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Question 2 MCQ · 0.0 marks
6406536344187.
Useful Data has been mentioned above.
6406536344188.
This data attachment is just for a reference & not for an evaluation.
Published solution
**1. Identify the entry:** This is a zero-mark acknowledgement attached to the exam’s useful-data sheet, which lists probability distributions, moments and inequalities. It is not a numerical statistics problem.
**2. Select the intended acknowledgement:** Choose the statement confirming that useful data has been provided above. In later questions, identify the appropriate formula and substitute the problem’s own parameters.
**Answer: A — Useful Data has been mentioned above..**
**Convention note:** This is an administrative response choice, not a claim that the second sentence is mathematically false.
A: **Correct:** This is the acknowledgement marked as the accepted response in the paper.
B: **Incorrect:** This describes the reference role of the sheet, but it is not the acknowledgement selected by the paper’s scoring key.
who make a reservation in a restaurant on a particular day is a random
variable with mean equal to 10 and variance equal to 5. Find a lower bound on the probability that
on a particular day, number of reservations made will lie in between 5 and 15 using Chebyshev's
inequality. Enter the answer correct to two decimal places.
Published solution
**1. Express the interval as a deviation:** The mean is \(\mu=10\), variance is \(\sigma^2=5\), and the endpoints are five units from the mean. Thus \(5<X<15\) is \(|X-10|<5\).
**2. Apply Chebyshev’s inequality:** \[P(|X-\mu|\ge a)\le\frac{\sigma^2}{a^2}=\frac5{25}=0.20.\] Taking complements gives \(P(|X-10|<5)\ge1-0.20=0.80\).
**3. Interpret the result:** The requested lower bound, to two decimal places, is 0.80. It is a guaranteed bound rather than the exact probability. Including the endpoints can only increase the probability, so the same bound remains valid for the closed interval.
**Answer: 0.80.**
**1. Expand covariance bilinearly:** With \(U=3X+2Y\) and \(V=Y-Z\), \[\operatorname{Cov}(U,V)=3\operatorname{Cov}(X,Y)-3\operatorname{Cov}(X,Z)+2\operatorname{Var}(Y)-2\operatorname{Cov}(Y,Z).\]
**2. Substitute the supplied moments:** \(3(6)-3(2)+2(5)-2(4)=18-6+10-8=14\). No independence assumption is needed; all cross-covariances are already specified.
**Answer: 14.**
(in minutes) that a customer waits to be served is a continuous random
variable with probability density function
Find the value of
.
Published solution
**1. Normalize the density:** The function is \(f_X(x)=k(10-x)\) on \(0<x<10\), and zero elsewhere. Its total integral must equal one.
**2. Evaluate the area:** \[1=k\int_0^{10}(10-x)\,dx=k\left[10x-\frac{x^2}{2}\right]_0^{10}=50k.\] Therefore \(k=1/50=0.02\).
**3. Check nonnegativity:** On the support, \(10-x>0\), and the computed \(k\) is positive, so the normalized function is a valid density.
**Answer: 0.02.**
A delivery company operates two independent service counters A and B. Let
and
denote the
number of parcels processed by counters A and B in one hour, respectively. The joint probability
mass function of
and
is given as:
To assess workload, the management defines the hourly load index as the maximum number of
parcels processed by counters A and B in a given hour. Find the probability that the hourly load
index is equal to the number of parcels processed by counter A.
Enter the correct option number for
.
Published solution
**1. Translate the load index:** If \(L=\max(X,Y)\), the phrase “load index equals counter A’s count” is the event \(L=X\), or \(\max(X,Y)=X\). Ties are included.
**2. Match blank A:** Option (3) states precisely \(\max(X,Y)=X\). Options (4) and (5) describe different equalities, and (1)/(2) are inequalities used to characterize the event in the next blank. Enter the option number 3.
**Answer: 3.**
**Convention note:** The passage calls the counters independent, but its joint table does not satisfy independence: \(P(X=0,Y=0)=0.10\ne P(X=0)P(Y=0)=0.20(0.30)=0.06\). Use the supplied joint PMF directly; the event calculations do not require independence.
A delivery company operates two independent service counters A and B. Let
and
denote the
number of parcels processed by counters A and B in one hour, respectively. The joint probability
mass function of
and
is given as:
To assess workload, the management defines the hourly load index as the maximum number of
parcels processed by counters A and B in a given hour. Find the probability that the hourly load
index is equal to the number of parcels processed by counter A.
Enter the correct option number for
.
Published solution
**1. Characterize the maximum event:** \(\max(X,Y)=X\) occurs exactly when \(X\ge Y\). Equality is allowed because both counts then attain the maximum.
**2. Match blank B:** Option (2) is \(X\ge Y\). Option (1), \(X\le Y\), instead characterizes the event that the maximum equals \(Y\). Enter 2.
