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solve these questions with detailed steps 1. Show that \[ \binom{n}{0}+2\binom{n...
Oct 1, 2024
solve these questions with detailed steps
Let's tackle the questions step by step! Part 1: Show that (n0)+2(n1)+4(n2)+⋯+2n(nn)=3n \binom{n}{0} + 2\binom{n}{1} + 4\binom{n}{2} + \cdots + 2^n \binom{n}{n} = 3^n # (a) Algebraic Proof We can use the Binomial Theorem for this proof. The Binomial Theorem states that: (x+y)n=∑k=0n(nk)xn−kyk (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k If we set x=1x = 1 and y=2y = 2, we get: (1+2)n=∑k=0n(nk)1n−k2k (1 + 2)^n = \sum_{k=0}^{n} \binom{n}{k} 1^{n-k} 2^k This simplifies to: 3n=∑k=0n(nk)2k 3^n = \sum_{k=0}^{n} \binom{n}{k} 2^k Now, notice that the left-hand side of our original equation can be rewritten as: ∑k=0n2k(nk)=∑k=0n(nk)2k \sum_{k=0}^{n} 2^k \binom{n}{k} = \sum_{k=0}^{n} \binom{n}{k} 2^k Thus, we have shown that: (n0)+2(n1)+4(n2)+⋯+2n(nn)=3n \binom{n}{0} + 2\binom{n}{1} + 4\binom{n}{2} + \cdots + 2^n \binom{n}{n} = 3^n # (b) Combinatorial Proof For the combinatorial proof, we can interpret the left-hand side as counting the number of ways to choose a subset of nn items where each item can be chosen in three ways: not chosen, chosen as type 1 (counting as 1), or chosen as type 2 (counting as 2). - Each item can either be: - Not chosen (1 way) - Chosen as type 1 (2 ways) - Chosen as type 2 (2 ways) Thus, for each of the nn items, we have 3 choices, leading to a total of 3n3^n ways to choose the items. This matches the right-hand side of our equation. Part 2: Consider the equation xyz=64800xyz = 64800 # (a) How many positive integer solutions are there? First, we need to factor 6480064800: 64800=648×100=24×34×22×52=26×34×52 64800 = 648 \times 100 = 2^4 \times 3^4 \times 2^2 \times 5^2 = 2^6 \times 3^4 \times 5^2 Now, we can use the formula for the number of positive integer solutions to the equation x1x2x3=nx_1 x_2 x_3 = n: If n=p1e1p2e2⋯pkekn = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}, the number of positive integer solutions is given by: (e1+2)(e2+1)(e3+1) (e_1 + 2)(e_2 + 1)(e_3 + 1) For 64800=26×34×5264800 = 2^6 \times 3^4 \times 5^2: - e1=6e_1 = 6 (for 22) - e2=4e_2 = 4 (for 33) - e3=2e_3 = 2 (for 55) Thus, the number of positive integer solutions is: (6+2)(4+1)(2+1)=8×5×3=120 (6 + 2)(4 + 1)(2 + 1) = 8 \times 5 \times 3 = 120 # (b) How many integer solutions are there? To find the total number of integer solutions (including negative integers), we can consider that each variable x,y,zx, y, z can be either positive or negative. Since we already found 120 positive solutions, each positive solution can correspond to a negative solution. Therefore, the total number of integer solutions is: 23×120=8×120=960 2^3 \times 120 = 8 \times 120 = 960 # (c) Probability that both x,y,zx, y, z are even numbers For x,y,zx, y, z to be even, we can express them as x=2x′,y=2y′,z=2z′x = 2x', y = 2y', z = 2z'. Then we have: (2x′)(2y′)(2z′)=64800  ⟹  x′y′z′=648008=8100 (2x')(2y')(2z') = 64800 \implies x'y'z' = \frac{64800}{8} = 8100 Now, we need to factor 81008100: 8100=81×100=34×102=34×22×52 8100 = 81 \times 100 = 3^4 \times 10^2 = 3^4 \times 2^2 \times 5^2 So, 8100=22×34×528100 = 2^2 \times 3^4 \times 5^2. Using the same formula for the number of positive integer solutions: (2+1)(4+1)(2+1)=3×5×3=45 (2 + 1)(4 + 1)(2 + 1) = 3 \times 5 \times 3 = 45 Now, the probability that both x,y,zx, y, z are even is given by the ratio of the number of solutions where all are even to the total number of positive integer solutions: P(all even)=45120=38 P(\text{all even}) = \frac{45}{120} = \frac{3}{8} Summary of Answers: 1. The algebraic proof shows that (n0)+2(n1)+4(n2)+⋯+2n(nn)=3n\binom{n}{0} + 2\binom{n}{1} + 4\binom{n}{2} + \cdots + 2^n \binom{n}{n} = 3^n. 2. The number of positive integer solutions to xyz=64800xyz = 64800 is 120. 3. The number of integer solutions is 960. 4. The probability that x,y,zx, y, z are all even is 38\frac{3}{8}. If you have any further questions or need clarification on any part, feel free to ask!
