What is the exam pattern and passing mark for GTU Probability and Statistics (3130006)?

The GTU theory examination carries 70 total marks with a duration of 2.5 Hours. Minimum passing criteria is 23 marks (33%). The paper consists of 5 main questions with official marking breakdown and internal choices.

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Gujarat Technological UniversitySummer 2026 examination

GTU 3130006 Probability and Statistics (PS) Summer 2026 Solved Paper & Solutions

B.E. · Computer Engineering · Semester 3 · Subject code 3130006

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Total: 70 marks
To pass: 23 marks (33%)
Duration: 2.5 Hours 2:30 PM – 5:00 PM
Questions: 5 with internal choice

2 of 25 questions on this paper were also asked in other years. 11 more are types of question GTU sets again with new data. Each is marked below with the years it appeared.See all 33 repeated 3130006 questions, most asked first

Q.1

14 Marks

(a)

3 marks

A committee of 4 people is to be appointed from 3 teachers, 4 health officers, 2 security personnel and 1 chartered accountant. Find the probability of forming the committee so that the chartered accountant is mandatorily present in the committee.

(b)

4 marks

A student doing his online homework on a regular basis has a chance of 83 percent to get a good grade; but the chance drops to 58 percent if he doesn’t do the homework regularly. The student figures out that he has only 69 percent chance of doing the homework regularly. What is his chance of not getting a good grade in the course?

(c)

7 marks

From a vessel containing 3 white and 5 black balls, 4 balls are transferred into an empty vessel. From this new vessel, a ball is drawn and is found to be white. What is the probability that out of the 4 balls transferred, 3 are white and 1 is black?

Step 1 Events.

Let $E_k$ be the event that $k$ white and $4 - k$ black balls are transferred, and let $W$ be the event that the ball drawn from the new vessel is white. The 8 balls give $\binom{8}{4} = 70$ ways to choose 4.

Step 2 Prior probabilities

(hypergeometric).

$k$ (white)	Ways	$P(E_k)$	$P(W \mid E_k) = \dfrac{k}{4}$
0	$\binom{3}{0}\binom{5}{4} = 5$	$\dfrac{5}{70}$	0
1	$\binom{3}{1}\binom{5}{3} = 30$	$\dfrac{30}{70}$	$\dfrac{1}{4}$
2	$\binom{3}{2}\binom{5}{2} = 30$	$\dfrac{30}{70}$	$\dfrac{2}{4}$
3	$\binom{3}{3}\binom{5}{1} = 5$	$\dfrac{5}{70}$	$\dfrac{3}{4}$

Step 3 Probability of drawing a white ball.

P(W) = \frac{1}{70}\left(5 \cdot 0 + 30 \cdot \tfrac{1}{4} + 30 \cdot \tfrac{2}{4} + 5 \cdot \tfrac{3}{4}\right) = \frac{1}{70}\cdot\frac{105}{4} = \frac{3}{8}

Step 4 Bayes' theorem.

We want the case of 3 white and 1 black, which is $k = 3$ .

P(E_3 \mid W) = \frac{P(E_3)\,P(W \mid E_3)}{P(W)} = \frac{\frac{5}{70}\cdot\frac{3}{4}}{\frac{3}{8}} = \frac{15/280}{105/280} = \frac{1}{7}

Answer

the probability is $\dfrac{1}{7} \approx 0.1429$ .

Q.2

14 Marks

(a)

3 marksSame type asked in 3 of 7 papers

Two variates have the regression lines as x + 4y + 3 = 0 and 4x + 9y + 5 = 0. Find the correlation coefficient.

Also asked in Winter 2025 Q3c OR, Summer 2024 Q3a OR

(b)

4 marks

If 𝑝(𝑥) = 𝑥/15; 𝑥 = 1, 2, 3, 4, 5 = 0, otherwise

iFind 𝑃(𝑋 = 1 or 2)

iiFind 𝑃(1/2 < 𝑋 < 5/2 | 𝑋 > 1)

(c)

7 marks

A car hire firm has two cars, which it hires out day by day. The number of demands for a car on each day is distributed as a Poisson distribution with mean 1.5. Calculate the proportion of days on which (i) neither car is used (ii) some demand is refused.

(c)

7 marks

The joint probability function of X and Y is given as $f(x, y) = e^{-(x+y)}, \ 0 \le x < \infty, \ 0 \le y < \infty$ . Evaluate $P(1 < X + Y < 2)$ .

Q.3

14 Marks

(a)

3 marksSame type asked in 2 of 7 papers

A random variable X with unknown probability distribution has a mean 8 and standard deviation 3. Evaluate the bounds on P (- 4 < X < 20).

