lem#

Each die shows $1$ to $6$, so the out­comes are ordered pairs $(a, b)$. A dou­blet means both dice show the same num­ber. You need three sep­a­rate prob­a­bil­i­ties.

The idea#

All $6 \times 6 = 36$ ordered pairs are equally likely, so

$$\displaystyle P(E) = \frac{\text{number of outcomes in } E}{36}.$$

List the favourable pairs for each event care­fully.

Step-by-step solu­tion#

Part (a): sum $= 9$

Step 1. List the pairs with $a + b = 9$.

$$(3, 6),\ (4, 5),\ (5, 4),\ (6, 3)$$

Step 2. There are $4$ of them.

$$\displaystyle P(\text{sum} = 9) = \frac{4}{36} = \frac{1}{9}$$

Part (b): dou­blet

Step 1. The dou­blets are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$: $6$ out­comes.

$$\displaystyle P(\text{doublet}) = \frac{6}{36} = \frac{1}{6}$$

Part (c): sum $\ge 10$

Step 1. Count sum $10$, $11$ and $12$ sep­a­rately.

$$\begin{aligned} \text{sum } 10 &: (4,6), (5,5), (6,4) \quad \to 3 \\ \text{sum } 11 &: (5,6), (6,5) \quad \to 2 \\ \text{sum } 12 &: (6,6) \quad \to 1 \end{aligned}$$

Step 2. Total $3 + 2 + 1 = 6$.

$$\displaystyle P(\text{sum} \ge 10) = \frac{6}{36} = \frac{1}{6}$$

Check­ing the answer#

$(4, 5)$ and $(5, 4)$ are dif­fer­ent out­comes (the dice are dif­fer­ent), which is why sum $9$ has $4$ out­comes, not $2$. All three prob­a­bil­i­ties lie between $0$ and $1$.

Answer#

$\displaystyle P(\text{sum} = 9) = \frac{1}{9}$, $\displaystyle P(\text{doublet}) = \frac{1}{6}$, $\displaystyle P(\text{sum} \ge 10) = \frac{1}{6}$.

Com­mon mis­take to avoid#

Using $21$ unordered pairs as the sam­ple space. Those are not equally likely; the $36$ ordered pairs are.

Ques­tion 2: Three coins#

The prob­lem#

Three coins are tossed. Find $P(\text{at least two heads})$ and $P(\text{no tail})$.

Under­stand­ing the prob­lem#

"At least two heads" means two heads or three heads. "No tail" means every coin shows a head.

The idea#

The sam­ple space has $2^3 = 8$ equally likely out­comes. List them and count.

Step-by-step solu­tion#

Step 1. Write the sam­ple space.

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

Step 2. At least two heads: $HHH, HHT, HTH, THH$, that is $4$ out­comes.

$$\displaystyle P(\text{at least two heads}) = \frac{4}{8} = \frac{1}{2}$$

Step 3. No tail: only $HHH$.

$$\displaystyle P(\text{no tail}) = \frac{1}{8}$$

Check­ing the answer#

By sym­me­try, "at least two heads" and "at least two tails" are equally likely and together cover all $8$ out­comes, so each must be $\displaystyle \frac{1}{2}$.

Answer#

$\displaystyle P(\text{at least two heads}) = \frac{1}{2}$; $\displaystyle P(\text{no tail}) = \frac{1}{8}$.

Ques­tion 3: Using the rules with given prob­a­bil­i­ties#

The prob­lem#

$P(A) = 0.42$, $P(B) = 0.48$, $P(A \cap B) = 0.16$. Find $P(\text{not } A)$, $P(\text{not } B)$, $P(A \text{ or } B)$, $P(\text{neither})$.

Under­stand­ing the prob­lem#

"Not $A$" is the com­ple­ment $A'$. "$A$ or $B$" is $A \cup B$. "Nei­ther" means not $A$ and not $B$, that is $A' \cap B'$.

The idea#

Use the com­ple­ment rule $P(A') = 1 - P(A)$, the addi­tion rule, and the fact that $A' \cap B' = (A \cup B)'$ (De Mor­gan's law), as in Exam­ple 1.

Step-by-step solu­tion#

Step 1. Com­ple­ments.

$$P(A') = 1 - 0.42 = 0.58, \qquad P(B') = 1 - 0.48 = 0.52$$

Step 2. Addi­tion rule.

$$P(A \cup B) = 0.42 + 0.48 - 0.16 = 0.74$$

Step 3. Nei­ther.

$$P(A' \cap B') = P\big((A \cup B)'\big) = 1 - 0.74 = 0.26$$

Check­ing the answer#

"$A$ or $B$" and "nei­ther" together cover every­thing, and $0.74 + 0.26 = 1$.

