What is the exam pattern and passing mark for GTU Probability and Statistics (BE03000251)?

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.

Where can I download the official GTU BE03000251 Summer 2026 question paper PDF?

The official GTU question paper PDF is linked directly from Gujarat Technological University's official examination archive. You can view the full paper online or download the original file.

How can I practice solutions for these GTU questions?

Every question on UnFoldPrep includes deep-linking and one-click AI prompt buttons for ChatGPT, Claude, and Perplexity that instantly formulate structured exam prompts with standard mark breakdowns and academic rubrics.

Gujarat Technological UniversitySummer 2026 examination

GTU BE03000251 Probability and Statistics Summer 2026 Solved Paper & Solutions

B.E. · Common Engineering · Semester 3 · Subject code BE03000251

Download official PDF View PDF here

Share with your class

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

Q.1

14 Marks

(a)

3 marks

The time between breakdown of a particular machine follows an exponential distribution with a mean of 17 days. Calculate the probability that a machine breaks down in a 15 days.

(b)

4 marks

A discrete random variable X has the following probability distributions:

Find (a) $k$ (b) $P(X<3)$ (c) $P(2<X<5)$ .

X	0	1	2	3	4	5
F(X=x)	0	k	0.2	2k	0.3	2k

Step 1 Find k

The total probability is 1:

0 + k + 0.2 + 2k + 0.3 + 2k = 1 \;\Rightarrow\; 5k + 0.5 = 1 \;\Rightarrow\; k = 0.1

So the distribution is:

X	0	1	2	3	4	5
P(X = x)	0	0.1	0.2	0.2	0.3	0.2

Step 2 $P(X < 3)$

P(X < 3) = P(0) + P(1) + P(2) = 0 + 0.1 + 0.2 = 0.3

Step 3 $P(2 < X < 5)$

P(2 < X < 5) = P(3) + P(4) = 0.2 + 0.3 = 0.5

Answer

(a) $k = 0.1$ ; (b) $P(X < 3) = 0.3$ ; (c) $P(2 < X < 5) = 0.5$ .

(c)

7 marks

Using method of least square, find the best fitting second degree curve to the following data:

x	1	2	3	4
y	6	11	18	27

Curve: $y = a + bx + cx^2$ . The normal equations are:

\sum y = na + b\sum x + c\sum x^2

\sum xy = a\sum x + b\sum x^2 + c\sum x^3

\sum x^2y = a\sum x^2 + b\sum x^3 + c\sum x^4

Step 1 Table

x	y	x²	x³	x⁴	xy	x²y
1	6	1	1	1	6	6
2	11	4	8	16	22	44
3	18	9	27	81	54	162
4	27	16	64	256	108	432
10	62	30	100	354	190	644

Step 2 Normal equations

( $n = 4$ )

4a + 10b + 30c = 62 \quad \text{(i)}

10a + 30b + 100c = 190 \quad \text{(ii)}

30a + 100b + 354c = 644 \quad \text{(iii)}

Step 3 Solve

(ii) × 2 − (i) × 5: $20a + 60b + 200c - 20a - 50b - 150c = 380 - 310$ , so

10b + 50c = 70 \;\Rightarrow\; b + 5c = 7 \quad \text{(iv)}

(iii) − (ii) × 3: $100b - 90b + 354c - 300c = 644 - 570$ , so

10b + 54c = 74 \quad \text{(v)}

(v) − 10 × (iv): $54c - 50c = 74 - 70$ , so $c = 1$ .

From (iv): $b = 7 - 5 = 2$ . From (i): $4a = 62 - 20 - 30 = 12$ , so $a = 3$ .

Check: at $x = 3$ , $y = 3 + 6 + 9 = 18$ , which matches the data exactly. In fact all four points lie on this curve.

Answer

The best fitting second degree curve is $y = 3 + 2x + x^2$ .

Q.2

14 Marks

(a)

3 marks

Three coins are tossed together and let random variable X be the number of heads in each outcome. Then find (a) probability distribution (b) mean.

Sample space for three coins: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT (8 outcomes). X = number of heads.

(a) Probability distribution

X	0	1	2	3
P(X)	1/8	3/8	3/8	1/8

(b) Mean

E(X) = \sum x\,P(x) = 0 \cdot \frac{1}{8} + 1 \cdot \frac{3}{8} + 2 \cdot \frac{3}{8} + 3 \cdot \frac{1}{8} = \frac{12}{8} = 1.5

Answer

The distribution is $\frac18, \frac38, \frac38, \frac18$ for $X = 0, 1, 2, 3$ , and the mean is $1.5$ .

(b)

4 marks

Find coefficient of correlation of price and demand of the commodity:

Price (Rs)	60	65	70	75	80	85	90	95	100
Demand (qty.)	35	30	25	25	23	21	20	20	18

(c)

7 marks

Three boxes contain 10%, 20% and 30% of defective finger joints. A finger joint is selected at random which is defective. Determine the probability that it comes from (a) first box (b) second box (c) third box.

