1. **State the problem:** We have 386 students signing up for trips to Thailand, Egypt, and Chile with given percentages for each country and their intersections, including those who did not sign up. 2. **Given data:** - Total students $= 386$ - $P(T) = 45\% = 0.45$ - $P(E) = 35\% = 0.35$ - $P(C) = 31\% = 0.31$ - $P(T \cap E \cap C^c) = 15\% = 0.15$ - $P(E \cap C \cap T^c) = 5\% = 0.05$ - $P(T \cap C \cap E^c) = 9\% = 0.09$ - $P(\text{no trip}) = 10.5\% = 0.105$ 3. **Find:** Number of students who signed up for all three trips, i.e., $P(T \cap E \cap C)$. 4. **Use the Inclusion-Exclusion Principle for three sets:** $$ P(T \cup E \cup C) = P(T) + P(E) + P(C) - P(T \cap E) - P(E \cap C) - P(T \cap C) + P(T \cap E \cap C) $$ 5. **Calculate $P(T \cup E \cup C)$:** Since $10.5\%$ did not sign up, those who signed up are: $$ P(T \cup E \cup C) = 1 - 0.105 = 0.895 $$ 6. **Express the pairwise intersections:** We know the percentages for exactly two trips (excluding the third): - $P(T \cap E \cap C^c) = 0.15$ - $P(E \cap C \cap T^c) = 0.05$ - $P(T \cap C \cap E^c) = 0.09$ The total pairwise intersections are: $$ P(T \cap E) = P(T \cap E \cap C^c) + P(T \cap E \cap C) = 0.15 + x $$ $$ P(E \cap C) = P(E \cap C \cap T^c) + P(T \cap E \cap C) = 0.05 + x $$ $$ P(T \cap C) = P(T \cap C \cap E^c) + P(T \cap E \cap C) = 0.09 + x $$ where $x = P(T \cap E \cap C)$ is the unknown we want to find. 7. **Substitute into Inclusion-Exclusion formula:** $$ 0.895 = 0.45 + 0.35 + 0.31 - (0.15 + x) - (0.05 + x) - (0.09 + x) + x $$ 8. **Simplify the right side:** $$ 0.895 = 1.11 - (0.15 + 0.05 + 0.09) - 3x + x $$ $$ 0.895 = 1.11 - 0.29 - 2x $$ $$ 0.895 = 0.82 - 2x $$ 9. **Solve for $x$:** $$ 0.895 - 0.82 = -2x $$ $$ 0.075 = -2x $$ $$ x = -\frac{0.075}{2} = -0.0375 $$ A negative probability is impossible, so re-check the algebra: Step 8 correction: $$ 0.895 = 1.11 - 0.29 - 2x $$ $$ 0.895 = 0.82 - 2x $$ Move $0.82$ to left: $$ 0.895 - 0.82 = -2x $$ $$ 0.075 = -2x $$ Divide both sides by -2: $$ x = -\frac{0.075}{2} = -0.0375 $$ Negative again, so check the sign in the Inclusion-Exclusion formula carefully. 10. **Re-examine the Inclusion-Exclusion formula:** The formula is: $$ P(T \cup E \cup C) = P(T) + P(E) + P(C) - P(T \cap E) - P(E \cap C) - P(T \cap C) + P(T \cap E \cap C) $$ Substitute pairwise intersections: $$ 0.895 = 0.45 + 0.35 + 0.31 - (0.15 + x) - (0.05 + x) - (0.09 + x) + x $$ Simplify the sum of pairwise intersections: $$ (0.15 + x) + (0.05 + x) + (0.09 + x) = 0.29 + 3x $$ So: $$ 0.895 = 1.11 - (0.29 + 3x) + x = 1.11 - 0.29 - 3x + x = 0.82 - 2x $$ This matches previous step. 11. **Solve for $x$ again:** $$ 0.895 = 0.82 - 2x $$ $$ 0.895 - 0.82 = -2x $$ $$ 0.075 = -2x $$ $$ x = -\frac{0.075}{2} = -0.0375 $$ Negative probability is impossible, so the problem data might be inconsistent or the percentages overlap differently. 12. **Alternative approach:** Since the sum of individual percentages is $0.45 + 0.35 + 0.31 = 1.11$, which is more than 1, overlaps exist. Sum of exactly two trips (excluding the third) is $0.15 + 0.05 + 0.09 = 0.29$. Sum of no trip is $0.105$. Total percentage accounted for: $$ P(\text{exactly one trip}) + P(\text{exactly two trips}) + P(\text{all three trips}) + P(\text{no trip}) = 1 $$ Let $x = P(T \cap E \cap C)$. Calculate $P(\text{exactly one trip})$: $$ P(T) + P(E) + P(C) - 2 \times P(\text{exactly two trips}) - 3x $$ But better to use: $$ P(\text{exactly one trip}) = P(T) + P(E) + P(C) - 2 \times (0.29 + x) - 3x $$ Simplify: $$ = 1.11 - 2(0.29 + x) - 3x = 1.11 - 0.58 - 2x - 3x = 0.53 - 5x $$ Sum all parts: $$ (0.53 - 5x) + 0.29 + x + 0.105 = 1 $$ $$ 0.53 - 5x + 0.29 + x + 0.105 = 1 $$ $$ 0.925 - 4x = 1 $$ $$ -4x = 1 - 0.925 = 0.075 $$ $$ x = -\frac{0.075}{4} = -0.01875 $$ Still negative. 13. **Conclusion:** The data is inconsistent as given, but assuming the problem expects the use of Inclusion-Exclusion directly, the negative result suggests the number of students who signed up for all three trips is 0. 14. **Calculate number of students:** $$ \text{Number} = x \times 386 = 0 \times 386 = 0 $$ **Final answer:** 0 students signed up for all three trips. --- **Slug:** trip overlap **Subject:** probability **Desmos:** {"latex":"","features":{"intercepts":true,"extrema":true}} **q_count:** 1