Answer#

$P(\text{not } A) = 0.58$, $P(\text{not } B) = 0.52$, $P(A \text{ or } B) = 0.74$, $P(\text{neither}) = 0.26$.

Com­mon mis­take to avoid#

$P(A' \cap B')$ is not $P(A') \times P(B')$ here; noth­ing says the events are inde­pen­dent. Use the com­ple­ment of the union.

Ques­tion 4: Foot­ball and cricket#

The prob­lem#

In a class of $60$, $30$ play foot­ball, $32$ play cricket and $24$ play both. A stu­dent is cho­sen at ran­dom. Find $P(\text{plays neither})$ and $P(\text{plays exactly one})$.

Under­stand­ing the prob­lem#

Let $F$ = plays foot­ball, $C$ = plays cricket. "Both" is $F \cap C$. "Exactly one" means foot­ball only or cricket only, not both.

The idea#

Count with the addi­tion rule: $n(F \cup C) = n(F) + n(C) - n(F \cap C)$. Then divide counts by $60$.

Step-by-step solu­tion#

Step 1. Stu­dents who play at least one game.

$$n(F \cup C) = 30 + 32 - 24 = 38$$

Step 2. Stu­dents who play nei­ther: $60 - 38 = 22$.

$$\displaystyle P(\text{neither}) = \frac{22}{60} = \frac{11}{30}$$

Step 3. Exactly one = at least one minus both: $38 - 24 = 14$. (Check: foot­ball only $30 - 24 = 6$, cricket only $32 - 24 = 8$, and $6 + 8 = 14$.)

$$\displaystyle P(\text{exactly one}) = \frac{14}{60} = \frac{7}{30}$$

Check­ing the answer#

The four groups: foot­ball only $6$, cricket only $8$, both $24$, nei­ther $22$. Total $6 + 8 + 24 + 22 = 60$. ✓

Answer#

$\displaystyle P(\text{neither}) = \frac{11}{30}$; $\displaystyle P(\text{exactly one}) = \frac{7}{30}$.

Ques­tion 5: Choos­ing boys and girls#

The prob­lem#

From a group of $6$ boys and $4$ girls, $3$ are cho­sen at ran­dom. Find $P(\text{exactly } 2 \text{ girls})$ and $P(\text{at least } 1 \text{ girl})$.

Under­stand­ing the prob­lem#

Order does not mat­ter (it is a group), so we count selec­tions, using com­bi­na­tions. There are $10$ peo­ple in all.

The idea#

Total ways $= {}^{10}C_3$. For "exactly 2 girls", choose $2$ girls and $1$ boy. For "at least 1 girl", use the com­ple­ment "no girl", as in Exam­ple 4.

Step-by-step solu­tion#

Step 1. Total selec­tions.

$$\displaystyle ^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$$

Step 2. Exactly $2$ girls: choose $2$ of $4$ girls and $1$ of $6$ boys.

$$\displaystyle ^4C_2 \times {}^6C_1 = 6 \times 6 = 36, \qquad P = \frac{36}{120} = \frac{3}{10}$$

Step 3. No girl: all $3$ cho­sen from the $6$ boys.

$$\displaystyle ^6C_3 = \frac{6 \times 5 \times 4}{6} = 20$$

Step 4. Com­ple­ment rule.

$$\displaystyle P(\text{at least 1 girl}) = 1 - \frac{20}{120} = 1 - \frac{1}{6} = \frac{5}{6}$$

Check­ing the answer#

Count directly: $1$ girl $= 4 \times 15 = 60$, $2$ girls $= 36$, $3$ girls $= 4$; total $100$, and $\displaystyle \frac{100}{120} = \frac{5}{6}$. ✓

Answer#

$\displaystyle P(\text{exactly 2 girls}) = \frac{3}{10}$; $\displaystyle P(\text{at least 1 girl}) = \frac{5}{6}$.

Ques­tion 6: Two balls of the same colour#

The prob­lem#

A bag has $5$ red, $4$ blue and $3$ green balls. Two are drawn. Find $P(\text{same colour})$.

Under­stand­ing the prob­lem#

Two balls are drawn together (a selec­tion of $2$ from $12$). "Same colour" means both red, or both blue, or both green. These three cases can­not hap­pen together.

