1. **Problem:** Find the probability that the 4 blocks selected are all of the same color when a baby pulls out a block 4 times with replacement from 5 colored blocks. 2. **Concepts:** This is a problem of independent events because each draw is with replacement, so the probability of each color remains constant. 3. **Formula:** For independent events, the probability of all events occurring is the product of their individual probabilities: $$P(A \cap B \cap C \cap D) = P(A) \times P(B) \times P(C) \times P(D)$$ 4. **Step-by-step:** - There are 5 colors, each with probability $\frac{1}{5}$. - Probability that all 4 draws are the same color is the sum over all colors of the probability of drawing that color 4 times: $$5 \times \left(\frac{1}{5}\right)^4 = 5 \times \frac{1}{5^4} = \frac{5}{5^4}$$ 5. **Answer:** $\boxed{\frac{5}{5^4}}$ 6. **Label:** Independent probability. --- 1. **Problem:** Probability of choosing a blue or red shirt from 10 shirts: 3 blue, 4 red, 3 other colors. 2. **Concepts:** Mutually exclusive events (choosing blue or red) since a shirt cannot be both. 3. **Formula:** For mutually exclusive events: $$P(A \cup B) = P(A) + P(B)$$ 4. **Step-by-step:** - $P(blue) = \frac{3}{10}$ - $P(red) = \frac{4}{10}$ - So, $P(blue \text{ or } red) = \frac{3}{10} + \frac{4}{10} = \frac{7}{10}$ 5. **Answer:** $\boxed{\frac{7}{10}}$ 6. **Label:** Mutually exclusive probability. --- 1. **Problem:** Probability of spinning an even number or a number less than 4 on an 8-section spinner numbered 1 to 8. 2. **Concepts:** Use the formula for union of two events with possible overlap. 3. **Formula:** $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ 4. **Step-by-step:** - Even numbers: 2,4,6,8 → 4 outcomes → $P(even) = \frac{4}{8} = \frac{1}{2}$ - Numbers less than 4: 1,2,3 → 3 outcomes → $P(<4) = \frac{3}{8}$ - Intersection (even and <4): 2 → 1 outcome → $P(even \cap <4) = \frac{1}{8}$ - So, $$P = \frac{1}{2} + \frac{3}{8} - \frac{1}{8} = \frac{4}{8} + \frac{3}{8} - \frac{1}{8} = \frac{6}{8} = \frac{3}{4}$$ 5. **Answer:** $\boxed{\frac{3}{4}}$ 6. **Label:** Probability of union of events. --- 1. **Problem:** Given calamansi is chosen, find probability pineapple is also chosen when selecting 3 cans from 4. 2. **Concepts:** Conditional probability. 3. **Formula:** $$P(B|A) = \frac{P(A \cap B)}{P(A)}$$ 4. **Step-by-step:** - Total ways to choose 3 cans from 4: $\binom{4}{3} = 4$ - Ways to choose 3 cans including calamansi: choose 2 from remaining 3 → $\binom{3}{2} = 3$ - Ways to choose 3 cans including both calamansi and pineapple: choose 1 from remaining 2 → $\binom{2}{1} = 2$ - So, $$P(pineapple | calamansi) = \frac{2}{3}$$ 5. **Answer:** $\boxed{\frac{2}{3}}$ 6. **Label:** Conditional probability. --- 1. **Problem:** Probability that a visit to the school clinic is due to both dental and medical reasons given probabilities of dental, medical, and neither. 2. **Concepts:** Use inclusion-exclusion principle. 3. **Formula:** $$P(D \cap M) = P(D) + P(M) - P(D \cup M)$$ 4. **Step-by-step:** - Given $P(neither) = 0.35$, so $P(D \cup M) = 1 - 0.35 = 0.65$ - $P(D) = 0.40$, $P(M) = 0.30$ - So, $$P(D \cap M) = 0.40 + 0.30 - 0.65 = 0.05$$ 5. **Answer:** $\boxed{0.05}$ 6. **Label:** Probability of intersection using inclusion-exclusion. --- 1. **Problem:** Probability that exactly two of the 5 selected dolphins are tagged from 24 dolphins with 6 tagged. 2. **Concepts:** Hypergeometric distribution (dependent probability). 3. **Formula:** $$P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$$ 4. **Step-by-step:** - $N=24$, $K=6$, $n=5$, $k=2$ - So, $$P = \frac{\binom{6}{2} \binom{18}{3}}{\binom{24}{5}}$$ 5. **Answer:** $\boxed{\frac{\binom{6}{2} \binom{18}{3}}{\binom{24}{5}}}$ 6. **Label:** Dependent probability (hypergeometric). --- **Summary:** - Independent events: multiply probabilities. - Mutually exclusive: add probabilities. - Conditional probability: use $P(A|B) = \frac{P(A \cap B)}{P(B)}$. - Inclusion-exclusion: $P(A \cap B) = P(A) + P(B) - P(A \cup B)$. - Hypergeometric: for dependent draws without replacement. These principles will help you solve similar problems on your test.