1. **Problem 1:** Subway station gates problem. (a) Define the sample space: - We have 6 inbound gates and 6 outbound gates. - Each gate is either open or closed, so for each direction, the number of open gates can be from 0 to 6. - The sample space is all pairs \((i,j)\) with \(i = 0,1,...,6\) and \(j=0,1,...,6\), representing \(i\) gates open inbound and \(j\) gates open outbound. - So the sample space size is \(7 \times 7 = 49\). (b) Probability that at most one gate is open in each direction: - At most one means \(i \leq 1\) and \(j \leq 1\). - Possible pairs are \((0,0), (0,1), (1,0), (1,1)\), total 4 outcomes. - Probability \(= \frac{4}{49}\). (c) Probability that at least one gate is open in each direction: - At least one means \(i \geq 1\) and \(j \geq 1\). - Complement is \(i=0\) or \(j=0\). - Number of pairs where \(i=0\): 7 pairs for \(j=0..6\). - Number of pairs where \(j=0\): 7 pairs for \(i=0..6\). - Overlap (where \(i=0, j=0\)) counted twice, subtract once. - Total where \(i=0\) or \(j=0\) is \(7 + 7 - 1 = 13\). - So number of favorable outcomes = \(49 - 13 = 36\). - Probability = \(\frac{36}{49}\). (d) Probability that the number of gates open is the same in both directions: - Favorable pairs \((i,i)\) for \(i=0...6\), total 7 outcomes. - Probability \(= \frac{7}{49} = \frac{1}{7}\). (e) Probability that total number of open gates is six: - We want \(i + j = 6\), with \(0 \leq i,j \leq 6\). - Possible pairs: \((0,6), (1,5), (2,4), (3,3), (4,2), (5,1), (6,0)\), total 7 outcomes. - Probability = \(\frac{7}{49} = \frac{1}{7}\). --- 2. **Problem 2:** Tossing coins and rolling dice. (1.i) Sample space for coin tossed three times with individual outcomes: - Each toss is H or T. - Sample space size = \(2^3 = 8\). - Sample space elements: \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}. (1.ii) Sample space when only number of heads (or tails) is of interest: - Possible number of heads \(= 0,1,2,3\). - Sample space size = 4. (2) Pair of six-sided dice rolled; total outcomes = \(6 \times 6 = 36\). (i) Probability sum of face values is 8: - Outcomes summing to 8: (2,6),(3,5),(4,4),(5,3),(6,2) - Count: - (2,6),(6,2): 2 outcomes - (3,5),(5,3): 2 outcomes - (4,4): 1 outcome - Total favorable outcomes = 5 - Probability = \(\frac{5}{36}\). (ii) Probability sum is 6 or 9: - Sum 6 outcomes: (1,5),(2,4),(3,3),(4,2),(5,1) = 5 outcomes - Sum 9 outcomes: (3,6),(4,5),(5,4),(6,3) = 4 outcomes - Total favorable = 5 + 4 = 9 - Probability = \(\frac{9}{36} = \frac{1}{4}\). (iii) Probability sum is 3, 8, or 12: - Sum 3: (1,2),(2,1) = 2 outcomes - Sum 8: 5 outcomes (from above) - Sum 12: (6,6) = 1 outcome - Total favorable = 2 + 5 + 1 = 8 - Probability = \(\frac{8}{36} = \frac{2}{9}\). (iv) Probability sum is not an even number: - Total outcomes = 36 - Number of outcomes with even sums (2,4,6,8,10,12) = 18 - So, not even = 36 - 18 = 18 - Probability = \(\frac{18}{36} = \frac{1}{2}\).