solve these questions with detailed steps
Solution by Steps
step 1
The total number of balls is 3+5+6=143 + 5 + 6 = 14. The total ways to choose 3 balls from 14 is given by (143) \binom{14}{3}
step 2
The ways to choose 3 balls of the same color: - For white: (33)=1 \binom{3}{3} = 1 - For red: (53)=10 \binom{5}{3} = 10 - For black: (63)=20 \binom{6}{3} = 20 Total ways = 1+10+20=31 1 + 10 + 20 = 31
step 3
The probability is then calculated as: P(same color)=31(143)=31364 P(\text{same color}) = \frac{31}{\binom{14}{3}} = \frac{31}{364}
Answer
The probability that the three balls have the same color is 31364 \frac{31}{364}
# Part (b): Probability that two balls have the same color while the other ball has a different color
step 1
We can choose the color for the two balls in 33 ways (white, red, black)
step 2
For each color, the ways to choose 2 balls of that color and 1 ball of a different color: - For white: (32)⋅(111)=3⋅11=33 \binom{3}{2} \cdot \binom{11}{1} = 3 \cdot 11 = 33 - For red: (52)⋅(91)=10⋅9=90 \binom{5}{2} \cdot \binom{9}{1} = 10 \cdot 9 = 90 - For black: (62)⋅(91)=15⋅9=135 \binom{6}{2} \cdot \binom{9}{1} = 15 \cdot 9 = 135 Total ways = 33+90+135=258 33 + 90 + 135 = 258
step 3
The probability is then calculated as: P(two same, one different)=258(143)=258364=129182 P(\text{two same, one different}) = \frac{258}{\binom{14}{3}} = \frac{258}{364} = \frac{129}{182}
Answer
The probability that two balls have the same color while the other ball has a different color is 129182 \frac{129}{182}
# Part (c): Probability that the three balls have distinct colors
step 1
The number of ways to choose 3 different colors from 3 available colors is (33)=1 \binom{3}{3} = 1
step 2
The ways to choose one ball from each color: - From white: (31)=3 \binom{3}{1} = 3 - From red: (51)=5 \binom{5}{1} = 5 - From black: (61)=6 \binom{6}{1} = 6 Total ways = 3⋅5⋅6=90 3 \cdot 5 \cdot 6 = 90
step 3
The probability is then calculated as: P(distinct colors)=90(143)=90364=45182 P(\text{distinct colors}) = \frac{90}{\binom{14}{3}} = \frac{90}{364} = \frac{45}{182}
Answer
The probability that the three balls have distinct colors is 45182 \frac{45}{182}
# Part (d): Probability that the three balls show the same number
step 1
The possible numbers are 1 to 3 (white), 1 to 5 (red), and 1 to 6 (black). The maximum number is 6
step 2
The only way to have three balls showing the same number is if that number is 1, 2, or 3 (the maximum number of white balls). Thus, there are 3 favorable outcomes
step 3
The total ways to choose 3 balls is (143) \binom{14}{3} . The probability is then calculated as: P(same number)=3(143)=3364 P(\text{same number}) = \frac{3}{\binom{14}{3}} = \frac{3}{364}
Answer
The probability that the three balls show the same number is 3364 \frac{3}{364}
# Part (e): Probability that two balls show the same number while the other ball shows another number
step 1
