Geometric Probability and Conditional Probability: practice solutions, Grade 10 – download the PDF

1 Spinner events ★★★

Prime numbers: 2, 3, 5, 7, so \(P(A)=\dfrac48=\dfrac12\). Greater than 5: 6, 7, 8, so \(P(B)=\dfrac38\).

Both: only 7, so \(P(A\cap B)=\dfrac18\). Union: \(\{2,3,5,6,7,8\}\), so \(P(A\cup B)=\dfrac68=\dfrac34\). Check: \(\dfrac48+\dfrac38-\dfrac18=\dfrac68\).

\(P(A^{c})=1-\dfrac12=\dfrac12\).

2 Two coins ★★★

(a) \(S=\{HH,HT,TH,TT\}\): 4 equally likely outcomes.

(b) Exactly one head: \(HT\) and \(TH\), so \(\dfrac24=\dfrac12\).

(c) At least one head: all but \(TT\), so \(\dfrac34\).

3 A rope cut at random ★★★

The favorable positions are the first 5 ft and the last 5 ft: \(5+5=10\) ft out of 30 ft.

\(P=\dfrac{10}{30}=\dfrac13\).

4 Rainy weekend ★★★

The complement rule gives \(1-0.35=0.65\).

“Rains or does not rain” is certain, so its probability is \(0.35+0.65=1\).

5 Lunch menu ★★★

(a) By the Fundamental Counting Principle, \(4\cdot3\cdot2=24\) meals.

(b) With the chicken sandwich fixed there are \(1\cdot3\cdot2=6\) meals, so \(P=\dfrac{6}{24}=\dfrac14\).

6 True or false: adding probabilities ★★★

False. There are 13 hearts and 4 kings, but the king of hearts is in both events. So \(P(A\cup B)=\dfrac{13+4-1}{52}=\dfrac{16}{52}\), not \(\dfrac{17}{52}\).

The formula without subtraction works only for mutually exclusive events.

7 Books on a shelf ★★★

(a) \(5!=120\).

(b) Treat the two books as one block: \(4!\) arrangements of the 4 objects, times \(2\) orders inside the block, so \(4!\cdot2=48\).

The probability of a random arrangement having them together is \(\dfrac{48}{120}=\dfrac25\).

8 Dartboard with a circle ★★★

Circle area: \(\pi\cdot6^2=36\pi\). Square area: \(16^2=256\).

\[ P=\dfrac{36\pi}{256}=\dfrac{9\pi}{64}\approx0.442. \]

9 Soccer and basketball ★★★

(a) \(22+15-7=30\) play at least one, so \(\dfrac{30}{40}=\dfrac34\).

(b) \(40-30=10\), so \(\dfrac{10}{40}=\dfrac14\).

(c) \(22-7=15\), so \(\dfrac{15}{40}=\dfrac38\).

10 Independent events ★★★

\(P(A\cap B)=0.4\cdot0.5=0.2\).

\(P(A\cup B)=0.4+0.5-0.2=0.7\).

Neither: \(1-0.7=0.3\). Check: \(0.6\cdot0.5=0.3\), because complements of independent events are independent.

11 Pet owners ★★★

Dog owners: \(8+17=25\). Cat owners: \(8+12=20\).

(a) \(P(\text{cat}\mid\text{dog})=\dfrac{8}{25}=0.32\).

(b) \(P(\text{dog}\mid\text{cat})=\dfrac{8}{20}=0.4\).

(c) \(P(\text{cat})=\dfrac{20}{50}=0.4\), which differs from \(0.32\). The events are not independent.

12 Team or roles ★★★

(a) Order does not matter: \({}_{10}C_4=\dfrac{10\cdot9\cdot8\cdot7}{4!}=\dfrac{5040}{24}=210\).

(b) Order matters: \({}_{10}P_4=10\cdot9\cdot8\cdot7=5040\).

Each team of 4 can be assigned the 4 roles in \(4!=24\) ways, and \(210\cdot24=5040\).

13 Flower bed in a garden ★★★

Triangle area: \(\dfrac12\cdot15\cdot8=60\text{ ft}^2\). Garden area: \(30\cdot20=600\text{ ft}^2\).

\(P(\text{flower bed})=\dfrac{60}{600}=\dfrac1{10}\) and \(P(\text{not})=1-\dfrac1{10}=\dfrac9{10}\).

14 Two dice ★★★

There are \(6\cdot6=36\) outcomes.

(a) Sum 9: \((3,6),(4,5),(5,4),(6,3)\), so \(\dfrac4{36}=\dfrac19\).

(b) Sum at most 4: \((1,1),(1,2),(2,1),(1,3),(3,1),(2,2)\), so \(\dfrac6{36}=\dfrac16\).

15 Given the sum is 8 ★★★

Sum 8: \((2,6),(3,5),(4,4),(5,3),(6,2)\), 5 outcomes. Those with a 6: \((2,6),(6,2)\), 2 outcomes.

\(P=\dfrac{2/36}{5/36}=\dfrac25\).

16 Marbles without replacement ★★★

(a) \(P=\dfrac58\cdot\dfrac47=\dfrac{20}{56}=\dfrac5{14}\).

(b) After one red is removed, 4 of the 7 marbles are red: \(\dfrac47\).

(c) Red then blue: \(\dfrac58\cdot\dfrac37=\dfrac{15}{56}\). Blue then red: \(\dfrac38\cdot\dfrac57=\dfrac{15}{56}\). Total \(\dfrac{30}{56}=\dfrac{15}{28}\).

17 Subscribers and age ★★★

Subscribers: \(45+30=75\), so \(P(\text{sub})=\dfrac{75}{200}=0.375\).

First way: \(P(\text{sub}\mid\text{under 30})=\dfrac{45}{120}=0.375=P(\text{sub})\).

Second way: \(P(\text{sub}\cap\text{under 30})=\dfrac{45}{200}=0.225\) and \(P(\text{sub})\cdot P(\text{under 30})=0.375\cdot0.6=0.225\).

The two numbers match, so the events are independent.

18 Photo lineup ★★★

(a) Glue Mia and Leo into one block: \(5!=120\) arrangements, times 2 orders inside the block, so \(240\).

(b) There are \(6!=720\) lineups in all, so \(P=\dfrac{240}{720}=\dfrac13\).

19 Committee probabilities ★★★

Total committees: \({}_9C_4=126\).

(a) \({}_5C_2\cdot{}_4C_2=10\cdot6=60\), so \(P=\dfrac{60}{126}=\dfrac{10}{21}\).

(b) \({}_5C_4=5\), so \(P=\dfrac5{126}\).

20 Meeting problem ★★★

Let \(x\) and \(y\) be the arrival times in minutes. They meet when \(|x-y|\le15\). The sample space is the square of area \(60^2=3600\).

The failure region is two right triangles with legs \(45\), each of area \(\dfrac12\cdot45^2=1012.5\), so \(2025\) in all.

\(P(\text{meet})=1-\dfrac{2025}{3600}=1-\dfrac9{16}=\dfrac7{16}=0.4375\).

21 Two machines ★★★

(a) \(P(D)=0.6\cdot0.02+0.4\cdot0.05=0.012+0.020=0.032\).

(b) \(P(B\mid D)=\dfrac{P(B\cap D)}{P(D)}=\dfrac{0.020}{0.032}=0.625=\dfrac58\).

So even though B makes fewer parts, it accounts for 62.5% of the defective ones.