GATE Breaker

$X$ is a uniformly distributed random variable that takes values between zero and one. The value of $E( X^3)$ will be  (A) 0  (B) $\dfrac{1}{8}$  (C) $\dfrac{1}{4}$  (D) $\dfrac{1}{2}$  Answer: (C) $\dfrac{1}{4}$  Explanation:  Here, $X$ is a uniformly distributed random variable that takes values between 0 and 1.  Therefore, its p.d.f. is $f(x) = \dfrac{1}{1 - 0} = 1, \mbox{for } 0 < x < 1.$  Hence,  $E(X^3) = \int\limits_{0}^{1}x^3f(x)dx = \int\limits_{0}^{1}x^3dx = \dfrac{1}{4}$  Previous Post Next Post

An examination consists of two papers, Paper-1 and Paper-2. The probability of failing in Paper-1 is 0.3 and that in Paper-2 is 0.2. Given that a student has failed in Paper-2, the probability of failing in Paper-1 is 0.6. The probability of a student failing in both the papers is  (A) 0.5  (B) 0.18  (C) 0.12  (D) 0.06  Answer: (C) 0.12  Explanation:  Let $E_1$ and $E_2$ be the two events of failing in Paper-1 and Paper-2 respectively.  Given, $P(E_1) = 0.3$, $P(E_2) = 0.2$ and $P(E_1 | E_2) = 0.6$  Now, $P(E_1 | E_2) = \dfrac{P(E_1 \cap E_2)}{P(E_2)}$  $\Rightarrow 0.6 = \dfrac{P(E_1 \cap E_2)}{0.2}$  $\Rightarrow P(E_1 \cap E_2) = 0.12$, which is the required probability. 

If X is a continuous random variable whose probability density function is given by \[f(x) = \left\{ \begin{array}{ll} k(5x - 2x^2) & \mbox{if } 0 \leq x \leq 2 \\ 0 & \mbox{otherwise,} \end{array} \right.\] then $P(X > 1)$ is  (A) $\dfrac{3}{14}$  (B) $\dfrac{4}{5}$  (C) $\dfrac{14}{7}$  (D) $\dfrac{17}{28}$  Answer: (D) $\dfrac{17}{28}$  Explanation:  $\int\limits_{-\infty}^{+\infty}f(x)dx = 1$  $\Rightarrow \int\limits_{0}^{2}k(5x-2x^2)dx = 1$  $\Rightarrow k = \dfrac{3}{14}$  Now, $P(X > 1)$  $= \int\limits_{1}^{\infty}f(x)dx$  $= \int\limits_{1}^{2} \dfrac{3}{14}(5x - 2x^2)dx$  $= \dfrac{17}{28}$

The random variable $X$ taken on the values 1, 2 or 3 with probabilities $\dfrac{2 + 5p}{5}$, $\dfrac{1 + 3p}{5}$, $\dfrac{1.5 + 2p}{5}$ respectively. The values of $p$ and $E(X)$ are respectively  (A) 0.05, 1.87  (B) 1.90, 5.87  (C) 0.05, 1.10  (D) 0.25, 1.40  (GATE 2007)  Answer: (A) 0.05, 1.87  Explaination:  $\dfrac{2 + 5p}{5} + \dfrac{1 + 3p}{5} + \dfrac{1.5 + 2p}{5} = 1$  $\Rightarrow  4.5 + 10p = 5$  $\Rightarrow p = 0.05$  Now, $E(X)$  $= \sum x \times P(X = x)$  $= 1 \times \left(\dfrac{2 + 5p}{5}\right) + 2 \times \left(\dfrac{1 + 3p}{5}\right) + 3 \times \left(\dfrac{1.5 + 2p}{5}\right)$  $= \dfrac{8.5 + 17p}{5}$  $= \dfrac{8.5 + 0.85}{5}$  $= \dfrac{9.35}{5}$  $= 1.87$ 

Two cards are drawn at random in succession with replacement from a deck of 52 well-shuffled cards. Probability of getting both ‘Aces’ is  (A) $\dfrac{1}{169}$  (B) $\dfrac{2}{169}$  (C) $\dfrac{1}{13}$  (D) $\dfrac{2}{13}$  (GATE 2007)  Answer: (A) $\dfrac{1}{169}$  Explanation:  Let $E_1$ and $E_2$ be the events of getting 'ace' in the first and the second drwans respectively.  Here, two cards are drawn with replacement.  Hence, the events $E_1$ and $E_2$ are independent.  Now, $P(E_1) = \dfrac{^4C_1}{^{52}C_1} = \dfrac{4}{52}$  and $P(E_2) = \dfrac{^4C_1}{^{52}C_1} = \dfrac{4}{52}$  Therefore, the required probability is  $= P(E_1) \times P(E_2)$  $= \dfrac{4}{52} \times \dfrac{4}{52}$  $= \dfrac{1}{169}$ 

