GATE Breaker

A two-faced fair coin has its faces designated as head (H) and tail (T). This coin is tossed three times in succession to record the following outcomes: H, H, H. If the coin is tossed one more time, find the probability (up to one decimal place) of obtaining H again, given the previous realizations of H, H, and H?  (GATE 2017)  Answer: 0.5  Explanation:  Here, the probability of getting one H is $= \dfrac{1}{2}$.  The probability of getting (H, H, H) is  $= \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}$  $= \dfrac{1}{16}$  The probability of getting (H, H, H, H) is  $= \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}$  $= \dfrac{1}{32}$  Therefore, the required probability is  $= \dfrac{\dfrac{1}{16}}{\dfrac{1}{8}}$  $= \dfrac{1}{2}$  $= 0.5$  Previous Post Next Post

If a random variable $X$ has a Poisson distribution with mean 5, then find the expectation $E[(X + 2)^2]$.  (GATE 2017)  Answer: 54  Explanation:  Here, $X$ is a random variable that follows a Poisson distribution with mean $E(X) = 5$.  Therefore, its variance $Var(X) = 5$.  Now, $Var(X) = E(X^2) - \big\{E(X)\big\}^2$   $\Rightarrow E(X^2) = 5 + 5^2$  $\Rightarrow E(X^2) = 30$  Therefore, $E[(X + 2)^2] $  $= E(X^2 + 4X + 4)$  $= E(X^2) + E(4X) + E(4)$  $= E(X^2) + 4E(X) + 4$  $= 30 + 4 \times 5 + 4$  $= 54$  Previous Post Next Post

For the function $f(x) = a + bx$, $0 \leq x \leq 1$, to be valid probability density function, which one of the following statements is correct?  (A) $a = 1$, $b = 4$  (B)  $a = 0.5$, $b = 1$  (C)  $a = 0$, $b = 1$  (D)  $a = 1$, $b = -1$  (GATE 2017)  Answer: ( B)  $a = 0.5$, $b = 1$  Explanation:  For the function $f(x)$ to be a valid probability density function (PDF), the following must hold:  $\int\limits_{0}^{1} f(x) = 1$  $\Rightarrow \int\limits_{0}^{1} (a + bx) = 1$  $\Rightarrow \Big[ax + \dfrac{bx^2}{2}\Big]_{0}^{1} = 1$  $\Rightarrow a + \dfrac{b}{2} = 1$  Only option (B)  $a = 0.5$, $b = 1$ satisfies the above equation.  Previous Post Next Post

In a game, two players $X$ and $Y$ toss a coin alternately. Whosoever gets a 'head' first wins the game and the game is terminated. Assuming that player $X$ starts the game. The probability of player $X$ winning the game is  (A) $\dfrac{1}{3}$  (B) $\dfrac{1}{2}$  (C) $\dfrac{2}{3}$  (D) $\dfrac{2}{4}$  Answer: (C) $\dfrac{2}{3}$  Explanation:  Here, the probability of getting a 'head' is, $p = \dfrac{1}{2}$, and  the probability of getting a 'tail' is, $q = \dfrac{1}{2}$.  Therefore, the required probability is  $p + q^2p + q^4p + \cdots$  $= p(1 + q^2 + q^4 + \cdots)$  $= p \times \dfrac{1}{1 - q^2}$  $= \dfrac{1}{2} \times \dfrac{1}{1 - \dfrac{1}{4}}$  $= \dfrac{2}{3}$ 

For a random variable $X$ ($-\infty < X < \infty$) following normal distribution, the mean is $\mu = 100$. If the $P(X \geq 110)$, the find $P(90 \leq X \leq 110)$.  (A) $1 - 2\alpha$  (B) $1 - \alpha$  (C) $1 - \dfrac{\alpha}{2}$  (D) $2\alpha$  Answer: (A) $1 - 2\alpha$  Explanation:  Here, $X$ is a random variable that follows a normal distribution with mean $\mu = 100$.  Given, $P(X \geq 110) = \alpha$  Hence, $P(X \leq 90) = \alpha$  Therefore, $P(90 \leq X \leq 110) = 1 - 2\alpha$. 

A coin is tossed 4 times. What is the probability of getting heads exactly 3 times?  (A) $\dfrac{1}{4}$  (B) $\dfrac{3}{8}$  (C) $\dfrac{1}{2}$  (D) $\dfrac{3}{4}$  Answer: (A) $\dfrac{1}{4}$  Explanation:  Let $X$ be a random variable that denotes the number of heads.  Here, $X$ follows the binomial distribution with the parameters $p = \dfrac{1}{2}$ and $n = 4$.  Hence, $P(X = r) = ^nC_r \times p^r \times (1 - p)^{n - r}$, where $r = 1, \ldots, n$.  Therefore, the required probability is  $P(X = 3) = ^4C_3 \times (\dfrac{1}{2})^3 \times (1 - \dfrac{1}{2})^{4 - 3}$  $= 4 \times \dfrac{1}{8} \times \dfrac{1}{2}$  $= \dfrac{1}{4}$ 

A random variable is uniformly distributed over the interval 2 to 10. Its variance will be  (A) $\dfrac{16}{3}$  (B) 6  (C) $\dfrac{256}{9}$  (D) 36  Answer: (A) $\dfrac{16}{3}$  Explanation:  The expected value (i.e. the mean) of a uniform random variable $X$ is:  $E(X) = \dfrac{b + a}{2}$.  The variance of a uniform random variable $X$ is:  $Var(X) = \dfrac{(b - a)^2}{12}$.  Here, $a$ is the minimum value in the distribution, and $b$ is the maximum value.  For the given problem, $a = 2$ and $b = 12$.  Therefore, $Var(X) = \dfrac{(10 - 2)^2}{12} = \dfrac{16}{3}$. 

$X$ 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  $P(E) = \dfrac{n(E)}{n(S)} = \dfrac{2}{4} = \dfrac{1}{2}$