**Answer: 2.**
**Convention note:** The passage calls the counters independent, but its joint table does not satisfy independence: \(P(X=0,Y=0)=0.10\ne P(X=0)P(Y=0)=0.20(0.30)=0.06\). Use the supplied joint PMF directly; the event calculations do not require independence.
A delivery company operates two independent service counters A and B. Let
and
denote the
number of parcels processed by counters A and B in one hour, respectively. The joint probability
mass function of
and
is given as:
To assess workload, the management defines the hourly load index as the maximum number of
parcels processed by counters A and B in a given hour. Find the probability that the hourly load
index is equal to the number of parcels processed by counter A.
Enter the correct option number for
.
Published solution
**1. Partition according to counter B:** The required event is \(X\ge Y\). Since \(Y\in\{0,1,2\}\), split it into the disjoint cases \(\{X\ge k,Y=k\}\), for \(k=0,1,2\).
**2. Sum the disjoint probabilities:** \[P(X\ge Y)=\sum_{k=0}^{2}P(X\ge k,Y=k).\] This is option (7), which fills blank C. Option (8) reverses the inequality; option (6) uses \(Y>k\) instead.
**Answer: 7.**
**Convention note:** The passage calls the counters independent, but its joint table does not satisfy independence: \(P(X=0,Y=0)=0.10\ne P(X=0)P(Y=0)=0.20(0.30)=0.06\). Use the supplied joint PMF directly; the event calculations do not require independence.
A delivery company operates two independent service counters A and B. Let
and
denote the
number of parcels processed by counters A and B in one hour, respectively. The joint probability
mass function of
and
is given as:
To assess workload, the management defines the hourly load index as the maximum number of
parcels processed by counters A and B in a given hour. Find the probability that the hourly load
index is equal to the number of parcels processed by counter A.
Enter the correct option number for
.
Published solution
**1. Enumerate the favorable cells:** The condition \(X\ge Y\) allows \((X,Y)=(0,0),(1,0),(2,0),(1,1),(2,1),(2,2)\). The first three cells are already written in the supplied expansion.
**2. Supply the remaining terms:** Blank D must be \(P(X=1,Y=1)+P(X=2,Y=1)+P(X=2,Y=2)\), which is option (10). Option (9) incorrectly adds \((1,2)\), while option (11) omits \((1,1)\).
**Answer: 10.**
**Convention note:** The passage calls the counters independent, but its joint table does not satisfy independence: \(P(X=0,Y=0)=0.10\ne P(X=0)P(Y=0)=0.20(0.30)=0.06\). Use the supplied joint PMF directly; the event calculations do not require independence.
A delivery company operates two independent service counters A and B. Let
and
denote the
number of parcels processed by counters A and B in one hour, respectively. The joint probability
mass function of
and
is given as:
To assess workload, the management defines the hourly load index as the maximum number of
parcels processed by counters A and B in a given hour. Find the probability that the hourly load
index is equal to the number of parcels processed by counter A.
Calculate the value for
. Enter the answer correct to one decimal place.
Published solution
**1. Identify the event:** The maximum equals counter A’s count exactly when \(X\ge Y\). Include the diagonal cells because ties satisfy this condition.
**2. Sum the supplied joint probabilities:** For \(Y=0\), the favorable sum is \(0.10+0.15+0.05=0.30\). For \(Y=1\), it is \(0.20+0.20=0.40\). For \(Y=2\), only \(X=2\) qualifies, contributing 0.10.
**3. Combine and round:** \(P(X\ge Y)=0.30+0.40+0.10=0.80\). Blank E is therefore 0.8 to one decimal place.
**Answer: 0.8.**
**Convention note:** The passage calls the counters independent, but its joint table does not satisfy independence: \(P(X=0,Y=0)=0.10\ne P(X=0)P(Y=0)=0.20(0.30)=0.06\). Use the supplied joint PMF directly; the event calculations do not require independence.
A box contains one red ball and one blue ball. Two balls are drawn one after another with
replacement and the colour of each ball drawn is recorded. Let
• A be the number of times a red ball is drawn in the two draws.
• B be the number of times a blue ball is drawn in the two draws.
Choose the correct statement(s) from the following.
6406536344196.
The possible values of
are {0, 1, 2}
6406536344197.
The possible values of
are {1, 2}
6406536344198.6406536344199.
Published solution
**1. Enumerate equally likely draws:** Replacement makes the outcomes RR, RB, BR and BB equally likely, each with probability \(1/4\). The count pairs \((A,B)\) are \((2,0),(1,1),(1,1),(0,2)\), respectively.