Assume that the duration in minutes of a telephone conversation follows the exponential distribution $f(x) = \frac{1}{5}e^{-\frac{x}{5}}, x \geq 0$. The probability that the conversation will exceed five minutes is  (A) $\dfrac{1}{e}$  (B) $1 - \dfrac{1}{e}$  (C) $\dfrac{1}{e^2}$  (D) $1 - \dfrac{1}{e^2}$  (GATE 2007)  Answer: (A) $\dfrac{1}{e}$  Explanation:  The required probability is  $P(5 < X <\infty)$  $= \int\limits_{5}^{\infty}f(x)dx$  $= \int\limits_{5}^{\infty}\frac{1}{5}e^{-\frac{x}{5}}dx$  $= \dfrac{1}{e}$ 

The life of a bulb (in hours) is random variable with an exponential distribution $f (t) =\alpha e^{- \alpha t}, 0 \leq t \leq \infty.$ The probability that its value lies between 100 and 200 hours is  (A) $e^{-100\alpha} - e^{-200\alpha}$  (B) $e^{-100} - e^{-200}$  (C) $e^{-100\alpha} + e^{-200\alpha}$  (D) $e^{-200\alpha} - e^{-100\alpha}$  (GATE 2005)  Answer: (A) $e^{-100\alpha} - e^{-200\alpha}$  Explanation:  The required probability is  $P(100 \leq X \leq 200)$   $= \int\limits_{100}^{200}f(t)dt$   $= \int\limits_{100}^{200}\alpha e^{-\alpha t}dt$  $= e^{-100\alpha} - e^{-200\alpha}$ 

A fair dice is rolled twice. The probability that an odd number will follow an even number is  (A) $\dfrac{1}{2}$  (B) $\dfrac{1}{6}$  (C) $\dfrac{1}{3}$  (D) $\dfrac{1}{4}$  (GATE 2005)  Answer: (D) $\dfrac{1}{4}$  Explanation:  Here, the sample space is  $S = \{(1, 1), \cdots, (1, 6), (2, 1), \cdots, (2, 6), (3, 1), \cdots, \\(3, 6), (4, 1), \cdots, (4, 6), (5, 1), \cdots, (5, 6), (6, 1), \cdots, (6, 6) \}.$  So, the number of sample points, $n(S) = 6 \times 6 = 36$.  Again, favorable event space,  $E = \{(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), \\~~~~~~~~~~(4, 5), (6, 1), (6, 3), (6, 5)\}.$   So, the number of event points, $n(E) = 9$.  Therefore, the required probability is  $= P(E) = \dfrac{n(E)}{n(S)} = \dfrac{9}{36} = \dfrac{1}{4}$. 

The probability that there are 53 Sundays in a randomly chosen leap year is  (A) $\dfrac{1}{7}$  (B) $\dfrac{1}{14}$  (C) $\dfrac{1}{28}$  (D) $\dfrac{2}{7}$  (GATE 2005)  Answer: (D) $\dfrac{2}{7}$  Explanation:  A leap year has 366 days.  Since, $\dfrac{364}{7} = 52$, there will 52 Sunday in 364 days. The remaining 2 days can be any of the following:  $\{(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), \\(Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), \\(Saturday, Sunday)\}.$  Therefore, the required probability is $\dfrac{2}{7}$. 

A single die is thrown two times. What is the probability that the sum is neither 8 nor 9?  (A) $\dfrac{1}{9}$  (B) $\dfrac{5}{36}$  (C) $\dfrac{1}{4}$  (D) $\dfrac{3}{4}$  (GATE 2005)  Answer:  (D) $\dfrac{3}{4}$  Explanation:  Here, the sample space,  $S = \{(1, 1), \cdots, (1, 6), (2, 1), \cdots, (2, 6), (3, 1), \cdots, \\(3, 6), (4, 1), \cdots, (4, 6), (5, 1), \cdots, (5, 6), (6, 1), \cdots, (6, 6) \}.$  So, the number of sample points, $n(S) = 6 \times 6 = 36$.  Again, favorable event space,  $E = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2), \\~~~~~~~~~~(3, 6), (4, 5), (5, 4), (6, 3)\}.$   So, the number of event points, $n(E) = 9$.  Therefore, the probability that the sum is either 8 or 9 is  $= P(E) = \dfrac{n(E)}{n(S)} = \dfrac{9}{36} = \dfrac{1}{4}$.  Therefore, the required probability is  $= 1 - P(E) = 1 - \dfrac{1}{4} = \dfrac{3}{4}$. 