Also asked in Winter 2025 Q3a OR

(b)

4 marks

Fit the curve $y = ax^b$ to the given data: x 61 26 7 2.6 y 350 400 500 600

For the curve $y = ax^b$ , take logarithms (base 10): $\log y = \log a + b \log x$ . Put $Y = \log y$ , $X = \log x$ and $A = \log a$ , so $Y = A + bX$ .

$x$	$y$	$X = \log x$	$Y = \log y$	$X^2$	$XY$
61	350	1.7853	2.5441	3.1873	4.5420
26	400	1.4150	2.6021	2.0022	3.6820
7	500	0.8451	2.6990	0.7142	2.2808
2.6	600	0.4150	2.7782	0.1722	1.1529
Sum		4.4604	10.6233	6.0759	11.6577

Normal equations ( $n = 4$ ):

10.6233 = 4A + 4.4604\,b, \qquad 11.6577 = 4.4604\,A + 6.0759\,b

b = \frac{4(11.6577) - (4.4604)(10.6233)}{4(6.0759) - (4.4604)^2} = \frac{-0.7536}{4.4084} = -0.1709

A = \frac{10.6233 + 0.1709(4.4604)}{4} = 2.8464, \qquad a = 10^{2.8464} \approx 702

Answer

the curve is $y = 702\,x^{-0.1709}$ .

(c)

7 marks

Find the coefficient of variation of the marks obtained by the students of a college: Marks : 20-30 30-40 40-50 50-60 60-70 No. of Students : 2 35 46 12

Step 1 Table.

Take the class marks $x$ , and the assumed mean $A = 45$ with class width $h = 10$ , $d = \dfrac{x - 45}{10}$ .

Class	$f$	$x$	$d$	$fd$	$fd^2$
20 - 30	2	25	-2	-4	8
30 - 40	35	35	-1	-35	35
40 - 50	46	45	0	0	0
50 - 60	12	55	1	12	12
60 - 70	5	65	2	10	20
Total	100			-17	75

Step 2 Mean.

\bar{x} = A + h\,\frac{\sum fd}{N} = 45 + 10\left(\frac{-17}{100}\right) = 43.3

Step 3 Standard deviation.

\sigma = h\sqrt{\frac{\sum fd^2}{N} - \left(\frac{\sum fd}{N}\right)^2} = 10\sqrt{0.75 - 0.0289} = 10\sqrt{0.7211} = 8.49

Step 4 Coefficient of variation.

CV = \frac{\sigma}{\bar{x}} \times 100 = \frac{8.49}{43.3} \times 100 = 19.61\%

Answer

the coefficient of variation is $19.61\%$ .

(a)

3 marksSame type asked in 5 of 7 papers

A sample of 12 fathers and their eldest sons gave the following data about their height in inches. Calculate coefficient of rank correlation. Father: 65 63 67 64 68 62 70 66 68 67 69 71 Son : 68 66 68 65 69 66 68 65 71 67 68 70

Also asked in Winter 2025 Q3b OR, Summer 2025 Q3b, Winter 2023 Q4c, Summer 2023 Q4b

(b)

4 marksSame type asked in 2 of 7 papers

An incomplete frequency distribution is given with N = 229 and median = 46. Determine the missing frequencies. Obs. 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Freq. 12 30 ? 65 ? 25 18

Also asked in Summer 2023 Q3c

(c)

7 marksAsked in 2 of 7 papers

Obtain the line of regression of monthly sales (Y) on advertisement expenditure (X) and estimate the monthly sales when the company will spend Rs. 50,000 on advertisement, if the data on X and Y are as follows: Y (in lac) : 74 76 60 68 79 70 71 94 X (in thousand) : 43 44 36 38 47 40 41 54

Also asked in Winter 2024 Q3c OR

Q.4

14 Marks

(a)

3 marks

The arrival rate of cars at a gas station is 40 customers per hour. What is the probability of having no arrival of cars in a 5 minute interval?

(b)

4 marksSame type asked in 2 of 7 papers

The standard deviations calculated from two random samples of sizes 9 and 13 are 2.1 and 1.8 respectively. May the samples be regarded as drawn from normal populations with the same standard deviation? (Test at 5% significance level)

Also asked in Winter 2025 Q2b

(c)

7 marksSame type asked in 3 of 7 papers

In a flu epidemic, 500 persons contracted the disease. 200 receives no serum treatment and of these 75 died. Of those who did receive serum treatment, 65 died. Was the serum treatment effective? Use test at 1% significance level.

Also asked in Winter 2025 Q5c, Summer 2025 Q4c OR

Step 1 Hypotheses.

$H_0$ : the serum has no effect (death and treatment are independent). $H_1$ : the serum is effective.

Step 2 Observed table.

The serum group is $500 - 200 = 300$ .

	Died	Survived	Total
No serum	75	125	200
Serum	65	235	300
Total	140	360	500

Step 3 Expected frequencies

$E = \dfrac{\text{row total} \times \text{column total}}{500}$ :

E_{11} = \frac{200 \times 140}{500} = 56, \quad E_{12} = 144, \quad E_{21} = \frac{300 \times 140}{500} = 84, \quad E_{22} = 216

Step 4 Chi-square.

