Probability — practice questions

We are given: $n(\text{Total}) = 30$, $n(C) = 15$, $n(P) = 20$, and $n(C' \cap P') = 6$. The number of students studying at least one subject is the total minus those studying neither: $n(C \cup P) = 30 - 6 = 24$. We can now use probabilities. $P(C) = 15/30 = 0.5$, $P(P) = 20/30 = 2/3$, $P(C \cup P) = 24/30 = 0.8$. We use the general addition rule: $P(C \cup P) = P(C) + P(P) - P(C \cap P)$. Rearranging to find the intersection: $P(C \cap P) = P(C) + P(P) - P(C \cup P)$. Substituting the values: $P(C \cap P) = 0.5 + \frac{2}{3} - 0.8$. $P(C \cap P) = \frac{1}{2} + \frac{2}{3} - \frac{4}{5} = \frac{15+20-24}{30} = \frac{11}{30}$. The probability that a student studies both subjects is $\frac{11}{30}$.

This is a 'without replacement' problem, so events are dependent. A tree diagram is a useful tool. Let $R_1$ be the event the first ball is red, and $B_1$ be the event the first is blue. Similarly for the second draw ($R_2, B_2$). $P(R_1) = 5/9$, $P(B_1) = 4/9$. The second-stage probabilities are conditional: $P(R_2|R_1) = 4/8$ (4 red left out of 8 total) $P(B_2|R_1) = 4/8$ (4 blue left out of 8 total) $P(R_2|B_1) = 5/8$ (5 red left out of 8 total) $P(B_2|B_1) = 3/8$ (3 blue left out of 8 total)

(i) Probability both are red is $P(R_1 \cap R_2)$. We multiply along the branches: $P(R_1 \cap R_2) = P(R_1) \times P(R_2|R_1) = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72} = \frac{5}{18}$.

(ii) The event 'the second ball is blue' can happen in two ways: Red then Blue, OR Blue then Blue. We find the probability of each path and add them (as they are mutually exclusive paths). Path 1: $P(R_1 \cap B_2) = P(R_1) \times P(B_2|R_1) = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72}$. Path 2: $P(B_1 \cap B_2) = P(B_1) \times P(B_2|B_1) = \frac{4}{9} \times \frac{3}{8} = \frac{12}{72}$. Total probability $P(B_2) = P(R_1 \cap B_2) + P(B_1 \cap B_2) = \frac{20}{72} + \frac{12}{72} = \frac{32}{72} = \frac{4}{9}$.