(c)

7 marks

The life in hours of a certain kind of radio tube has the probability density

f(x) = \begin{cases} \frac{100}{x^2} & ; \text{for } x \ge 100 \\ 0 & ; \text{elsewhere} \end{cases}

Find the distribution function and use it to determine the probability that the life of tube is more than 150 hrs.

Q.3

14 Marks

(a)

3 marks

Find the rank coefficient of correlation:

x	12	10	17	14	13	18	20
y	110	210	108	135	160	104	70

x	y	$R_x$	$R_y$	$d$	$d^2$
12	110	6	4	2	4
10	210	7	1	6	36
17	108	3	5	−2	4
14	135	4	3	1	1
13	160	5	2	3	9
18	104	2	6	−4	16
20	70	1	7	−6	36

(b)

4 marks

Determine the equation of a straight line which best fits the following date:

x	2003	2004	2005	2006	2007
y	35	56	79	80	40

Let $X = x - 2005$ (so $\sum X = 0$ ), and fit $y = a + bX$ .

x	X	y	XY	X²
2003	−2	35	−70	4
2004	−1	56	−56	1
2005	0	79	0	0
2006	1	80	80	1
2007	2	40	80	4
	0	290	34	10

Normal equations (with $\sum X = 0$ ):

a = \frac{\sum y}{n} = \frac{290}{5} = 58, \qquad b = \frac{\sum XY}{\sum X^2} = \frac{34}{10} = 3.4

y = 58 + 3.4X = 58 + 3.4(x - 2005)

Answer

The best fitting line is $y = 58 + 3.4(x - 2005)$ ; for example, for 2008 it gives $y = 68.2$ .

(c)

7 marks

Find the Karl Pearson’s coefficient of skewness for the following frequency distribution:

Class	50-55	55-60	60-65	65-70	70-75
Frequency	8	10	15	17	8

Karl Pearson's coefficient of skewness: $S_k = \dfrac{\text{Mean} - \text{Mode}}{\sigma}$ .

Step 1 Table

(mid-values $m$ , $u = \dfrac{m - 62.5}{5}$ )

Class	f	m	u	fu	fu²
50-55	8	52.5	−2	−16	32
55-60	10	57.5	−1	−10	10
60-65	15	62.5	0	0	0
65-70	17	67.5	1	17	17
70-75	8	72.5	2	16	32
	58			7	91

Step 2 Mean

\bar{x} = 62.5 + 5 \times \frac{7}{58} = 62.5 + 0.603 = 63.10

Step 3 Standard deviation

\sigma = 5\sqrt{\frac{91}{58} - \left(\frac{7}{58}\right)^2} = 5\sqrt{1.5690 - 0.0146} = 5 \times 1.2468 = 6.234

Step 4 Mode

The modal class is 65-70 (highest frequency 17). $L = 65$ , $f_1 = 17$ , $f_0 = 15$ , $f_2 = 8$ , $h = 5$ :

\text{Mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 65 + \frac{17 - 15}{34 - 15 - 8} \times 5 = 65 + \frac{2}{11} \times 5 = 65.91

Step 5 Coefficient of skewness

S_k = \frac{63.10 - 65.91}{6.234} = \frac{-2.81}{6.234} = -0.45

The mean is less than the mode, so the distribution is negatively skewed (longer tail on the left).

Answer

Karl Pearson's coefficient of skewness is about $-0.45$ .

(a)

3 marks

Find standard deviation for the following distribution:

$x_i$	5	10	15	20	25
$f_i$	7	4	6	3	5

$N = \sum f = 25$ .

x	f	fx	fx²
5	7	35	175
10	4	40	400
15	6	90	1350
20	3	60	1200
25	5	125	3125
	25	350	6250

\bar{x} = \frac{350}{25} = 14, \qquad \sigma = \sqrt{\frac{6250}{25} - 14^2} = \sqrt{250 - 196} = \sqrt{54} = 7.35

Answer

The standard deviation is about 7.35.

(b)

4 marks

Fit the curve

y = ax^b

to the following data:

x	61	26	7	2.6
y	350	400	500	600

Linearise: $y = ax^b$ gives $\log y = \log a + b\log x$ , that is, $Y = A + bX$ with $Y = \log_{10} y$ , $X = \log_{10} x$ , $A = \log_{10} a$ .

x	y	X = log x	Y = log y	X²	XY
61	350	1.7853	2.5441	3.1874	4.5420
26	400	1.4150	2.6021	2.0021	3.6819
7	500	0.8451	2.6990	0.7142	2.2809
2.6	600	0.4150	2.7782	0.1722	1.1529
		4.4604	10.6232	6.0759	11.6576

Normal equations:

10.6232 = 4A + 4.4604\,b, \qquad 11.6576 = 4.4604\,A + 6.0759\,b

Solving:

b = \frac{4(11.6576) - 4.4604(10.6232)}{4(6.0759) - 4.4604^2} = \frac{-0.7533}{4.4086} = -0.1709

A = \frac{10.6232 + 0.1709 \times 4.4604}{4} = 2.8463 \;\Rightarrow\; a = 10^{2.8463} = 702

Answer

$y = 702\,x^{-0.171}$ .