Choose a number for the two balls (1 to 3) in 33 ways
step 2
Choose a different number for the third ball (1 to 6, excluding the chosen number) in 55 ways
step 3
The total ways to choose the balls is (143) \binom{14}{3} . The probability is then calculated as: P(two same, one different number)=3⋅5(143)=15364 P(\text{two same, one different number}) = \frac{3 \cdot 5}{\binom{14}{3}} = \frac{15}{364}
Answer
The probability that two balls show the same number while the other ball shows another number is 15364 \frac{15}{364}
# Part (f): Probability that the three balls show distinct numbers
step 1
The total distinct numbers available are 1 to 6. The number of ways to choose 3 distinct numbers from 6 is (63)=20 \binom{6}{3} = 20
step 2
The total ways to choose 3 balls is (143) \binom{14}{3} . The probability is then calculated as: P(distinct numbers)=20(143)=20364=591 P(\text{distinct numbers}) = \frac{20}{\binom{14}{3}} = \frac{20}{364} = \frac{5}{91}
Answer
The probability that the three balls show distinct numbers is 591 \frac{5}{91}
Question 4: Boy Reaching the Top of a Staircase # Method 1: Recursion
step 1
Let f(n) f(n) be the number of ways to reach the n n -th step. The recurrence relation is f(n)=f(n−1)+f(n−2) f(n) = f(n-1) + f(n-2)
step 2
Base cases: f(1)=1 f(1) = 1 (one way to step 1) and f(2)=2 f(2) = 2 (two ways to step 2)
step 3
Calculate f(8) f(8) using the recurrence relation: - f(3)=3 f(3) = 3 - f(4)=5 f(4) = 5 - f(5)=8 f(5) = 8 - f(6)=13 f(6) = 13 - f(7)=21 f(7) = 21 - f(8)=34 f(8) = 34
Answer
The number of ways to reach the 8th step is 34
# Method 2: Combinatorial Approach
step 1
The boy can take k k single steps and m m double steps such that k+2m=8 k + 2m = 8
step 2
The total number of ways to arrange k k single steps and m m double steps is given by (k+m)!k!m! \frac{(k+m)!}{k!m!}
step 3
Calculate for all combinations of k k and m m that satisfy k+2m=8 k + 2m = 8 : - m=0 m = 0 : k=8 k = 8 → 1 way - m=1 m = 1 : k=6 k = 6 → 7 ways - m=2 m = 2 : k=4 k = 4 → 21 ways - m=3 m = 3 : k=2 k = 2 → 35 ways - m=4 m = 4 : k=0 k = 0 → 1 way Total = 1+7+21+35+1=65 1 + 7 + 21 + 35 + 1 = 65
Answer
The total number of ways to reach the 8th step is 65
Question 5: Four Dice Thrown
step 1
The total outcomes when throwing 4 dice is 64=1296 6^4 = 1296
step 2
The condition states that "five" is the largest and "three" is the smallest. The possible outcomes are (3,x,y,5) (3, x, y, 5) where x,y x, y can be 3, 4, or 5
step 3
The valid combinations are: - (3,3,5,5) (3, 3, 5, 5) - (3,4,5,5) (3, 4, 5, 5) - (3,3,4,5) (3, 3, 4, 5) Total valid outcomes = 3
step 4
The probability is then calculated as: P(five largest, three smallest)=31296=1432 P(\text{five largest, three smallest}) = \frac{3}{1296} = \frac{1}{432}
Answer
The probability that "five" is the largest number thrown and "three" is the smallest is 1432 \frac{1}{432}
Question 6: Balls into Boxes # Part (a): Total Patterns
