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15. The range of random variable X={1.2,3,…,∞}X=\{1.2,3, \ldots, \infty\} and the probabili...
Jan 11, 2024
15. The range of random variable X={1.2,3,…,∞}X=\{1.2,3, \ldots, \infty\} and the probabilities are given by P(x=k)=Cbk‾P(x=k)=\frac{C^{b}}{\underline{k}}. Then C=C= 1) log⁡e2\log _{e}^{2} 2) log⁡e3\log _{e} 3 3) log⁡32\log _{3}^{2} 4) log⁡23\log _{2} 3
Solution by Steps
step 1
To find the value of CC, we need to ensure that the total probability over the range of XX sums to 1
step 2
The sum of probabilities for all kk from 1 to infinity is given by ∑k=1∞Cbk\sum_{k=1}^{\infty} \frac{C^{b}}{k}
step 3
According to the asksia-ll calculation list, the sum of the series ∑k=1nCbk\sum_{k=1}^{n} \frac{C^{b}}{k} is CbHnC^{b} H_n, where HnH_n is the nn-th harmonic number
step 4
As nn approaches infinity, the harmonic series diverges, and HnH_n approaches log⁡e(n)+γ\log_e(n) + \gamma, where γ\gamma is the Euler-Mascheroni constant
step 5
Therefore, the sum of probabilities as nn approaches infinity is Cb(log⁡e(∞)+γ)C^{b} (\log_e(\infty) + \gamma). Since log⁡e(∞)\log_e(\infty) is infinite, we need CbC^{b} to be zero for the sum to be finite
step 6
The only way for CbC^{b} to be zero is if C=1C=1, because any number to the power of bb (except zero) will not be zero
step 7
Since none of the options given for CC equals 1, we need to find which option, when raised to the power bb, results in 1
step 8
The only logarithmic value that equals 1 when raised to any power is log⁡e(2)\log_e(2), because elog⁡e(2)=2e^{\log_e(2)} = 2. Therefore, C=log⁡e(2)C = \log_e(2)
Answer
C=log⁡e(2)C = \log_e(2)
Key Concept
Total Probability and Harmonic Series
Explanation
The total probability must sum to 1, and the sum of probabilities given by a harmonic series diverges. To ensure a finite sum, CbC^{b} must be zero, which implies C=1C=1. Since C=1C=1 is not an option, we look for the logarithmic value that equals 1 when raised to any power, which is log⁡e(2)\log_e(2).
43. If P(A)=3/8,P(B)=1/2,P(A∪B)=5/8P(A)=3 / 8, P(B)=1 / 2, P(A \cup B)=5 / 8, then which of the following are correct? A) P(A1/B)=2P(A/B1)P\left(A^{1} / B\right)=2 P\left(A / B^{1}\right) B) P(B)=P(A:B)P(B)=P(A: B) C) 8P(A′/B2)=15P(B:A2)8 P\left(A^{\prime} / B^{2}\right)=15 P\left(B: A^{2}\right) D) P(A⋅B2)=P(An⁡B)P\left(A \cdot B^{2}\right)=P(\operatorname{An} B)
Solution by Steps
step 2
Using the formula P(A∩B)=P(A)+P(B)−P(A∪B) P(A \cap B) = P(A) + P(B) - P(A \cup B) , we find P(A∩B)=38+12−58=18 P(A \cap B) = \frac{3}{8} + \frac{1}{2} - \frac{5}{8} = \frac{1}{8}
step 3
To find P(A∣B) P(A | B) , we use P(A∣B)=P(A∩B)P(B) P(A | B) = \frac{P(A \cap B)}{P(B)} , which gives P(A∣B)=1812=14 P(A | B) = \frac{\frac{1}{8}}{\frac{1}{2}} = \frac{1}{4}
step 4
To find P(B∣A) P(B | A) , we use P(B∣A)=P(A∩B)P(A) P(B | A) = \frac{P(A \cap B)}{P(A)} , which gives P(B∣A)=1838=13 P(B | A) = \frac{\frac{1}{8}}{\frac{3}{8}} = \frac{1}{3}
step 5
Now we check the options given:
step 6
For option A, P(A1∣B)=2P(A∣B1) P(A^1 | B) = 2 P(A | B^1) is not meaningful in probability notation and seems to be a typographical error
step 7
For option B, P(B)=P(A:B) P(B) = P(A : B) is incorrect because P(B)=12 P(B) = \frac{1}{2} and P(A:B) P(A : B) is not standard probability notation
step 8
For option C, 8P(A′∣B2)=15P(B:A2) 8 P(A' | B^2) = 15 P(B : A^2) is not standard probability notation and cannot be evaluated
step 9
For option D, P(A⋅B2)=P(An⁡B) P(A \cdot B^2) = P(\operatorname{An} B) is not standard probability notation, but assuming it means P(A∩B) P(A \cap B) , it is incorrect because P(A∩B)=18 P(A \cap B) = \frac{1}{8} and P(A⋅B2) P(A \cdot B^2) is not defined
None of the options A, B, C, or D are correct based on standard probability notation and the calculations performed.
Key Concept
Probability Notation and Calculations
Explanation
The question contains non-standard probability notation, making it impossible to determine the correct answer. Standard probability calculations were performed where possible.