A lot had 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly two of the chosen items are defective is  (A) 0.0036  (B) 0.1937  (C) 0.2234  (D) 0.3874  (GATE 2005)  Answer: (B) 0.1937  Explanation:  Let $X$ be a random variable that denotes the number of defective items chosen.   Here, the number of 'trials', $n = 10$ and  the probability of 'success', $p = \dfrac{10}{100} = 0.10$  Therefore, the required probability is  $= P(X = 2)$ $ = ^{10}C_2 \times (0.10)^2 \times (1 - 0.10)^{10 - 2}$  $= 0.1937$ 

Two dice are thrown simultaneously. The probability that the sum of numbers on both exceeds 8 is  (A) $\dfrac{4}{36}$  (B) $\dfrac{7}{36}$  (C) $\dfrac{9}{36}$  (D) $\dfrac{10}{36}$  (GATE 2005)  Answer: (D) $\dfrac{10}{36}$  Explanation:  Here, the sample space,  $S = \{(1, 1), \cdots, (1, 6), (2, 1), \cdots, (2, 6), (3, 1), \cdots, \\(3, 6), (4, 1), \cdots, (4, 6), (5, 1), \cdots, (5, 6), (6, 1), \cdots, (6, 6) \}.$  So, the number of sample points, $n(S) = 6 \times 6 = 36$.  Again, favorable event space,  $E = \{(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), \\(5, 6), (6, 3), (6, 4), (6, 5), (6, 6)\}.$   So, the number of event points, $n(E) = 10$.  Therefore, the required probability is  $= P(E) = \dfrac{n(E)}{n(S)} = \dfrac{10}{36}$. 

A fair coin is tossed three times in succession. If the first toss produces a head, then the probability of getting exactly two heads in three tosses is  (A) $\dfrac{1}{8}$  (B)  $\dfrac{1}{2}$  (C)  $\dfrac{3}{8}$  (D)  $\dfrac{3}{4}$  (GATE 2005)  Answer:  (B)  $\dfrac{1}{2}$  Explanation:  Here, the first produces a head.  Hence, the sample space, $S = \{HHH, HHT, HTH, HTT\}$.  So, the number of sample points, $n(S) = 4$.  Let, event $E = \{HHT, HTH\}$.  So, the number of event points, $n(E) = 2$.  Therefore, the required probability is  $P(E) = \dfrac{n(E)}{n(S)} = \dfrac{2}{4} = \dfrac{1}{2}$ 

A bag contains 10 blue marbles, 20 black marbles and 30 red marbles. A marble is drawn from the bag, its color recorded and it is put back in the bag. This process is repeated 3 times. The probability that no 2 of the marbles drawn have the same color is (A) $\dfrac{1}{36}$  (B) $\dfrac{1}{6}$  (C) $\dfrac{1}{4}$  (D) $\dfrac{1}{3}$  (GATE 2005)  Answer: (B) $\dfrac{1}{6}$  Explanation:  Let $E_1$, $E_2$ and $E_3$ be the events of drawing a blue marble, a black marble and a red marble, respectively.  Here, three marbles are drawn from the bag one-by-one with replacements.  Therefore, $P(E_1) = \dfrac{10}{60}$, $P(E_2) = \dfrac{20}{60}$, $P(E_3) = \dfrac{30}{60}$.  Therefore, the required probability is  $P(E_1 \cap E_2 \cap E_3) + P(E_1 \cap E_3 \cap E_2) + P(E_2 \cap E_1 \cap E_3) \\+ P(E_2 \cap E_3 \cap E_1) + P(E_3 \cap E_1 \cap E_2) + P(E_3 \cap E_2 \cap E_1)$  $= 6 \times \left[P(E_1) \times P(E_2) \times P(E_3)\right]$, s...

