Exercise 6A: Combining Events

1. A bag contains 8 blue and 6 red balls. Total = 14.(a) Find the probability of picking a blue ball: * \(P(\text{Blue}) = \frac{\text{Number of Blue Balls}}{\text{Total Balls}} = \frac{8}{14} = \frac{4}{7}\)

2. Find the probability of picking a red ball: * \(P(\text{Red}) = \frac{\text{Number of Red Balls}}{\text{Total Balls}} = \frac{6}{14} = \frac{3}{7}\)

2. If \(P(A)=0.4\), \(P(B)=0.5\), and A and B are independent events, find \(P(A \cap B)\).

For independent events: \(P(A \cap B) = P(A) \times P(B)\)\(P(A \cap B) = 0.4 \times 0.5 = 0.2\)

3. In a group of 40 students, 25 like Mathematics (M), 20 like Science (S), and 10 like both.

1. Find the probability that a student chosen at random likes Mathematics or Science:Use the Addition Rule: \(P(M \cup S) = P(M) + P(S) - P(M \cap S)\)\(P(M \cup S) = \frac{25}{40} + \frac{20}{40} - \frac{10}{40} = \frac{35}{40} = \frac{7}{8}\)

2. Find the probability that a student likes neither subject:\(P(\text{Neither}) = 1 - P(M \cup S) = 1 - \frac{35}{40} = \frac{5}{40} = \frac{1}{8}\)

Exercise 6B: Conditional Events

1.Given \(P(A)=0.7\), \(P(B)=0.4\), and \(P(A \cap B)=0.28\), are A and B independent?Check if \(P(A) \times P(B) = P(A \cap B)\)\(0.7 \times 0.4 = 0.28\).

Answer: Yes, they are independent because the product equals the intersection.

2. If \(P(A)=0.6\) and \(P(B|A)=0.5\),

find \(P(A \cap B)\).Use the Multiplication Rule: \(P(A \cap B) = P(A) \times P(B|A)\)\(P(A \cap B) = 0.6 \times 0.5 = 0.3\)

3. A box contains 5 white (W) and 3 black (B) marbles. Two marbles are drawn without replacement.

1. Find the probability that both are white:\(P(W_1 \cap W_2) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\)

2. Find the probability that the second is black, given that the first was white:\(P(B_2 | W_1) = \frac{3}{7}\) (Only 7 marbles remain after one white is taken).

Exercise 6C: Tree Diagrams

1. A coin is tossed twice.

1. Probability of two heads (HH): \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)

2. Probability of at least one tail: \(1 - P(HH) = 1 - \frac{1}{4} = \frac{3}{4}\)ShutterstockExplore

2. The probability it rains is 0.3. If it rains, Ahmed goes to market with prob 0.4. If no rain, he goes with prob 0.8.\(P(\text{Market}) = P(\text{Rain} \cap \text{Market}) + P(\text{No Rain} \cap \text{Market})\)\(P(\text{Market}) = (0.3 \times 0.4) + (0.7 \times 0.8) = 0.12 + 0.56 = 0.68\)

3. A bag has 3 red and 7 blue pens. Two pens are selected without replacement.Probability of different colors: \(P(RB) + P(BR)\)\(P = (\frac{3}{10} \times \frac{7}{9}) + (\frac{7}{10} \times \frac{3}{9}) = \frac{21}{90} + \frac{21}{90} = \frac{42}{90} = \frac{7}{15}\)

Exercise 6D: Counting Principles and Permutations

1. In how many ways can 4 people be seated in a row of 4 chairs?\(4! = 4 \times 3 \times 2 \times 1 = 24\) ways.

2. 3-letter codes from {A, B, C, D, E} without repetition:\(^5P_3 = \frac{5!}{(5-3)!} = 5 \times 4 \times 3 = 60\) codes.

3. Find the value of:\(6! = 720\)\(^{10}P_2 = 10 \times 9 = 90\)

4. How many ways can the letters of the word “MOGADISHU” be arranged?The word has 9 unique letters.\(9! = 362,880\) ways.

Exercise 6E: Combinations

1. A committee of 4 members from 9 people:\(^9C_4 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126\) ways.

2. Find the value of:\(^7C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\)\(^5C_5 = 1\)

3. Box with 6 red and 4 green balls. Select 3 (2 red, 1 green):\(^6C_2 \times ^4C_1 = 15 \times 4 = 60\) ways.

4. In a test, a student must choose 5 questions out of 8:\(^8C_5 = \frac{8 \times 7 \times 6 \times 5 \times 4}{5 \times 4 \times 3 \times 2 \times 1} = 56\) choices.