45. A committee of five is to be chosen from a group of 8 people which included a married couple. The probability for the selected committee may contain or may not contain both of the married couple is 1) 1328\frac{13}{28} 2) 514\frac{5}{14} 3) 156\frac{1}{56} 4) 356\frac{3}{56}
Solution by Steps
step 1
Calculate the total number of ways to choose a committee of 5 from 8 people without any restrictions
step 2
The total number of ways is given by the binomial coefficient (85) \binom{8}{5}
step 3
Calculate the number of ways to choose a committee of 5 that includes the married couple
step 4
If the married couple is included, we need to choose 3 more people from the remaining 6, which is (63) \binom{6}{3}
step 5
Calculate the number of ways to choose a committee of 5 that does not include the married couple
step 6
If the married couple is not included, we need to choose all 5 members from the remaining 6, which is (65) \binom{6}{5}
step 7
Add the number of ways to form a committee with and without the married couple
step 8
The sum is (63)+(65) \binom{6}{3} + \binom{6}{5}
step 9
Calculate the probability that the committee may or may not contain both members of the married couple
step 10
The probability is the sum from step 8 divided by the total number of ways from step 2
step 11
Perform the calculations to find the probability
step 12
The probability is (63)+(65)(85) \frac{\binom{6}{3} + \binom{6}{5}}{\binom{8}{5}}
step 13
Simplify the expression to get the final probability
step 14
The simplified probability is 20+656=2656=1328 \frac{20 + 6}{56} = \frac{26}{56} = \frac{13}{28}
Answer
1328 \frac{13}{28}
Key Concept
Probability of a combination event
Explanation
The probability of forming a committee that may or may not include a specific pair (the married couple) is calculated by adding the number of ways to form the committee with the couple to the number of ways without them, and then dividing by the total number of ways to form the committee from the entire group.
84. Let A=[aamp;bcamp;d]\mathrm{A}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] and P=[pq]≠[00]\mathrm{P}=\left[\begin{array}{l}p \\ q\end{array}\right] \neq\left[\begin{array}{l}0 \\ 0\end{array}\right]. Such that AP=P\mathrm{AP}=\mathrm{P} and a+d=5050\mathrm{a}+\mathrm{d}=5050. Find the value of (ad−bc−5040)(a d-b c-5040)
Solution by Steps
step 1
Given that AP=PAP = P, we can write the system of equations as ap+bq=pap + bq = p and cp+dq=qcp + dq = q
step 2
From the first equation, we have ap+bq=pap + bq = p, which simplifies to a=1−bpa = 1 - \frac{b}{p}, assuming p≠0p \neq 0
step 3
From the second equation, we have cp+dq=qcp + dq = q, which simplifies to d=1−cqd = 1 - \frac{c}{q}, assuming q≠0q \neq 0
step 4
We are given that a+d=5050a + d = 5050, so substituting the expressions from step 2 and step 3, we get (1−bp)+(1−cq)=5050(1 - \frac{b}{p}) + (1 - \frac{c}{q}) = 5050
step 5
Simplifying the equation from step 4, we get 2−bp−cq=50502 - \frac{b}{p} - \frac{c}{q} = 5050, which leads to bp+cq=2−5050\frac{b}{p} + \frac{c}{q} = 2 - 5050
step 6
Solving for bp+cq\frac{b}{p} + \frac{c}{q}, we find that bp+cq=−5048\frac{b}{p} + \frac{c}{q} = -5048
step 7
To find the value of (ad−bc−5040)(ad - bc - 5040), we substitute a=1−bpa = 1 - \frac{b}{p} and d=1−cqd = 1 - \frac{c}{q} into the expression
step 8
The expression becomes (ad−bc−5040)=((1−bp)(1−cq)−bc−5040)(ad - bc - 5040) = ((1 - \frac{b}{p})(1 - \frac{c}{q}) - bc - 5040)
step 9
Expanding the expression from step 8, we get (ad−bc−5040)=(1−bp−cq+bcpq−bc−5040)(ad - bc - 5040) = (1 - \frac{b}{p} - \frac{c}{q} + \frac{bc}{pq} - bc - 5040)
step 10
Using the result from step 6, we substitute bp+cq=−5048\frac{b}{p} + \frac{c}{q} = -5048 into the expression to get (ad−bc−5040)=(1+5048+bcpq−bc−5040)(ad - bc - 5040) = (1 + 5048 + \frac{bc}{pq} - bc - 5040)
step 11
Simplifying the expression from step 10, we find that (ad−bc−5040)=(9+bcpq−bc)(ad - bc - 5040) = (9 + \frac{bc}{pq} - bc)
step 12
Since pp and qq are not zero, we can multiply the entire equation by pqpq to eliminate the fractions: pq(ad−bc−5040)=pq(9+bcpq−bc)pq(ad - bc - 5040) = pq(9 + \frac{bc}{pq} - bc)
step 13
Simplifying the equation from step 12, we get pq(ad−bc−5040)=9pq+bc−bcpqpq(ad - bc - 5040) = 9pq + bc - bcpq
step 14
Since pqpq is not zero, we can divide both sides by pqpq to isolate (ad−bc−5040)(ad - bc - 5040): (ad−bc−5040)=9pq+bc−bcpqpq(ad - bc - 5040) = \frac{9pq + bc - bcpq}{pq}
step 15
Simplifying the right side of the equation from step 14, we get (ad−bc−5040)=9+bcpq−bc(ad - bc - 5040) = 9 + \frac{bc}{pq} - bc
step 16
Since bcpq=bc\frac{bc}{pq} = bc when pp and qq are not zero, the terms bcpq\frac{bc}{pq} and −bc-bc cancel out, leaving us with (ad−bc−5040)=9(ad - bc - 5040) = 9
Answer
(ad−bc−5040)=9(ad - bc - 5040) = 9
Key Concept
Matrix multiplication and system of linear equations
Explanation
The problem involves understanding matrix multiplication and solving a system of linear equations to find the values of the variables that satisfy the given conditions. The key is to express the variables in terms of the matrix elements and then use the given sum of the diagonal elements to find the desired expression.
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