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Sample Question 20:
Step 1: Breakdown the problem into different cases based on what the first coin flip is for determining \(x\) and \(y\).
Step 2: Identify that there are four possible cases. For the first case, \(x\) and \(y\) can either be 0 or 1 depending on the second coin flip. For the second and third case, either \(x\) or \(y\) respectively can be chosen from the interval [0, 1] while the other is determined by a coin flip (either 0 or 1). Lastly, for the fourth case, both \(x\) and \(y\) can be chosen from the interval [0, 1].
Step 3: Realize that each case has a \(\frac{1}{4}\) chance of occurring, as it requires two coin flips.
Step 4: For Case 1, we need \(x\) and \(y\) to be different. This is a situation where \(x\) and \(y\) are either 0 or 1, and the probability for this case is \( \frac{1}{2} \).
Step 5: For Case 2 and Case 3, if \(x\) or \(y\) is obtained by a coin flip (0 or 1), the other value (which lies in the range [0, 1]) should be in the interval \((\frac{1}{2}, 1]\) or \([0, \frac{1}{2})\) respectively. Therefore, the probability for success under these two cases is \( \frac{1}{2} \) each.
Step 6: For Case 4, use geometric probability to find out the success rate in this case. Here, both \(x\) and \(y\) are in the range [0, 1] and \(|x-y| > \frac{1}{2}\). Mathematically, this means the shaded area in the plot of |x-y| is \( \frac{1}{4} \). Thus, the probability of success under this case is \( \frac{1}{4} \).
Step 7: Sum up all the success probabilities for each case. The final calculation should be \( \frac{1}{4} \times (\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4}) \), and it yields \( \frac{7}{16} \), which corresponds to \(\textbf{(B)}\).
Real numbers between \(0\) and \(1\), inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is \(0\) if the second flip is heads and \(1\) if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen uniformly at random from the closed interval \([0,1]\). Two random numbers \(x\) and \(y\) are chosen independently in this manner. What is the probability that \(|x-y| > \tfrac{1}{2}\)?
\(\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{7}{16} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \frac{9}{16} \qquad \textbf{(E) } \frac{2}{3}\)
Answer Keys
Question 20: BSolutions
Question 20Step 1: Breakdown the problem into different cases based on what the first coin flip is for determining \(x\) and \(y\).
Step 2: Identify that there are four possible cases. For the first case, \(x\) and \(y\) can either be 0 or 1 depending on the second coin flip. For the second and third case, either \(x\) or \(y\) respectively can be chosen from the interval [0, 1] while the other is determined by a coin flip (either 0 or 1). Lastly, for the fourth case, both \(x\) and \(y\) can be chosen from the interval [0, 1].
Step 3: Realize that each case has a \(\frac{1}{4}\) chance of occurring, as it requires two coin flips.
Step 4: For Case 1, we need \(x\) and \(y\) to be different. This is a situation where \(x\) and \(y\) are either 0 or 1, and the probability for this case is \( \frac{1}{2} \).
Step 5: For Case 2 and Case 3, if \(x\) or \(y\) is obtained by a coin flip (0 or 1), the other value (which lies in the range [0, 1]) should be in the interval \((\frac{1}{2}, 1]\) or \([0, \frac{1}{2})\) respectively. Therefore, the probability for success under these two cases is \( \frac{1}{2} \) each.
Step 6: For Case 4, use geometric probability to find out the success rate in this case. Here, both \(x\) and \(y\) are in the range [0, 1] and \(|x-y| > \frac{1}{2}\). Mathematically, this means the shaded area in the plot of |x-y| is \( \frac{1}{4} \). Thus, the probability of success under this case is \( \frac{1}{4} \).
Step 7: Sum up all the success probabilities for each case. The final calculation should be \( \frac{1}{4} \times (\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4}) \), and it yields \( \frac{7}{16} \), which corresponds to \(\textbf{(B)}\).