Worksheet Ch 9: Probability of Combined Events

Section A • Bahagian A [12 Marks]

Question 1 [4 Marks]

A box contains 5 red balls, 3 blue balls, and 2 green balls. A ball is picked at random, its color noted, and replaced. A second ball is then picked.

(a) State whether the two selections are independent or dependent events. [1 mark]
(b) Calculate the probability that both balls chosen are red. [3 marks]

(a) Because the first ball is replaced, the sample space remains 10

Independent events (Peristiwa tak bersandar)

(b) $P(R_1) = \frac{5}{10} = \frac{1}{2}$, $P(R_2) = \frac{5}{10} = \frac{1}{2}$

$$P(R_1 \cap R_2) = \frac{1}{2} \times \frac{1}{2} = \mathbf{\frac{1}{4}}$$

K1 N2

Question 2 [4 Marks]

In a sports carnival, events $A$ and $B$ are such that $P(A) = \frac{2}{5}$ and $P(B) = \frac{1}{3}$.

(a) If $A$ and $B$ are mutually exclusive events, find $P(A \cup B)$. [2 marks]
(b) If $A$ and $B$ are independent events, find $P(A \cup B)$. [2 marks]

(a) Mutually exclusive: $P(A \cup B) = P(A) + P(B)$

$$P(A \cup B) = \frac{2}{5} + \frac{1}{3} = \mathbf{\frac{11}{15}}$$

K1 N1

(b) Independent: $P(A \cap B) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}$

$$P(A \cup B) = \frac{2}{5} + \frac{1}{3} - \frac{2}{15} = \frac{11}{15} - \frac{2}{15} = \mathbf{\frac{9}{15} = \frac{3}{5}}$$

K1 N1

Question 3 [4 Marks]

A letter is chosen at random from the word $\text{"SUCCESS"}$ and another letter is chosen at random from the word $\text{"MATHS"}$. Find the probability that both letters chosen are consonants.

In "SUCCESS": Consonants are S, C, C, S, S (5 consonants out of 7 letters) $\implies P(C_1) = \frac{5}{7}$

In "MATHS": Consonants are M, T, H, S (4 consonants out of 5 letters) $\implies P(C_2) = \frac{4}{5}$

Independent multiplication: $\frac{5}{7} \times \frac{4}{5}$

$$P = \mathbf{\frac{4}{7}}$$

Section B • Bahagian B (HOTS / KBAT) [18 Marks]

Question 4 (Lucky Draw Voucher Without Replacement) [9 Marks]

In a supermarket lucky draw box, there are 4 vouchers of RM50 and 6 vouchers of RM20. A customer draws two vouchers consecutively without replacement.

(a) Draw or describe a complete tree diagram showing the probabilities on each branch. [3 marks]
(b) Calculate the probability that the customer wins a total value of:

(i) Exactly RM100. [2 marks]

(ii) At least RM70. [4 marks]

(a) Tree branches without replacement (Total 10 $\to$ 9):

Branch 1: $P(50) = \frac{4}{10} \implies P(50|50) = \frac{3}{9}, P(20|50) = \frac{6}{9}$
Branch 2: $P(20) = \frac{6}{10} \implies P(50|20) = \frac{4}{9}, P(20|20) = \frac{5}{9}$

(b)(i) Winning RM100 means two RM50 vouchers ($50, 50$):

$$P(50, 50) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \mathbf{\frac{2}{15}}$$

K1 N1

(b)(ii) At least RM70 means outcomes: $(50, 50)$, $(50, 20)$, or $(20, 50)$ [or $1 - P(20, 20)$]:

$$P(20, 20) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$$

$$P(\text{At least RM70}) = 1 - \frac{1}{3} = \mathbf{\frac{2}{3}}$$

K2 N2

Question 5 (Badminton Championship Tournament) [9 Marks]

School $A$ and School $B$ compete in a best-of-three badminton doubles series. The probability that School $A$ wins any single match is $\frac{3}{5}$.

(a) Find the probability that School $A$ wins the championship in straight sets (winning the first 2 matches). [3 marks]
(b) Calculate the probability that the championship series goes to a deciding third match. [3 marks]
(c) Find the overall probability that School $A$ emerges as the champion. [3 marks]

(a) Straight sets: Wins match 1 AND match 2 ($A, A$)

$$P(A, A) = \frac{3}{5} \times \frac{3}{5} = \mathbf{\frac{9}{25}}$$

K1 N2

(b) 3rd match happens if score after 2 matches is 1-1 ($A, B$ or $B, A$):

$$P(A, B) + P(B, A) = \left(\frac{3}{5} \times \frac{2}{5}\right) + \left(\frac{2}{5} \times \frac{3}{5}\right) = \frac{6}{25} + \frac{6}{25}$$

$$= \mathbf{\frac{12}{25}}$$

K1 N2

$$P = \frac{9}{25} + \left(\frac{3}{5} \times \frac{2}{5} \times \frac{3}{5}\right) + \left(\frac{2}{5} \times \frac{3}{5} \times \frac{3}{5}\right) = \frac{9}{25} + \frac{18}{125} + \frac{18}{125} = \frac{45 + 36}{125}$$

$$= \mathbf{\frac{81}{125}}$$

K1 N2