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Question 2 MCQ · 2.0 marks
In an examination, there are
multiple-choice questions. Each question has
options, out of
which exactly one option is correct. Every student is required to attempt all the questions. Suppose
that a student selects an option randomly for each question. What is the probability that exactly
Each independent random guess succeeds with probability 1/4 and fails with probability 3/4. Exactly three correct answers among five follow the binomial formula: \[P(X=3)=\binom53(1/4)^3(3/4)^2=\frac{\binom53\,1^3\,3^2}{4^5}=\frac{90}{1024}.\] Choose which three questions are correct, then multiply their success and failure probabilities.
**Answer: A.**
A: **Correct:** Correct binomial coefficient and powers for three successes and two failures.
B: **Incorrect:** Interchanges the success and failure counts and thus describes two correct answers.
C: **Incorrect:** Uses three failures as well as three successes, inconsistent with five questions.
D: **Incorrect:** Uses an incorrect denominator and success/failure powers.
. For these values, the cumulative
distribution function is defined by
Find the value of
.
Published solution
Since X takes values only in 1 through 12, all probability has accumulated by 12: F(12)=1. Substituting in the supplied CDF gives \((12^2+k)/212=1\), so \(144+k=212\) and k=68. Check validity: F(1)=69/212 is positive, subsequent jumps are \((2x-1)/212>0\), and the final value is 1.
**Answer: 68.**
The correlation coefficient between the number of dresses sold and the daily revenue of a clothing
store is
. The standard deviations of the number of dresses sold and daily revenue are 5
dresses and ₹3000 , respectively. If the covariance between the two variables is
, find the value of
.
Published solution
Pearson correlation is \(r=\operatorname{Cov}(X,Y)/(\sigma_X\sigma_Y)\). Rearranging, \(\operatorname{Cov}(X,Y)=r\sigma_X\sigma_Y=0.65(5)(3000)=9750\). Covariance retains the product of the variables' units, unlike dimensionless correlation.
**Answer: 9750.**
Use \(\operatorname{Var}(aX+b)=a^2\operatorname{Var}(X)\): \(25\operatorname{Var}(X)=200\), so Var(X)=8; \(9\operatorname{Var}(Y)=72\), so Var(Y)=8. Independence makes covariance zero, hence \(\operatorname{Var}(X+Y)=8+8=16\). The requested standard deviation is the square root: \(SD(X+Y)=\sqrt{16}=4\), not 16.
**Answer: 4.**
A group of 6 friends is seated around a circular table for a dinner. In how many different ways can
the 6 friends be seated around the table such that Anu and Meera are not sitting next to each
other?
Published solution
Count circular arrangements up to rotation, treating mirror orders as different. Six distinct friends have \((6-1)!=120\) circular arrangements. When Anu and Meera sit together, combine them into one block: five units have \((5-1)!=24\) circular orders, and the pair has two internal orders. Adjacent arrangements number 48. Thus non-adjacent arrangements \(=120-48=72\).
**Answer: 72.**
A clothing store has 6 red dresses, 2 blue dresses, and 3 black dresses in a rack. Two dresses are
selected at random without replacement. What is the probability that both selected dresses are of
different colours? Enter the answer correct to two decimal places.
Published solution
There are 11 dresses and \(\binom{11}2=55\) unordered selections. Different-colour pairs can be red-blue, red-black or blue-black: \(6\cdot2+6\cdot3+2\cdot3=12+18+6=36\). These cases are disjoint. Therefore the probability is \(36/55\approx0.654545\), rounding to 0.65. Without replacement means the two selected dresses must be distinct.
**Answer: 0.65.**
The expected number of food orders during the lunch hour is 4.15.
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The expected number of food orders during the lunch hour is 3.95.
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The expected value of the delivery incentive during lunch hour is 79.8 rupees.
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The expected value of the delivery incentive during lunch hour is 49.8 rupees.
Published solution
Compute the weighted average of the possible order counts: \[E[X]=2(0.10)+3(0.20)+4(0.30)+5(0.25)+6(0.15)=4.15.\] The incentive is g(X)=12X+30. By linearity, \(E[g(X)]=12E[X]+30=12(4.15)+30=79.8\) rupees. Do not omit the fixed 30-rupee component.
**Answer: A, C.**
A: **Correct:** The probability-weighted sum is 4.15.
B: **Incorrect:** The stated 3.95 does not equal the weighted sum.
C: **Correct:** Includes both the variable payment 49.8 and the fixed payment 30.
D: **Incorrect:** 49.8 includes only 12E[X] and omits the fixed payment.
The operating time of a backup generator is exponentially distributed with a mean of
hours. A
data centre needs the generator to run continuously for more than
hours.
Based on the above data, answer the given subquestions.
What is the probability that the generator will run for more than
hours?(Give the answer
correct to two decimal places)
Published solution
Mean lifetime 120 hours gives rate \(\lambda=1/120\). For an exponential variable, \(P(T>t)=e^{-\lambda t}\). Thus \(P(T>60)=e^{-60/120}=e^{-0.5}\approx0.606531\). Correct to two decimal places, enter 0.61.
**Answer: 0.61.**
The operating time of a backup generator is exponentially distributed with a mean of
hours. A
data centre needs the generator to run continuously for more than
hours.
Based on the above data, answer the given subquestions.
