1. **Stating the problem:** We have a sports club with members practicing Volleyball (V), Basketball (B), and Tennis (T). We want to find the probability that a randomly chosen member attends the reunion (R), verify the president's claim about half not attending, and analyze the payment variable $X$. 2. **Given data:** - $p(V) = 0.30$ - $p(B) = 0.20$ (initially given, but problem states "The rest practice basketball" which is contradictory; assuming typo and correcting below) - $p(T) = 0.20$ - The rest practice basketball, so $p(B) = 1 - (0.30 + 0.20) = 0.50$ 3. **Reunion participation rates:** - $p(R|V) = 0.25$ - $p(R|B) = 0.30$ - $p(R|T) = 0.70$ 4. **Calculate $p(R)$ using total probability formula:** $$ p(R) = p(R|V) p(V) + p(R|B) p(B) + p(R|T) p(T) $$ Substitute values: $$ p(R) = 0.25 \times 0.30 + 0.30 \times 0.50 + 0.70 \times 0.20 $$ Calculate each term: $$ 0.25 \times 0.30 = 0.075 $$ $$ 0.30 \times 0.50 = 0.15 $$ $$ 0.70 \times 0.20 = 0.14 $$ Sum: $$ p(R) = 0.075 + 0.15 + 0.14 = 0.365 $$ 5. **Verify president's claim that half do not participate:** Probability not attending reunion: $$ p(R^c) = 1 - p(R) = 1 - 0.365 = 0.635 $$ This is about 63.5%, which is more than half, so the claim that half do not participate is justified. 6. **Define random variable $X$ as total payment:** - Tennis members pay 100,000 LL - Volleyball or Basketball members pay 60,000 LL - Additional 15,000 LL if attending reunion 7. **Possible values of $X$:** - $X_1 = 60,000$ (V or B, no reunion) - $X_2 = 75,000$ (V or B, reunion) - $X_3 = 100,000$ (T, no reunion) - $X_4 = 115,000$ (T, reunion) 8. **Calculate $p(X=75,000)$:** This corresponds to members practicing V or B and attending reunion: $$ p(X=75,000) = p(R \cap (V \cup B)) = p(R \cap V) + p(R \cap B) $$ Calculate each: $$ p(R \cap V) = p(R|V) p(V) = 0.25 \times 0.30 = 0.075 $$ $$ p(R \cap B) = p(R|B) p(B) = 0.30 \times 0.50 = 0.15 $$ Sum: $$ p(X=75,000) = 0.075 + 0.15 = 0.225 $$ The problem states $p(X=75,000) = 0.11$, which conflicts with data; assuming typo or different data, but based on given data, $p(X=75,000) = 0.225$. 9. **Determine probability law of $X$:** - $p(X=60,000) = p(V \cup B \cap R^c) = p(V \cap R^c) + p(B \cap R^c)$ Calculate: $$ p(V \cap R^c) = p(V) - p(R \cap V) = 0.30 - 0.075 = 0.225 $$ $$ p(B \cap R^c) = p(B) - p(R \cap B) = 0.50 - 0.15 = 0.35 $$ Sum: $$ p(X=60,000) = 0.225 + 0.35 = 0.575 $$ - $p(X=100,000) = p(T \cap R^c) = p(T) - p(R \cap T) = 0.20 - (0.70 \times 0.20) = 0.20 - 0.14 = 0.06$ - $p(X=115,000) = p(R \cap T) = 0.14$ Check sum: $$ 0.575 + 0.225 + 0.06 + 0.14 = 1.00 $$ 10. **Calculate expected value $E(X)$:** $$ E(X) = 60,000 \times 0.575 + 75,000 \times 0.225 + 100,000 \times 0.06 + 115,000 \times 0.14 $$ Calculate each term: $$ 60,000 \times 0.575 = 34,500 $$ $$ 75,000 \times 0.225 = 16,875 $$ $$ 100,000 \times 0.06 = 6,000 $$ $$ 115,000 \times 0.14 = 16,100 $$ Sum: $$ E(X) = 34,500 + 16,875 + 6,000 + 16,100 = 73,475 $$ 11. **Estimate club revenue for 200 members:** $$ \text{Revenue} = 200 \times E(X) = 200 \times 73,475 = 14,695,000 $$ **Final answers:** - $p(R) = 0.365$ - President's claim justified since $p(R^c) = 0.635 > 0.5$ - Possible values of $X$: 60,000; 75,000; 100,000; 115,000 - $p(X=75,000) = 0.225$ - Probability law of $X$ as above - $E(X) = 73,475$ - Estimated revenue = 14,695,000