\chi^2 = \sum\frac{(O - E)^2}{E} = \frac{19^2}{56} + \frac{19^2}{144} + \frac{19^2}{84} + \frac{19^2}{216} = 6.446 + 2.507 + 4.298 + 1.671 = 14.92

Step 5 Decision.

The degrees of freedom are $(2-1)(2-1) = 1$ . The table value at 1% is $\chi^2_{0.01}(1) = 6.635$ . Since $14.92 > 6.635$ , $H_0$ is rejected. (With Yates' correction the value is $14.15$ , and the result is the same.)

Answer

yes, the serum treatment was effective.

(a)

3 marks

In a distribution exactly normal, 10.03% of the items are under 25 kg weight and 89.97% of the items are under 70 kg weight. What are the mean and standard deviation of the distribution?

(b)

4 marksSame type asked in 4 of 7 papers

A machine produced 20 defective articles in a batch of 400. After overhauling, it produced 10 defectives in a batch of 300. Has the machine improved? (Test at 5% level of significance)

Also asked in Winter 2025 Q5a OR, Winter 2024 Q4c, Summer 2023 Q5b OR

(c)

7 marksAsked in 2 of 7 papers

Fit a second degree parabola to the data:

x 1 2 3 4 5 6 7 8 9 y 2 6 7 8 10 11 11 10

Also asked in Summer 2024 Q5c OR

Step 1 Method.

The parabola is $y = a + bx + cx^2$ . To make the sums small, put $u = x - 5$ , so $u = -4, \dots, 4$ and $y = A + Bu + Cu^2$ . Then $\sum u = 0$ and $\sum u^3 = 0$ .

Step 2 Table.

$x$	$u$	$y$	$u^2$	$u^4$	$uy$	$u^2y$
1	-4	2	16	256	-8	32
2	-3	6	9	81	-18	54
3	-2	7	4	16	-14	28
4	-1	8	1	1	-8	8
5	0	10	0	0	0	0
6	1	11	1	1	11	11
7	2	11	4	16	22	44
8	3	10	9	81	30	90
9	4	9	16	256	36	144
Sum	0	74	60	708	51	411

Step 3 Normal equations

( $n = 9$ ):

\sum y = nA + C\sum u^2 \Rightarrow 74 = 9A + 60C

\sum uy = B\sum u^2 \Rightarrow 51 = 60B \Rightarrow B = 0.85

\sum u^2y = A\sum u^2 + C\sum u^4 \Rightarrow 411 = 60A + 708C

Step 4 Solve.

From the first and third: $A = 10.0043$ and $C = -0.2673$ .

Step 5 Back to $x$ .

Put $u = x - 5$ :

y = 10.0043 + 0.85(x - 5) - 0.2673(x - 5)^2 = -0.9286 + 3.5232\,x - 0.2673\,x^2

Answer

the second degree parabola is $y = -0.929 + 3.523\,x - 0.267\,x^2$ .

Q.5

14 Marks

(a)

3 marksSame type asked in 4 of 7 papers

The mean weekly sales of soap bars was 146.3 bars per store. After an advertising campaign, the mean weekly sales in 22 stores for a typical week increased to 153.7 with standard deviation of 17.2. Was the advertising campaign successful? ( Test at 5% level)

Also asked in Winter 2024 Q4a OR, Winter 2023 Q5a OR, Summer 2023 Q5a OR

(b)

4 marksSame type asked in 2 of 7 papers

A test is carried out to check the efficiency of both the factories ascertaining the variability of the life of the bulbs produced by each factory. The results are as follows: Factory A Factory B Standard deviation of lifetime (hours) 240 220 No. of bulbs in sample 100 200 Determine whether the difference between the variability of life of light bulbs from the two factories is significant (Test at 1% level).

Also asked in Winter 2025 Q2b

(c)

7 marksSame type asked in 4 of 7 papers

In his experiments on pea-breeding, Mendel obtained the following frequencies of seeds: Class Observed Frequency Round and yellow 315 Wrinkled and yellow 101 Round and green 108 Wrinkled and green 32 Total 556 Theory predicts that the frequencies should be in the proportions 9 : 3 : 3 : 1. Use test to examine the correspondence between theory and observations at 5% significance level.

Also asked in Winter 2025 Q5c OR, Summer 2024 Q4c, Winter 2023 Q2b

Step 1 Hypotheses.

$H_0$ : the data follow the ratio $9 : 3 : 3 : 1$ . $H_1$ : they do not.

Step 2 Expected frequencies.

The total is $556$ and $9 + 3 + 3 + 1 = 16$ , so $E = 556 \times \dfrac{\text{part}}{16}$ .