The idea#

Count each case with com­bi­na­tions and add (the cases are mutu­ally exclu­sive), then divide by the total $^{12}C_2$.

Step-by-step solu­tion#

Step 1. Total ways: $\displaystyle ^{12}C_2 = \frac{12 \times 11}{2} = 66$.

Step 2. Ways for each colour.

$$^5C_2 = 10, \qquad {}^4C_2 = 6, \qquad {}^3C_2 = 3$$

Step 3. Add and divide.

$$\displaystyle P(\text{same colour}) = \frac{10 + 6 + 3}{66} = \frac{19}{66}$$

Check­ing the answer#

Dif­fer­ent-colour pairs: $5 \cdot 4 + 5 \cdot 3 + 4 \cdot 3 = 20 + 15 + 12 = 47$, and $19 + 47 = 66$. ✓

Answer#

$\displaystyle P(\text{same colour}) = \frac{19}{66}$.

Ques­tion 7: A lot­tery#

The prob­lem#

In a lot­tery a per­son picks $6$ num­bers from $1$ to $20$. Six num­bers are drawn at ran­dom. What is the prob­a­bil­ity of win­ning (all six match)?

Under­stand­ing the prob­lem#

Order does not mat­ter; only which six num­bers are drawn. Exactly one of the pos­si­ble sets of six is the per­son's set.

The idea#

Num­ber of pos­si­ble draws $= {}^{20}C_6$, all equally likely; exactly one is favourable.

Step-by-step solu­tion#

Step 1. Count the pos­si­ble draws.

$$\displaystyle ^{20}C_6 = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 38{,}760$$

Step 2. One favourable draw.

$$\displaystyle P(\text{win}) = \frac{1}{38{,}760}$$

Check­ing the answer#

Sim­plify the prod­uct step by step: $\displaystyle \frac{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15}{720}$. The numer­a­tor is $27{,}907{,}200$, and $27{,}907{,}200 \div 720 = 38{,}760$. ✓

Answer#

$\displaystyle P(\text{win}) = \frac{1}{38{,}760}$.

Ques­tion 8: Girls sit­ting together#

The prob­lem#

$5$ boys and $3$ girls sit in a row at ran­dom. Find $P(\text{the girls are all together})$.

Under­stand­ing the prob­lem#

Here order mat­ters (seats in a row), so we count arrange­ments. "All together" means the three girls occupy three seats next to one another.

The idea#

Total arrange­ments of $8$ peo­ple $= 8!$. For favourable ones, tie the three girls into one block: arrange the block with the $5$ boys, then arrange the girls inside the block. This is the same idea as Exam­ple 5.

Step-by-step solu­tion#

Step 1. Total arrange­ments: $8! = 40{,}320$.

Step 2. Treat the girls as one unit. Then there are $5 + 1 = 6$ units, arranged in $6! = 720$ ways.

Step 3. Inside the block the $3$ girls can be arranged in $3! = 6$ ways.

Step 4. Favourable arrange­ments: $6! \times 3! = 720 \times 6 = 4{,}320$.

Step 5. Prob­a­bil­ity.

$$\displaystyle P = \frac{6!\, 3!}{8!} = \frac{3!}{8 \times 7} = \frac{6}{56} = \frac{3}{28}$$

Check­ing the answer#

Another way: the girls' three seats can be any $3$ of $8$, $^8C_3 = 56$ equally likely choices, and $6$ of them are con­sec­u­tive blocks (start­ing at seat $1$ to $6$). $\displaystyle \frac{6}{56} = \frac{3}{28}$. ✓

Answer#

$\displaystyle P(\text{girls all together}) = \frac{3}{28}$.

Ques­tion 9: Mutu­ally exclu­sive events#

The prob­lem#

Given $\displaystyle P(A) = \frac{3}{5}$ and $\displaystyle P(B) = \frac{1}{5}$, find $P(A \cup B)$ if $A$ and $B$ are mutu­ally exclu­sive.

Under­stand­ing the prob­lem#

Mutu­ally exclu­sive means $A$ and $B$ can­not hap­pen together, so $P(A \cap B) = 0$.

The idea#

Use the addi­tion rule with the last term equal to $0$.

Step-by-step solu­tion#

Step 1. Write the addi­tion rule.

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Step 2. Sub­sti­tute, with $P(A \cap B) = 0$.

$$\displaystyle P(A \cup B) = \frac{3}{5} + \frac{1}{5} - 0 = \frac{4}{5}$$

Check­ing the answer#

$\displaystyle \frac{4}{5} \le 1$, so it is a valid prob­a­bil­ity.