(c)

7 marks

Find the coefficient of skewness for the following:

Years under	10	20	30	40	50	60
No. of persons	15	32	51	78	97	109

"Years under" gives cumulative frequencies (less than type). Convert to class frequencies:

Age (years)	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	15	17	19	27	19	12
c.f.	15	32	51	78	97	109

$N = 109$ . Use Karl Pearson's coefficient $S_k = \dfrac{\text{Mean} - \text{Mode}}{\sigma}$ .

Step 1 Mean and SD

(mid-values 5, 15, ..., 55; $u = \dfrac{m - 35}{10}$ )

f	u	fu	fu²
15	−3	−45	135
17	−2	−34	68
19	−1	−19	19
27	0	0	0
19	1	19	19
12	2	24	48
109		−55	289

\bar{x} = 35 + 10 \times \frac{-55}{109} = 35 - 5.046 = 29.95

\sigma = 10\sqrt{\frac{289}{109} - \left(\frac{-55}{109}\right)^2} = 10\sqrt{2.6514 - 0.2546} = 10 \times 1.5481 = 15.48

Step 2 Mode

Modal class 30-40 ( $f_1 = 27$ , $f_0 = 19$ , $f_2 = 19$ , $h = 10$ ):

\text{Mode} = 30 + \frac{27 - 19}{54 - 19 - 19} \times 10 = 30 + \frac{8}{16} \times 10 = 35

Step 3 Coefficient of skewness

S_k = \frac{29.95 - 35}{15.48} = \frac{-5.05}{15.48} = -0.33

Check using quartiles (Bowley): $Q_1 = 17.21$ , median $= 31.30$ , $Q_3 = 41.97$ , so $S_k = \dfrac{Q_3 + Q_1 - 2M}{Q_3 - Q_1} = \dfrac{41.97 + 17.21 - 62.59}{24.77} = -0.14$ . Both show negative skewness.

Answer

Karl Pearson's coefficient of skewness is about $-0.33$ (negatively skewed).

Q.4

14 Marks

(a)

3 marks

There are 5 black balls and 4 red balls. Find the number of ways in which 6 balls can be selected so that there are at least 2 red balls in that selection.

Select 6 balls from 5 black (B) and 4 red (R) with at least 2 red:

Red	Black	Ways
2	4	$\binom{4}{2}\binom{5}{4} = 6 \times 5 = 30$
3	3	$\binom{4}{3}\binom{5}{3} = 4 \times 10 = 40$
4	2	$\binom{4}{4}\binom{5}{2} = 1 \times 10 = 10$

Total $= 30 + 40 + 10 = 80$ .

Check: all selections $\binom{9}{6} = 84$ , minus those with 1 red ( $\binom{4}{1}\binom{5}{5} = 4$ ) and 0 red (impossible, only 5 black), gives $84 - 4 = 80$ .

Answer

80 ways.

(b)

4 marks

Assume that on the average one telephone number out of fifteen called between 1 pm and 2 pm on week days is busy. What is the probability that if 6 randomly selected telephone numbers are called (a) not more than 3 (b) at least three of them would be busy?

Given: binomial with $n = 6$ , $p = \dfrac{1}{15}$ , $q = \dfrac{14}{15}$ .

P(X = x) = \binom{6}{x}p^x q^{6 - x}

x	P(X = x)
0	$(14/15)^6 = 0.66103$
1	$6(1/15)(14/15)^5 = 0.28330$
2	$15(1/15)^2(14/15)^4 = 0.05059$
3	$20(1/15)^3(14/15)^3 = 0.00482$

(a) Not more than 3 busy

P(X \le 3) = 0.66103 + 0.28330 + 0.05059 + 0.00482 = 0.99973

(b) At least 3 busy

P(X \ge 3) = 1 - P(X \le 2) = 1 - (0.66103 + 0.28330 + 0.05059) = 0.00508

Answer

(a) $\approx 0.9997$ ; (b) $\approx 0.0051$ .

