Here’s how we can solve each part of the problems provided in the image:

Problem 18:

Using the table provided:

(a) $P(\text{likes school} \mid \text{student is female}) = \frac{\text{Females who like school}}{\text{Total females}} = \frac{10}{14} = 0.714$.

(b) $P(\text{dislikes school OR student is male})$:
We use the formula for union: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ Where:

• $A = \text{dislikes school}$, $P(A) = \frac{8}{33}$
• $B = \text{student is male}$, $P(B) = \frac{19}{33}$
• $P(A \cap B) = \frac{\text{Males who dislike school}}{\text{Total}} = \frac{5}{33}$

$P(\text{dislikes school OR student is male}) = \frac{8}{33} + \frac{19}{33} - \frac{5}{33} = \frac{22}{33} = 0.667$

Problem 19:

Tree Diagram & Probability of at least one boy or one girl:
We can solve this by finding the complement of the probability that all four selected students are boys.

• Total: $19$ males, $14$ females.
• Probability of selecting all boys: $P(\text{all boys}) = \frac{19}{33} \cdot \frac{18}{32} \cdot \frac{17}{31} \cdot \frac{16}{30}$

Let’s compute it: $P(\text{all boys}) = \frac{19 \cdot 18 \cdot 17 \cdot 16}{33 \cdot 32 \cdot 31 \cdot 30} = 0.159$

Probability of at least one boy or one girl: $P(\text{at least one boy or one girl}) = 1 - P(\text{all boys}) = 1 - 0.159 = 0.841$

Problem 20:

Probability of bad luck:
Joe has:

• 5 shorts: $2$ grey, $1$ green, $2$ blue.
• 5 shirts: $2$ green, $1$ black, $2$ blue.

Bad luck occurs if both the selected shirt and shorts are the same color:

• Probability of grey: $P(\text{grey}) = \frac{2}{5} \cdot \frac{2}{5} = \frac{4}{25}$.
• Probability of green: $P(\text{green}) = \frac{1}{5} \cdot \frac{2}{5} = \frac{2}{25}$.
• Probability of blue: $P(\text{blue}) = \frac{2}{5} \cdot \frac{2}{5} = \frac{4}{25}$.

Total probability: $P(\text{bad luck}) = \frac{4}{25} + \frac{2}{25} + \frac{4}{25} = \frac{10}{25} = 0.4$

Problem 21:

Probability of a striker NOT being selected for an interview:

• Probability of being selected for an interview: $0.37$.
• Given that a player is selected:
  • Probability of being a striker: $0.55$.

Probability of NOT being selected as a striker: $P(\text{NOT striker}) = 1 - P(\text{striker}) = 1 - 0.55 = 0.45$

Total probability of a striker NOT being selected for an interview: $P(\text{striker NOT selected}) = P(\text{not interviewed}) + P(\text{interviewed AND not a striker})$

Using: $P(\text{not interviewed}) = 1 - 0.37 = 0.63$

$P(\text{interviewed AND not a striker}) = 0.37 \cdot 0.45 = 0.1665$

$P(\text{striker NOT selected}) = 0.63 + 0.1665 = 0.7965$

Do you want detailed calculations for any part?

Expanded Questions:

1. How would you use a tree diagram to visualize Problem 19?
2. How can conditional probability principles simplify Problem 18(b)?
3. Can you express Problem 20 using combinatorics instead of probability fractions?
4. What assumptions do we make about independence in Problem 21?
5. How does the union formula in probability simplify Problem 18(b)?

Tip:

Always start by breaking problems into smaller, independent parts—especially for conditional and joint probabilities!