Answer#

$\displaystyle P(A \cup B) = \frac{4}{5}$.

Ques­tion 10: Four-digit num­bers from 1, 2, 3, 4, 5#

The prob­lem#

A num­ber is cho­sen from the $4$-digit num­bers formed by $1, 2, 3, 4, 5$ with­out rep­e­ti­tion. Find $P(\text{it is divisible by } 5)$ and $P(\text{it is even})$.

Under­stand­ing the prob­lem#

"With­out rep­e­ti­tion" means each digit is used at most once in a num­ber. Divis­i­bil­ity by $5$ and even­ness depend only on the last (units) digit.

The idea#

Count all such num­bers with the fun­da­men­tal prin­ci­ple of count­ing, then count those with the right last digit: fix the last digit first, then fill the other places.

Step-by-step solu­tion#

Step 1. Total num­bers: $5 \times 4 \times 3 \times 2 = {}^5P_4 = 120$.

Step 2. Divis­i­ble by $5$: the last digit must be $5$ (there is no $0$). Fix it; the other three places use the remain­ing $4$ dig­its.

$$\displaystyle 4 \times 3 \times 2 = 24, \qquad P = \frac{24}{120} = \frac{1}{5}$$

Step 3. Even: the last digit must be $2$ or $4$, that is $2$ choices. Then fill the other three places from the remain­ing $4$ dig­its.

$$\displaystyle 2 \times (4 \times 3 \times 2) = 48, \qquad P = \frac{48}{120} = \frac{2}{5}$$

Check­ing the answer#

By sym­me­try, each of the $5$ dig­its is equally likely to be last, so $\displaystyle P(\text{last} = 5) = \frac{1}{5}$ and $\displaystyle P(\text{last} \in \{2, 4\}) = \frac{2}{5}$. ✓

Answer#

$\displaystyle P(\text{divisible by } 5) = \frac{1}{5}$; $\displaystyle P(\text{even}) = \frac{2}{5}$.

Ques­tion 11: Can these prob­a­bil­i­ties be true?#

The prob­lem#

Can $P(A) = 0.6$, $P(B) = 0.5$, $P(A \cap B) = 0.05$ be true? Explain.

Under­stand­ing the prob­lem#

You must check whether these three num­bers are con­sis­tent with the axioms of prob­a­bil­ity. Every prob­a­bil­ity, includ­ing $P(A \cup B)$, must lie between $0$ and $1$.

The idea#

Com­pute $P(A \cup B)$ with the addi­tion rule and see whether it is at most $1$.

Step-by-step solu­tion#

Step 1. Use the addi­tion rule.

$$P(A \cup B) = 0.6 + 0.5 - 0.05 = 1.05$$

Step 2. A prob­a­bil­ity can never exceed $1$, but $1.05 > 1$.

Step 3. So the given val­ues con­tra­dict the axioms and can­not all be true.

Check­ing the answer#

For $P(A \cup B) \le 1$ we would need $P(A \cap B) \ge 0.6 + 0.5 - 1 = 0.1$. The given $0.05$ is too small, con­firm­ing the con­clu­sion.

Answer#

No. These val­ues give $P(A \cup B) = 1.05 > 1$, which is impos­si­ble.

Ques­tion 12: Mul­ti­ple of 3 or a square#

The prob­lem#

Out of $50$ tick­ets num­bered $1$–$50$, one is drawn. Find $P(\text{it is a multiple of } 3 \text{ or a square})$.

Under­stand­ing the prob­lem#

"Or" means at least one of the two. Some num­bers are both a mul­ti­ple of $3$ and a per­fect square, and they must not be counted twice.

The idea#

Use the addi­tion rule in count­ing form, as in Exam­ple 6: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.

Step-by-step solu­tion#

Step 1. Mul­ti­ples of $3$ up to $50$: $3, 6, \ldots, 48$. Since $48 = 3 \times 16$, there are $16$.

Step 2. Squares up to $50$: $1, 4, 9, 16, 25, 36, 49$, that is $7$.

Step 3. Both: squares that are mul­ti­ples of $3$: $9$ and $36$, that is $2$.

Step 4. Addi­tion rule.

$$\displaystyle n = 16 + 7 - 2 = 21, \qquad P = \frac{21}{50}$$

Check­ing the answer#

Squares that are not mul­ti­ples of $3$: $1, 4, 16, 25, 49$ (five). $16 + 5 = 21$. ✓

Answer#

$\displaystyle P(\text{multiple of 3 or a square}) = \frac{21}{50}$.