**2. Determine the supports and joint mass:** Both counts can be 0, 1 or 2. The pair \((1,1)\) occurs in RB and BR, so \(f_{AB}(1,1)=2/4=1/2\).
**3. Check the conditional statement:** Since \(A+B=2\), conditioning on \(A=0\) forces \(B=2\). Thus \(P(B=1\mid A=0)=0\), not \(1/4\).
**Answer: A, C — A and C.**
A: **Correct:** RR, mixed draws and BB give red counts 2, 1 and 0.
B: **Incorrect:** RR gives \(B=0\), which the proposed support omits.
C: **Correct:** The two mixed-color outcomes together have probability \(1/2\).
D: **Incorrect:** Given no red draw, both draws are blue, so the conditional probability of exactly one blue is zero.
) passing a signal in a day follows a Poisson distribution with parameter
. Each vehicle independently violates traffic rules with probability 0.1. Let
denote the
number of vehicles violating traffic rules in a day.
Based on the above data, answer the given subquestions.
Which of the following option(s) is/are true?
6406536344201.
Binomial(15, 0.1)
6406536344202.
Binomial(30, 0.1)
6406536344203.
Poisson(0.1)
6406536344204.
Published solution
**1. Condition on the total vehicles:** Given \(X=n\), each of the \(n\) vehicles independently violates with probability 0.1. Hence \(Y\mid X=n\sim\operatorname{Binomial}(n,0.1)\), and at \(n=15\) the trial count is 15.
**2. Find the unconditional distribution:** Thinning a \(\operatorname{Poisson}(30)\) count with probability 0.1 gives \(Y\sim\operatorname{Poisson}(3)\). Equivalently, the generating function is \(E[t^Y]=E[(0.9+0.1t)^X]=e^{30(0.1t-0.1)}=e^{3(t-1)}\).
**3. Read the mean:** \(E[Y]=3\). Thus A and D hold; the unconditional Poisson parameter is 3, not 0.1.
**Answer: A, D — A and D.**
A: **Correct:** The condition fixes the number of trials to 15.
B: **Incorrect:** The mean arrival count 30 is not the number of trials after conditioning on \(X=15\).
C: **Incorrect:** The thinning parameter is \(30(0.1)=3\), not 0.1.
D: **Correct:** Total expectation gives \(E[Y]=E[0.1X]=3\).
) passing a signal in a day follows a Poisson distribution with parameter
. Each vehicle independently violates traffic rules with probability 0.1. Let
denote the
number of vehicles violating traffic rules in a day.
Based on the above data, answer the given subquestions.
Find the value of Var(2X)−3Var(Y ) + 4.
Published solution
**1. Identify the variances:** \(X\sim\operatorname{Poisson}(30)\) gives \(\operatorname{Var}(X)=30\). Independent thinning gives \(Y\sim\operatorname{Poisson}(3)\), so \(\operatorname{Var}(Y)=3\).
**2. Scale the variance correctly:** \(\operatorname{Var}(2X)=2^2\operatorname{Var}(X)=120\). The scalar is squared when it is inside the variance.
**3. Evaluate the requested expression:** \(120-3(3)+4=115\). This is an arithmetic combination of variances; no covariance term belongs in it.
**Answer: 115.**
A small bookstore tracks the number of books sold in an hour. Let a discrete random variable
denote the number of books sold in an hour. The probability mass function of
is given by:
The store defines a discount indicator
as:
Based on the above data, answer the given subquestions.
Find the value of
. Enter the answer correct to two decimal places.
Published solution
**1. Determine the missing probability:** The masses at \(x=0,1,2,3,4\) are \(0.10,a,0.35,0.25,0.10\). Normalization gives \(a=1-0.80=0.20\).
**2. Compute the weighted mean:** \[E[X]=\sum_x xP(X=x)=0(0.10)+1(0.20)+2(0.35)+3(0.25)+4(0.10)=2.05.\] The mean is a probability-weighted average, not the unweighted average of the support.
**Answer: 2.05.**
A small bookstore tracks the number of books sold in an hour. Let a discrete random variable
denote the number of books sold in an hour. The probability mass function of
is given by:
The store defines a discount indicator
as:
Based on the above data, answer the given subquestions.
Find the value of
. Enter the answer correct to two decimal places.
Published solution
**1. Translate the indicator:** \(Y=1\) exactly when \(X\ge2\), which is contained in the conditioning event \(X\ge1\). Thus the numerator is \(P(X\ge2)=0.35+0.25+0.10=0.70\).
**2. Calculate the conditional denominator:** \(P(X\ge1)=1-P(X=0)=0.90\). Therefore \[P(Y=1\mid X\ge1)=\frac{0.70}{0.90}=\frac79=0.777\ldots.\]
**3. Round at the end:** To two decimal places, enter 0.78.
**Answer: 0.78.**
**1. Account for the shared first toss:** Let \(H_1,H_2,H_3\) be head indicators. Then \(X=H_1\), \(Y=H_1+H_2+H_3\), and \(X+Y=2H_1+H_2+H_3\). The variables \(X,Y\) are not independent.