From a pack of regular playing cards, two cards are drawn at random. What is the probability that both cards will be kings if the card is NOT replaced?  (A) $\dfrac{1}{26}$  (B) $\dfrac{1}{52}$  (C) $\dfrac{1}{169}$  (D) $\dfrac{1}{221}$  Answer: (D) $\dfrac{1}{221}$  Explanation:  Let $E_1$ and $E_2$ be the events of getting the first king and the second king respectively.  Here, two cards are drawn without replacement. In this case, the probabilities for the second pick are affected by the result of the first pick.   Now, the probability of getting the first king is  $= P(E_1) = \dfrac{4}{52}$  and the probability of getting the second king is  $P(E_2) = \dfrac{3}{51}$  Therefore, the required probability is  $= P(E_1) \times P(E_2)$  $= \dfrac{4}{52} \times \dfrac{3}{51}$  $= \dfrac{1}{221}$ 

A hydraulic structure has four gates which operate independently. The probability of failure of each gate is 0.2. given that gate 1 has failed, the probability that both gates 2 and 3 will fail is  (A) 0.240  (B) 0.200  (C) 0.040  (D) 0.008  (GATE 2004)  Answer: (C) 0.040  Explanation:  Let $G_1$, $G_2$, $G_3$ and $G_4$ be the events of failure of the four gate respectively.  Here, $G_1$, $G_2$, $G_3$ and $G_4$ are independent and $P(G_1) = 0.2$, $P(G_2) = 0.2$, $P(G_3) = 0.2$ and $P(G_4) = 0.2$.  Therefore, the probability  that both gates 2 and 3 will fail is  $= P(G_1 \cap G_3)$  $= P(G_1) \times P(G_3)$, since $G_2$ and $G_3$ are independent events  $= 0.2 \times 0.2$  $= 0.04$ 

In a class of 200 students, 125 students have taken Programming Language course, 85 students have taken Data Structures course, 65 students have taken Computer Organization course, 50 students have taken both Programming Language and Data Structures, 35 students have taken both Data Structures and Computer Organization 30 students have taken both Data Structures and Computer Organization, 15 students have taken all the three courses. How many students have not taken any of the three courses? (A) 15  (B) 20  (C) 25  (D) 35  (GATE 2004)  Answer: (C) 25  Explanation:  Let $P$, $D$ and $C$ be the sets of students taking programming language, data structure and computer organization courses, respectively.  Here, $n(P) = 125$, $n(D) = 85$, $n(C) = 65$, $n(P \cap D) = 50$, $n(P \cap C) = 35$, $n(D \cap C) = 30$ and $n(P \cap D \cap C) = 15$  [$n(X)$ denotes the number of elements present in the set $X$.]  Now, $n(P \cup D \cup C)$  $= n(P)...

Αn exam paper has 150 multiple choice questions of 1 mark each, with each question having four choices. Each incorrect answer fetches -0.25 marks. Suppose 1000 students choose all their answers randomly with uniform probability. Find the sum total of the expected marks obtained by all the students.  (A) 0  (B) 2550  (C) 7525  (D) 9375  (GATE 2004)  Answer: (D) 9375  Explanation:  Let $X$ be a random variable that denotes the marks obtained for each question.  Here, the probability distribution of $X$ is as follows:   \[P(X = x) = \left\{ \begin{array}{ll} 0.25 & \mbox{if } x = 1 \\ 0.75 & \mbox{if } x = -0.25 \end{array} \right.\] Now, the expected marks for one question $= E(X) = 1 \times 0.25 + (-0.25) \times 0.75$  $= 0.0625$  Therefore, the expected marks for 150 questions   $= 150 \times E(X)$  $= 150 \times 0.0625$  $= 9.375$  Therefore, the sum total of the expected marks obtained by...

If a fair coin is tossed 4 times, what is the probability that two heads and two tails will result?  (A) $\dfrac{3}{8}$  (B) $\dfrac{1}{2}$  (C) $\dfrac{5}{8}$  (D) $\dfrac{3}{4}$  (GATE 2004)  Answer: (A) $\dfrac{3}{8}$  Explanation:  Let $X$ be a random variable that denotes the number of heads.  Here, the probability of 'success', $p = \dfrac{1}{2}$, and number of trials, $n = 4$.  Therefore, by the binomial distribution, we get the required probability  $= P(X = 2)$  $= {}^{4}\textrm{C}_{2}\left(\dfrac{1}{2}\right)^{2}\left(\dfrac{1}{2}\right)^{2}$  $= \dfrac{3}{8}$ 

How many distinct binary search trees can be created out of 4 distinct keys?  (GATE 2005)  (A) 5  (B) 14   (C) 24   (D) 42  Answer: (B) 14  Explanation:  Here, number of keys, $k = 4$.  Therefore, the number of distinct binary search trees that can be formed out of 4 distinct keys is  $= \dfrac{^{2k}C_k}{k + 1}$  $ = \dfrac{^{8}C_4}{5}$  $= 14$