When unbiased hree coins are tossed, what is the probability of getting (i) 3 heads, \( \frac{\text { (ii) }}{2 \text { heads and a tail, }} \) (iii) at least 2 tails?

When three unbiased (fair) coins are tossed, each coin has two possible outcomes: Heads (H) or Tails (T). The total number of possible outcomes is \(2^3 = 8\). These outcomes are: 1. HHH 2. HHT 3. HTH 4. THH 5. HTT 6. THT 7. TTH 8. TTT Let's calculate the probabilities for each scenario: ### (i) Probability of Getting **3 Heads** **Favorable Outcome:** Only one outcome has all three coins showing heads: **HHH**. **Probability:** \[ P(3 \text{ Heads}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Outcomes}} = \frac{1}{8} \] ### (ii) Probability of Getting **2 Heads and 1 Tail** **Favorable Outcomes:** There are three outcomes where exactly two coins show heads and one shows a tail: - HHT - HTH - THH **Probability:** \[ P(2 \text{ Heads and } 1 \text{ Tail}) = \frac{3}{8} \] ### (iii) Probability of Getting **At Least 2 Tails** "At least 2 tails" includes scenarios with exactly 2 tails or all 3 tails. **Favorable Outcomes:** - Exactly 2 tails: HTT, THT, TTH - Exactly 3 tails: TTT Total favorable outcomes: **4** **Probability:** \[ P(\text{At least } 2 \text{ Tails}) = \frac{4}{8} = \frac{1}{2} \] ### Summary 1. **3 Heads:** \(\frac{1}{8}\) 2. **2 Heads and 1 Tail:** \(\frac{3}{8}\) 3. **At Least 2 Tails:** \(\frac{1}{2}\) **Final Answer:** (i) Probability = 1⁄8; (ii) Probability = 3⁄8; (iii) Probability = ½.