(c)

7 marks

The lifeline of a certain kind of batteries has a mean life of 400 hours and the standard deviation as 45 hours. Assuming the distribution of lifetime to be normal. Find the percentage of batteries with lifetime (a) at least 490 hours, (b) between 385 and 490 hours. Also find the minimum life of the best 5% of batteries. (Use $P(0 < z < 2) = 0.4772$ , $P(0 < z < 0.33) = 0.1293$ and $P(0 < z < 1.65) = 0.45$ )

(a)

3 marks

Let A and B be two events with

P(A) = 3/8

P(B) = 7/8

and

P(A\text{ or }B) = 3/4

. Find

P(A|B)

and

P(B|A)

(b)

4 marks

Suppose that the time it takes to get service in a restaurant follows a gamma distribution with mean 8 minutes and variance 32 minutes. Suppose that you went to this restaurant at 6:30 pm. What is the probability that you will receive service before 6:36 pm?

(c)

7 marks

On an average one in 400 times items is defective follows poisson distribution. If the items are packed in boxes of 100, what is the probability, that any given box of items will contain (a) no defective (b) less than two defectives (c) one or more defectives (d) more than three defective.

Q.5

14 Marks

(a)

3 marks

A coin was tossed 960 times and returned heads 183 times. Test the hypothesis that the coin is unbiased. Use 5% level of significant. (Use

Z_{0.05} = 1.96

)

(b)

4 marks

Manager of a diet plan advertise that the mean weight loss for people on their plan is at least 45 pounds in 6 months. A sample of 28 people on this plan lose an average of 35 pounds with a standard deviation of 20 pounds. Test at 1% of significance level, if the claim is to high. (Use

t_{0.01} = -2.473

)

(c)

7 marks

If 57 out of 150 patients suffering from certain disease are cured by allopathy and 33 out of 100 patients with the same disease are cured by homeopathy, is there reason to believe that allopathy is better than homeopathy at 0.05 level of significance? (Use

Z_{0.05} = 1.645

)

(a)

3 marks

A random sample size of 20 from a normal population has mean 40 and standard deviation of 4. Test the hypothesis that the population mean is 45.(Use

t_{0.05,19} = 2.09

)

(b)

4 marks

In a coal mining it is observed that at least 90% of detonators will ignite the explosive when charged. From a random sample of 200 detonators it is found that 174 are upto standards. Test the level of significance at 5%. (Use $Z_{0.05} = -1.645$ )

(c)

7 marks

Two random samples are drawn from two populations and the following results are obtained:

Find the variances of the two samples and test whether the two populations have same variances. (Use $F_{0.05}(5,4) = 6.26$ )

Sample-1	21	24	25	26	27	----
Sample-2	22	27	28	30	31	36

Sample 1: 21, 24, 25, 26, 27 ( $n_1 = 5$ ). Sample 2: 22, 27, 28, 30, 31, 36 ( $n_2 = 6$ ).

Step 1 Means

\bar{x}_1 = \frac{123}{5} = 24.6, \qquad \bar{x}_2 = \frac{174}{6} = 29

Step 2 Sums of squared deviations

Sample 1: deviations −3.6, −0.6, 0.4, 1.4, 2.4, so $\sum(x - \bar{x}_1)^2 = 12.96 + 0.36 + 0.16 + 1.96 + 5.76 = 21.2$ .

Sample 2: deviations −7, −2, −1, 1, 2, 7, so $\sum(x - \bar{x}_2)^2 = 49 + 4 + 1 + 1 + 4 + 49 = 108$ .

Step 3 Unbiased sample variances

s_1^2 = \frac{21.2}{5 - 1} = 5.3, \qquad s_2^2 = \frac{108}{6 - 1} = 21.6

Step 4 F test

$H_0$ : $\sigma_1^2 = \sigma_2^2$ . $H_1$ : $\sigma_1^2 \ne \sigma_2^2$ . Put the larger variance in the numerator:

F = \frac{s_2^2}{s_1^2} = \frac{21.6}{5.3} = 4.08

with $(n_2 - 1, n_1 - 1) = (5, 4)$ degrees of freedom.

Step 5 Decision

$F = 4.08 < F_{0.05}(5, 4) = 6.26$ , so $H_0$ is accepted.

Answer

The sample variances are $5.3$ and $21.6$ (or 4.24 and 18 with divisor $n$ ), and since $F = 4.08 < 6.26$ , the two populations may be taken to have the same variance.

More Probability and Statistics papers

Questions repeat across sessions. Working through two or three papers shows you which topics GTU asks most.

Winter 202570 marks · 5 questions · Solved

Other Semester 3 subjects

BE Common

2130002

Advance Engineering Mathematics

2130003

Mechanics of Solids

3130005

Complex Variables and Partial Differential Equations

BE03000051

Professional Communication and Ethics

Preparing with friends?

Send this paper to your class group. Every question has its own link.

About this paper

Official Gujarat Technological University (GTU) examination paper and step-by-step solutions for Probability and Statistics (Summer 2026, B.E. · Common 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.

Download official PDF All Probability and Statistics Papers (2) →Report a typo or correction