If the generator has already been operating for more than
hours, what is the probability that it
will continue to operate for the next
hours without requiring maintenance?(Give the answer
correct to two decimal places)
Published solution
The exponential distribution is memoryless: \(P(T>s+t\mid T>s)=P(T>t)\). Already lasting 50 hours does not alter the model's distribution of remaining life. For the next 10 hours, \[P(T>60\mid T>50)=\frac{e^{-60/120}}{e^{-50/120}}=e^{-10/120}\approx0.920044.\] Round to 0.92. The question asks for ten additional hours, not a new 60-hour lifetime from zero.
**Answer: 0.92.**
A quality-control engineer studies the number of defective items in a randomly selected batch. The
number of defective items
is assumed to follow a binomial distribution. It is known that the
probability of an item being defective is
and that the variance of
is
. Based on this
information, choose the correct option(s) from the following.
6406537436048.
The number of trials in the binomial experiment is
.
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The mean of
is
.
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The cumulative distribution function of
is given by
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The random variable
can take values from
to
.
Published solution
For a binomial variable, Var(X)=np(1-p). Given p=0.7 and variance 2.1, \(n=2.1/(0.7\cdot0.3)=10\). Its mean is np=7. Its PMF is \(P(X=k)=\binom{10}k0.7^k0.3^{10-k}\); summing from k=0 to x gives the displayed CDF at integer x=0,...,10. The possible counts include zero successes, so the support is 0 through 10, not 1 through 10.
**Answer: A, B, C.**
A: **Correct:** Solving the variance equation gives 10 trials.
B: **Correct:** The mean is 10 times 0.7, or 7.
C: **Correct:** The displayed sum accumulates the binomial masses through x.
D: **Incorrect:** Zero successes has probability 0.3 to the tenth power and cannot be excluded.
The battery life, in hours, of a certain electronic device is uniformly distributed between
and
hours, inclusive. Let
denote the battery life of a randomly selected device.
Based on the above data, answer the given subquestions.
Find the value of
if the probability that the battery life is more than
hours is 0.25.
Published solution
For a continuous uniform distribution on [a,12], probability equals interval length divided by total length. Since P(X>10)=0.25, \((12-10)/(12-a)=0.25\). Thus 2=0.25(12-a), giving 12-a=8 and a=4. The density is consequently 1/8 on [4,12].
**Answer: 4.**
The battery life, in hours, of a certain electronic device is uniformly distributed between
and
hours, inclusive. Let
denote the battery life of a randomly selected device.
Based on the above data, answer the given subquestions.
What is the probability that battery life is less than 10 hours and more than 6 hours?(Give the
answer correct to one decimal place)
Published solution
The shared condition gives a=4, since 2/(12-a)=0.25. The desired interval is 6<X<10, of length 4, within the support [4,12], of length 8. Uniform probability is therefore \((10-6)/(12-4)=4/8=0.5\). Endpoints have zero probability for a continuous distribution.
**Answer: 0.5.**
A bakery records information about its daily orders, including the type of product purchased
(bread, cake, pastry, etc.), the size of the cake (small, medium, large), the number of items sold per
day, and the customer's order number. Which of the following statements are correct?
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The type of product purchased is measured on the nominal scale.
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The size of the cake is measured on the ordinal scale.
6406537436056.
The number of items sold is measured on the interval scale.
6406537436057.
The customer’s order number is measured on the ratio scale.
Published solution
Product types are unordered category labels, so they use a nominal scale. Small, medium and large have a natural order without equal numerical gaps, so cake size is ordinal. Item count has a meaningful zero and meaningful ratios, so it is ratio-scale. An order number identifies an order rather than measuring its size, so it is nominal even when written with digits.
**Answer: A, B.**
A: **Correct:** Product labels have no inherent quantitative order.
B: **Correct:** Size categories are ordered.
C: **Incorrect:** Item counts are ratio-scale, not merely interval-scale.
D: **Incorrect:** Order numbers are identifiers; twice an order number does not mean twice as much order.
Both branches have the same median number of dresses sold.
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Branch B has a larger range of daily sales than Branch A.
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The mean number of dresses sold is not the same for both branches.
6406537436061.
The IQR of Branch B is greater than the IQR of Branch A.
Published solution
The sorted sales are A=(12,16,20,24,28) and B=(8,14,20,26,32). Both medians are 20. Their means are also 20 because each sums to 100. Ranges are 28-12=16 and 32-8=24, so B has the larger range. Using median-of-halves excluding the centre, A has Q1=14, Q3=26, IQR=12; B has Q1=11, Q3=29, IQR=18. Hence B also has the larger IQR. Common alternative quartile conventions preserve this IQR comparison.
**Answer: A, B, D.**
A: **Correct:** Both middle observations are 20.
B: **Correct:** B's range 24 exceeds A's range 16.
C: **Incorrect:** The means are equal, both 100/5=20.
D: **Correct:** B has a larger middle-50% spread under the stated quartile convention.
A bakery has two branches, Branch A and Branch B. Branch A prepares 40% of the total cakes,
while Branch B prepares the remaining 60%. The probability that a cake is defective is 2% for
Branch A and 5% for Branch B. A randomly selected cake is found to be defective. What is the
probability that the defective cake was prepared by Branch A? Enter the answer correct to two
decimal places.
Published solution
Let D mean defective. The joint probabilities are \(P(A\cap D)=0.4(0.02)=0.008\) and \(P(B\cap D)=0.6(0.05)=0.03\). Therefore P(D)=0.038. Bayes' rule gives \[P(A\mid D)=\frac{0.008}{0.038}=\frac4{19}\approx0.210526.\] Round to 0.21. The defect rate for Branch A alone is not the answer; the observation that the cake is defective changes the branch probabilities.
**Answer: 0.21.**