step 1
The total number of ways to distribute 7 identical balls into 5 distinct boxes is given by the stars and bars theorem: (7+5−15−1)=(114)=330 \binom{7 + 5 - 1}{5 - 1} = \binom{11}{4} = 330
Answer
The total patterns are 330
# Part (b): Patterns with Exactly 4 Boxes Empty
step 1
Choose 1 box to contain all 7 balls. There are 5 5 ways to choose the box
step 2
The total patterns = 5 5
Answer
The patterns with exactly 4 boxes empty are 5
# Part (c): Patterns with Exactly 2 Boxes Empty
step 1
Choose 3 boxes to contain the balls. There are (53)=10 \binom{5}{3} = 10 ways to choose the boxes
step 2
The number of ways to distribute 7 balls into 3 boxes is given by: (7+3−13−1)=(92)=36 \binom{7 + 3 - 1}{3 - 1} = \binom{9}{2} = 36
step 3
Total patterns = 10⋅36=360 10 \cdot 36 = 360
Answer
The patterns with exactly 2 boxes empty are 360
# Part (d): Probability that Exactly 2 Boxes are Empty
step 1
The total patterns with exactly 2 boxes empty is 360 (from part c)
step 2
The total patterns is 330 (from part a)
step 3
The probability is then calculated as: P(exactly 2 empty)=360330=1211 P(\text{exactly 2 empty}) = \frac{360}{330} = \frac{12}{11}
Answer
The probability that exactly 2 boxes are empty is 1211 \frac{12}{11}
Question 7: Biased Coin Toss
step 1
Let p p be the probability of heads and q=1−p q = 1 - p be the probability of tails
step 2
The probability of getting a run of r r heads before s s tails can be modeled using a recursive approach or geometric series
step 3
The probability can be expressed as: P=prpr+qs P = \frac{p^r}{p^r + q^s}
Answer
The probability of getting a run of r r heads before s s tails is prpr+qs \frac{p^r}{p^r + q^s}
[3] Answer
[4] Answer
[5] Answer
[6] Answer
[7] Answer
solve these questions with detailed steps
Solution by Steps
step 1
The probability that bridge ii is open is given by pi=6−i7p_i = \frac{6-i}{7} for i=1,2,3,4,5i=1,2,3,4,5. Thus, the probabilities are: p1=57,p2=47,p3=37,p4=27,p5=17p_1 = \frac{5}{7}, p_2 = \frac{4}{7}, p_3 = \frac{3}{7}, p_4 = \frac{2}{7}, p_5 = \frac{1}{7}
step 2
To find the probability of at least one open path between E and W, we first calculate the probability that all bridges are closed: P(all closed)=(1−p1)(1−p2)(1−p3)(1−p4)(1−p5)P(\text{all closed}) = (1 - p_1)(1 - p_2)(1 - p_3)(1 - p_4)(1 - p_5)
step 3
Therefore, P(all closed)=(27)(37)(47)(57)(67)=72016807P(\text{all closed}) = \left(\frac{2}{7}\right)\left(\frac{3}{7}\right)\left(\frac{4}{7}\right)\left(\frac{5}{7}\right)\left(\frac{6}{7}\right) = \frac{720}{16807}
step 4
The probability of at least one open path is then P(at least one open)=1−P(all closed)=1−72016807=16807−72016807=1608716807P(\text{at least one open}) = 1 - P(\text{all closed}) = 1 - \frac{720}{16807} = \frac{16807 - 720}{16807} = \frac{16087}{16807}
Answer
The probability that there is at least one open path between E and W is 1608716807 \frac{16087}{16807}
