91Ó°ÊÓ

We would look for the probability of getting two heads and the probability of getting three heads. Then, we'll add those probabilities together. The possible outcomes for throwing a coin three times are: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. In total, there are 2³ = 8 possible outcomes. i. Probability of getting 2 heads: There are 3 combinations of getting 2 heads: {HHT, HTH, THH}. So, the probability of getting 2 heads is: P(2 heads) = \(\frac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}} = \frac{3}{8}\) ii. Probability of getting 3 heads: There is only 1 combination of getting 3 heads: {HHH}. Hence, the probability of getting 3 heads is: P(3 heads) = \(\frac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}} = \frac{1}{8}\) The probability of getting at least two heads is the sum of the probabilities of getting 2 heads and 3 heads: P(at least 2 heads) = P(2 heads) + P(3 heads) = \(\frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}\)

Since we know that heads were thrown on the first toss, we have narrowed down our sample space to {HHH, HHT, HTH, HTT}. Now, there are only 2² = 4 possible outcomes. We want to find the probability of getting heads on the second toss, given that heads were thrown on the first toss. Among our new sample space, there are 2 favorable outcomes: {HHH, HHT}. Using conditional probability, we have: P(heads on 2nd toss | heads on 1st toss) = \(\frac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}} = \frac{2}{4} = \frac{1}{2}\)

Since we know that tails were thrown on the first toss, we have narrowed down our sample space to {THH, THT, TTH, TTT}. Now, there are only 2² = 4 possible outcomes. We want to find the probability of getting heads on the third toss, given that tails were thrown on the first toss. Among our new sample space, there are 2 favorable outcomes: {THH, THT}. Using conditional probability, we have: P(heads on 3rd toss | tails on 1st toss) = \(\frac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}} = \frac{2}{4} = \frac{1}{2}\) To sum up, the probabilities are: a. At least two heads: \(\frac{1}{2}\) b. Heads on the second toss, given that heads were thrown on the first toss: \(\frac{1}{2}\) c. Heads on the third toss, given that tails were thrown on the first toss: \(\frac{1}{2}\)