Class	$O$	$E$	$\dfrac{(O - E)^2}{E}$
Round and yellow	315	312.75	0.0162
Wrinkled and yellow	101	104.25	0.1013
Round and green	108	104.25	0.1349
Wrinkled and green	32	34.75	0.2176
Total	556	556	0.4700

Step 3 Chi-square.

$\chi^2 = \sum\dfrac{(O - E)^2}{E} = 0.470$ .

Step 4 Decision.

The degrees of freedom are $4 - 1 = 3$ . The table value is $\chi^2_{0.05}(3) = 7.815$ . Since $0.470 < 7.815$ , $H_0$ is accepted.

Answer

the observations agree well with the $9 : 3 : 3 : 1$ theory at the 5% level.

(a)

3 marks

The correlation coefficient between income and food expenditure for sample of 7 household from a low income group is 0.9. Using 1% level of significance, test whether the correlation coefficient between incomes and food expenditure is positive. Assume that the population of both variables are normally distributed.

(b)

4 marksSame type asked in 3 of 7 papers

A machine part was designed to withstand an average pressure of 120 units. A random sample of size 100 from a large batch was tested and it was found that the average pressure which these parts can withstand is 105 units with a standard deviation of 20 units. Test at 1% significance level whether the batch meets the specification.

Also asked in Winter 2025 Q5a, Summer 2023 Q5a

(c)

7 marks

The following table gives the joint distribution of X and Y:

(rows are the values of X, columns are the values of Y) Y: 0 1 2 X = 1: 0.3 0.2 0.1 X = 2: 0.1 0 0.3 Are X and Y independent?

$X$ and $Y$ are independent only if $P(X = x, Y = y) = P(X = x)\,P(Y = y)$ for every cell.

Step 1 Marginal probabilities

(row and column totals).

	$Y = 0$	$Y = 1$	$Y = 2$	$P(X = x)$
$X = 1$	0.3	0.2	0.1	0.6
$X = 2$	0.1	0	0.3	0.4
$P(Y = y)$	0.4	0.2	0.4	1

Step 2 Compare each cell with the product of its marginals.

Cell $(x, y)$	Given $P$	$P(x)\,P(y)$
(1, 0)	0.3	$0.6 \times 0.4 = 0.24$
(1, 1)	0.2	$0.6 \times 0.2 = 0.12$
(1, 2)	0.1	$0.6 \times 0.4 = 0.24$
(2, 0)	0.1	$0.4 \times 0.4 = 0.16$
(2, 1)	0	$0.4 \times 0.2 = 0.08$
(2, 2)	0.3	$0.4 \times 0.4 = 0.16$

Step 3 Conclusion.

The two columns differ in every cell. For example, $P(X = 2, Y = 1) = 0$ , but $P(X = 2)\,P(Y = 1) = 0.08$ . One cell that fails is enough.

Answer

$X$ and $Y$ are not independent.

More Probability and Statistics papers

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About this paper

Official Gujarat Technological University (GTU) examination paper and step-by-step solutions for Probability and Statistics (PS) (Summer 2026, B.E. · Computer Engineering, Sem 3). Features complete 70-mark regular & remedial examination pattern, official marking distribution across all 5 questions, and direct 1-click official PDF download.

Transcribed for student exam preparation from Gujarat Technological University official examination archives. Questions, syllabus guidelines, and curriculum marking schemes remain the intellectual property of Gujarat Technological University.

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$x$	$u$	$y$	$u^2$	$u^4$	$uy$	$u^2y$
1	-4	2	16	256	-8	32
2	-3	6	9	81	-18	54
3	-2	7	4	16	-14	28
4	-1	8	1	1	-8	8
5	0	10	0	0	0	0
6	1	11	1	1	11	11
7	2	11	4	16	22	44
8	3	10	9	81	30	90
9	4	9	16	256	36	144
Sum	0	74	60	708	51	411

$x$	$u$	$y$	$u^2$	$u^4$	$uy$	$u^2y$
1	-4	2	16	256	-8	32
2	-3	6	9	81	-18	54
3	-2	7	4	16	-14	28
4	-1	8	1	1	-8	8
5	0	10	0	0	0	0
6	1	11	1	1	11	11
7	2	11	4	16	22	44
8	3	10	9	81	30	90
9	4	9	16	256	36	144
Sum	0	74	60	708	51	411

$x$	$u$	$y$	$u^2$	$u^4$	$uy$	$u^2y$
1	-4	2	16	256	-8	32
2	-3	6	9	81	-18	54
3	-2	7	4	16	-14	28
4	-1	8	1	1	-8	8
5	0	10	0	0	0	0
6	1	11	1	1	11	11
7	2	11	4	16	22	44
8	3	10	9	81	30	90
9	4	9	16	256	36	144
Sum	0	74	60	708	51	411