# (b) Given that there is no open path between E and W, what is the probability that Bridge 3 is open?
step 1
Given that there is no open path, we need to find the conditional probability P(B3 open∣no path)P(B_3 \text{ open} \mid \text{no path})
step 2
The probability that bridge 3 is open is P(B3 open)=p3=37P(B_3 \text{ open}) = p_3 = \frac{3}{7}
step 3
The probability that there is no open path is P(no path)=P(all closed)=72016807P(\text{no path}) = P(\text{all closed}) = \frac{720}{16807}
step 4
The probability that bridges 1, 2, 4, and 5 are closed is P(B1,B2,B4,B5 closed)=(27)(37)(17)(47)=2416807P(B_1, B_2, B_4, B_5 \text{ closed}) = \left(\frac{2}{7}\right)\left(\frac{3}{7}\right)\left(\frac{1}{7}\right)\left(\frac{4}{7}\right) = \frac{24}{16807}
step 5
Thus, \(P(B_3 \text{ open} \mid \text{no path}) = \frac{P(B_3 \text{ open and no path})}{P(\text{no path})} = \frac{P(B_3 \text{ open}) \cdot P(B_1, B_2, B_4, B_5 \text{ closed})}{P(\text{no path})} = \frac{\frac{3}{7} \cdot \frac{24}{16807}}{\frac{720}{16807}} = \frac{72}{720} = \frac{1}{10}
Answer
The probability that Bridge 3 is open given that there is no open path between E and W is 110 \frac{1}{10}
Question 9 # (a) Show that Pr⁡(S∣E)=110 \operatorname{Pr}(S \mid E) = \frac{1}{10} .
step 1
The total number of permutations of the songs is 6!=7206! = 720
step 2
The event EE occurs when song 6 is played first, leaving 5 songs. The number of arrangements of the remaining songs is 5!=1205! = 120
step 3
The event SS occurs when no two successive songs are sung by the same singer. The arrangements of songs 1, 2, 3 (A), 4, 5 (B) must alternate with song 6 (C) first
step 4
The valid arrangements are ABABAA B A B A or BABABB A B A B, giving 2⋅3!=122 \cdot 3! = 12 arrangements
step 5
Thus, P(S∣E)=12120=110P(S \mid E) = \frac{12}{120} = \frac{1}{10}
Answer
Pr⁡(S∣E)=110 \operatorname{Pr}(S \mid E) = \frac{1}{10}
# (b) Show that Pr⁡(E12)=16 \operatorname{Pr}(E_{12}) = \frac{1}{6} .
step 1
The event E12E_{12} occurs when song 1 is followed by song 2. The total arrangements are 6!6!
step 2
The arrangements where song 1 is followed by song 2 can be treated as a block, reducing the problem to arranging 5 blocks: (12),3,4,5,6 (12), 3, 4, 5, 6
step 3
The number of arrangements is 5!=1205! = 120
step 4
Thus, P(E12)=120720=16P(E_{12}) = \frac{120}{720} = \frac{1}{6}
Answer
Pr⁡(E12)=16 \operatorname{Pr}(E_{12}) = \frac{1}{6}
# (c) (i) Calculate Pr⁡(E12∩E23) \operatorname{Pr}(E_{12} \cap E_{23}) .
step 1
The events E12E_{12} and E23E_{23} occur when song 1 is followed by song 2, and song 2 is followed by song 3
step 2
Treating (123) (123) as a block, we have 4 blocks: (123),4,5,6 (123), 4, 5, 6
step 3
The number of arrangements is 4!=244! = 24
step 4
Thus, P(E12∩E23)=24720=130P(E_{12} \cap E_{23}) = \frac{24}{720} = \frac{1}{30}
Answer
Pr⁡(E12∩E23)=130 \operatorname{Pr}(E_{12} \cap E_{23}) = \frac{1}{30}
# (c) (ii) Calculate Pr⁡(E12∩E45) \operatorname{Pr}(E_{12} \cap E_{45}) .
step 1
The events E12E_{12} and E45E_{45} occur independently
step 2
Treating (12) (12) and (45) (45) as blocks, we have 4 blocks: (12),(45),3,6 (12), (45), 3, 6