**2. Enumerate the successful outcomes:** If \(H_1=0\), both later tosses must be heads, giving THH. If \(H_1=1\), both must be tails, giving HTT. Each of these two outcomes has probability \(1/8\).
**3. Add their probabilities:** \(P(X+Y=2)=2/8=1/4\).
**Answer: B — 1/4.**
A: **Incorrect:** \(1/8\) counts only one of the two qualifying outcomes.
B: **Correct:** THH and HTT each contribute \(1/8\), totaling \(1/4\).
C: **Incorrect:** There are eight equally likely three-toss outcomes; the favorable probability is \(2/8\), not \(1/16\).
D: **Incorrect:** \(3/8\) is \(P(Y=2)\), which differs from the requested event involving \(X+Y\).
be the
total number of heads obtained across all three tosses. Based on this, answer the given sub
questions:
Find the value of
. Enter the answer correct to three decimal places.
Published solution
**1. Translate the event:** \(Y\) is the total number of heads in three fair independent tosses. The event \(Y=3\) requires HHH.
**2. Multiply the three head probabilities:** \(P(Y=3)=(1/2)^3=1/8=0.125\). This already has the requested three decimal places.
**Answer: 0.125.**
Based on the above data, answer the given subquestions.
Find the value of
.
Published solution
**1. Normalize over all nine pairs:** The joint PMF is \((2i+j)/k\) for \(i,j\in\{1,2,3\}\). Therefore \(k=\sum_{i=1}^3\sum_{j=1}^3(2i+j)\).
**2. Sum each row of weights:** At fixed \(i\), the sum is \(6i+(1+2+3)=6i+6\). The three sums are 12, 18 and 24, totaling \(k=54\).
**3. Check:** Every weight is positive and the normalized weights sum to \(54/54=1\).
**Answer: 54.**
**1. Determine the normalizer:** Summing \(2i+j\) over \(i,j=1,2,3\) gives \(k=12+18+24=54\).
**2. Marginalize over \(Y\):** \[P(X=1)=\sum_{j=1}^3\frac{2+j}{54}=\frac{3+4+5}{54}=\frac{12}{54}=\frac29.\] Fix \(X=1\), but include every possible value of \(Y\).
**Answer: D — 2/9.**
A: **Incorrect:** \(3/54\) is only the cell \(P(X=1,Y=1)\), not the full marginal.
B: **Incorrect:** The marginal numerator is 12, not 15, so the result is \(4/18\), not \(5/18\).
C: **Incorrect:** \(1/3=18/54\) is the marginal for \(X=2\), not \(X=1\).
D: **Correct:** Adding the three cells at \(X=1\) yields \(12/54=2/9\).
**1. Differentiate the CDF on its support:** On \(0<x<5\), \(F_X(x)=x^2/25\), hence \(f_X(x)=F_X^{\prime}(x)=2x/25\). Outside this interval the CDF is constant, so the density is zero.
**2. Verify the total probability:** \(\int_0^5(2x/25)\,dx=[x^2/25]_0^5=1\). The CDF is continuous at the endpoints, so there are no point masses. Values assigned to the density at individual endpoints do not affect probabilities.
**Answer: D — f(x)=2x/25 on [0,5], and 0 elsewhere.**
A: **Incorrect:** This uniform density would give CDF \(x/5\), rather than \(x^2/25\).
B: **Incorrect:** Its integral on the support is \(25/10=2.5\), not one.
C: **Incorrect:** It omits the derivative factor 2 and integrates to \(1/2\).
D: **Correct:** It is the derivative \(2x/25\) on the support and integrates to one.
**1. Intersect with the condition:** Given \(X<4\), the event \(X>3\) becomes \(3<X<4\). Because \(X\) is continuous, endpoint inclusions do not change probabilities.
**2. Use the CDF:** \(P(3<X<4)=F(4)-F(3)=(16-9)/25=7/25\), and \(P(X<4)=F(4)=16/25\).
**3. Form the conditional ratio:** \[P(X>3\mid X<4)=\frac{7/25}{16/25}=\frac7{16}.\]
**Answer: A — 7/16.**
A: **Correct:** It divides the intersection probability \(7/25\) by the conditioning probability \(16/25\).
B: **Incorrect:** It uses the probability at the lower cutoff in the denominator rather than \(P(X<4)\).
C: **Incorrect:** It gives the complementary conditional probability \(P(X\le3\mid X<4)\).
D: **Incorrect:** \(7/25\) is the intersection probability before conditioning.