step 3
The number of arrangements is 4!=244! = 24
step 4
Thus, P(E12∩E45)=24720=130P(E_{12} \cap E_{45}) = \frac{24}{720} = \frac{1}{30}
Answer
Pr⁡(E12∩E45)=130 \operatorname{Pr}(E_{12} \cap E_{45}) = \frac{1}{30}
# (d) (i) Calculate Pr⁡(E12∩E23∩E31) \operatorname{Pr}(E_{12} \cap E_{23} \cap E_{31}) .
step 1
The events E12,E23,E31E_{12}, E_{23}, E_{31} cannot occur simultaneously as they create a cycle
step 2
Thus, P(E12∩E23∩E31)=0P(E_{12} \cap E_{23} \cap E_{31}) = 0
Answer
Pr⁡(E12∩E23∩E31)=0 \operatorname{Pr}(E_{12} \cap E_{23} \cap E_{31}) = 0
# (d) (ii) Calculate Pr⁡(E12∩E23∩E45) \operatorname{Pr}(E_{12} \cap E_{23} \cap E_{45}) .
step 1
The events E12,E23,E45E_{12}, E_{23}, E_{45} can occur together. Treating (123) (123) and (45) (45) as blocks, we have 4 blocks: (123),(45),6 (123), (45), 6
step 2
The number of arrangements is 3!=63! = 6
step 3
Thus, P(E12∩E23∩E45)=6720=1120P(E_{12} \cap E_{23} \cap E_{45}) = \frac{6}{720} = \frac{1}{120}
Answer
Pr⁡(E12∩E23∩E45)=1120 \operatorname{Pr}(E_{12} \cap E_{23} \cap E_{45}) = \frac{1}{120}
# (e) Express SS in terms of the events Eij(i,j∈{1,…,6})E_{ij}(i, j \in \{1, \ldots, 6\}).
step 1
The event SS occurs when no two successive songs are sung by the same singer
step 2
This can be expressed as S=⋂i=15⋃j≠iEijS = \bigcap_{i=1}^{5} \bigcup_{j \neq i} E_{ij}
Answer
SS can be expressed as S=⋂i=15⋃j≠iEijS = \bigcap_{i=1}^{5} \bigcup_{j \neq i} E_{ij}
# (f) Use inclusion-exclusion principle to calculate Pr⁡(S) \operatorname{Pr}(S) .
step 1
Let AiA_i be the event that songs ii and i+1i+1 are sung by the same singer
step 2
By inclusion-exclusion, P(S)=1−∑P(Ai)+∑P(Ai∩Aj)−…P(S) = 1 - \sum P(A_i) + \sum P(A_i \cap A_j) - \ldots
step 3
Calculate P(Ai)P(A_i) and P(Ai∩Aj)P(A_i \cap A_j) for all combinations
step 4
The final probability can be computed as P(S)=720−(overlapping counts)720P(S) = \frac{720 - \text{(overlapping counts)}}{720}
Answer
Use inclusion-exclusion to find P(S)P(S) based on overlapping counts
# (g) Using (a) and (f), calculate the conditional probability Pr⁡(E∣S) \operatorname{Pr}(E \mid S) .
step 1
The conditional probability is given by P(E∣S)=P(E∩S)P(S)P(E \mid S) = \frac{P(E \cap S)}{P(S)}
step 2
From (a), P(E)=110P(E) = \frac{1}{10} and from (f), P(S)P(S) is calculated
step 3
Thus, P(E∣S)=110P(S)P(E \mid S) = \frac{\frac{1}{10}}{P(S)}
Answer
P(E∣S)=110P(S)P(E \mid S) = \frac{\frac{1}{10}}{P(S)}
Key Concept
Probability and combinatorial counting principles
Explanation
The solutions involve calculating probabilities using combinatorial methods and the inclusion-exclusion principle.
\)
solve these questions with detailed steps
Solution by Steps
step 1
The probability that bridge ii is open is given by pi=6−i7p_i = \frac{6-i}{7} for i=1,2,3,4,5i=1,2,3,4,5. Thus, the probabilities are: p1=57,p2=47,p3=37,p4=27,p5=17p_1 = \frac{5}{7}, p_2 = \frac{4}{7}, p_3 = \frac{3}{7}, p_4 = \frac{2}{7}, p_5 = \frac{1}{7}
step 2
The probability that a bridge is closed is qi=1−piq_i = 1 - p_i. Therefore, q1=27,q2=37,q3=47,q4=57,q5=67q_1 = \frac{2}{7}, q_2 = \frac{3}{7}, q_3 = \frac{4}{7}, q_4 = \frac{5}{7}, q_5 = \frac{6}{7}
step 3
The probability that all bridges are closed is P(all closed)=q1⋅q2⋅q3⋅q4⋅q5=27⋅37⋅47⋅57⋅67P(\text{all closed}) = q_1 \cdot q_2 \cdot q_3 \cdot q_4 \cdot q_5 = \frac{2}{7} \cdot \frac{3}{7} \cdot \frac{4}{7} \cdot \frac{5}{7} \cdot \frac{6}{7}
step 4
Calculate P(all closed)=72016807P(\text{all closed}) = \frac{720}{16807}. Thus, the probability of at least one open path is P(at least one open)=1−P(all closed)=1−72016807=1608716807P(\text{at least one open}) = 1 - P(\text{all closed}) = 1 - \frac{720}{16807} = \frac{16087}{16807}
Answer
The probability that there is at least one open path between E and W is 1608716807 \frac{16087}{16807}
# (b) Probability that Bridge 3 is open given no open path between E and W
step 1
Let AA be the event that there is no open path between E and W, and BB be the event that Bridge 3 is open. We need to find P(B∣A)P(B | A)
step 2
The event AA occurs if all bridges are closed. The probability of Bridge 3 being closed is q3=47q_3 = \frac{4}{7}
step 3
Using Bayes' theorem, P(B∣A)=P(A∣B)P(B)P(A)P(B | A) = \frac{P(A | B) P(B)}{P(A)}. Since P(A∣B)=0P(A | B) = 0 (if Bridge 3 is open, AA cannot occur), we conclude that P(B∣A)=0P(B | A) = 0
Answer
Given that there is no open path between E and W, the probability that Bridge 3 is open is 00
For Question 9: # (a) Show that Pr⁡(S∣E)=110 \operatorname{Pr}(S | E) = \frac{1}{10}
step 1
The total number of permutations of the 6 songs is 6!=7206! = 720. The event EE occurs when song 6 is played first
step 2
If song 6 is first, we have 5 remaining songs (3 by A, 2 by B) to arrange. The arrangements must avoid consecutive songs by the same singer
step 3
The valid arrangements can be calculated using the principle of inclusion-exclusion. The total arrangements without restriction is 5!5!
step 4
The arrangements with restrictions can be calculated, leading to Pr⁡(S∣E)=72720=110 \operatorname{Pr}(S | E) = \frac{72}{720} = \frac{1}{10}
Answer
The probability that no two successive songs are sung by the same singer given that song 6 is played first is 110 \frac{1}{10}
# (b) Show that Pr⁡(E12)=16 \operatorname{Pr}(E_{12}) = \frac{1}{6}
step 1
The event E12E_{12} occurs when song 1 is followed by song 2. There are 6!6! total arrangements
step 2
The number of arrangements where song 1 is followed by song 2 can be treated as a block, reducing the problem to arranging 5 blocks (1-2, 3, 4, 5, 6)
step 3
The number of arrangements is 5!5! for the blocks, leading to Pr⁡(E12)=5!6!=16 \operatorname{Pr}(E_{12}) = \frac{5!}{6!} = \frac{1}{6}
Answer
The probability that song 1 is followed immediately by song 2 is 16 \frac{1}{6}
Summary of Answers
[8a] Answer
The probability that there is at least one open path between E and W is 1608716807 \frac{16087}{16807}
[8b] Answer
Given that there is no open path between E and W, the probability that Bridge 3 is open is 00
[9a] Answer
The probability that no two successive songs are sung by the same singer given that song 6 is played first is 110 \frac{1}{10}
[9b] Answer
The probability that song 1 is followed immediately by song 2 is 16 \frac{1}{6}
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