GATE Breaker

If the matrix $A = \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & -1 & 0 & -1 \\ 0 & 0 & i & i \\ 0 & 0 & 0 & -i \end{bmatrix}$, then find the value of $A^4$ using Cayley-Hamilton theorem.  (GATE 1993)  Answer: $A^4 = I$.  Explanation:  Since the given matrix $A$ is an upper triangular matrix, the eigenvalues of $A$ are the main diagonal values.  Therefore, the eigenvalues of $A$ are $1$, $-1$, $i$, $-i$.  Hence, the characteristic equation of of $A$ is  $(\lambda - 1)(\lambda + 1)(\lambda - i)(\lambda + i) = 0$   $\Rightarrow  (\lambda^2 - 1)(\lambda^2 + 1) = 0$  $\Rightarrow  (\lambda^4 - 1) = 0$  Therefore, by Calley-Hamilton theorem, we have $A^4 - I = 0$, i.e., $A^4 = I$. 

Let $P(E)$ denote probability of an event $E$. Given $P(A) = 1$, $P(B) = \dfrac{1}{2}$. The values of $P(A|B)$ and $P(B|A)$ respectively are  (A) $\dfrac{1}{4}$, $\dfrac{1}{2}$  (B) $\dfrac{1}{2}$, $\dfrac{1}{4}$  (C) $\dfrac{1}{2}$, 1  (D) 1, $\dfrac{1}{2}$  (GATE 2003)  Answer: (D) 1, $\dfrac{1}{2}$  Explanation:  Since, $P(A) = 1$ and $P(B) = \dfrac{1}{2}$, hence, $P(A \cap B) = P(B) = \dfrac{1}{2}$.  Now, $P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{\dfrac{1}{2}}{\dfrac{1}{2}} = 1$,  and $P(B|A) = \dfrac{P(A \cap B)}{P(A)} = \dfrac{\dfrac{1}{2}}{1} = \dfrac{1}{2}$. 

Four fair coins are tossed simultaneously. The probability that at least one heads and at least one tails turn up is?  (A) $\dfrac{1}{16}$,     (B)  $\dfrac{1}{8}$,     (C)  $\dfrac{7}{8}$,     (D)  $\dfrac{15}{16}$.  (GATE 2002)  Answer: (C) $\dfrac{7}{8}$  Explanation:  Probability of getting head, $p = \dfrac{1}{2}$.  Probability of getting tail, $q = \dfrac{1}{2}$ .  Therefore, the probability of getting at least one head and one tail is  $= {}^{4}\textrm{C}_{1} p^{1} q^{3} + {}^{4}\textrm{C}_{2} p^{2} q^{2} + {}^{4}\textrm{C}_{3} p^{3} q^{1}$, from binomial distribution  $= {}^{4}\textrm{C}_{1}\left(\dfrac{1}{2}\right)^{1}\left(\dfrac{1}{2}\right)^{3} + {}^{4}\textrm{C}_{2}\left(\dfrac{1}{2}\right)^{2}\left(\dfrac{1}{2}\right)^{2} + {}^{4}\textrm{C}_{3}\left(\dfrac{1}{2}\right)^{3}\left(\dfrac{1}{2}\right)^{1}$  $= \dfrac{7}{8}$ 

Seven car accidents occurred in a week, what is the probability that they all occurred on the same day?  (A) $\dfrac{1}{7^7}$,     (B) $\dfrac{1}{7^6}$,     (C)  $\dfrac{1}{2^7}$,     (D) $\dfrac{7}{2^7}$  (GATE 2001)  Answer:  (B) $\dfrac{1}{7^6}$  Explanation:  The probability that all accidents occurred on Monday $= \dfrac{1}{7^7}$  The probability that all accidents occurred on Tuesday $= \dfrac{1}{7^7}$  $\cdots$      $\cdots$      $\cdots$      $\cdots$      $\cdots$      $\cdots$ $\cdots$    $\cdots$    $\cdots$    $\cdots$    $\cdots$    $\cdots$ The probability that all accidents occurred on Sunday $= \dfrac{1}{7^7}$  Therefore, the probability that they all occurred on the same day is  $7 \times \dfrac{1}{7^7} = \dfrac{1}{7^6}$. 

In a manufacturing plant, the probability of making a defective bolt is 0.1. The mean and standard deviation of defective bolts in a total of 900 bolts are respectively  (A) 90 and 9,     (B) 9 and 90,     (C) 81 and 9,     (D) 9 and 81.  (GATE 2000)  Answer:  (A) 90 and 9  Explanation:  Let $X$ be a random variable that denotes the number of defective bolts.  Here, $X$ follows a binomial distribution with the parameters $p = 0.1$ and $n = 900$.  Therefore, mean $= E(X) = n \times p = 0.1 \times 900 = 90$, and  standard deviation $= \sqrt{n \times p \times (1 -p)} = \sqrt{900 \times 0.1 \times (1 - 0.1)} = 9$. 

$E_1$ and $E_2$ are events in a probability space satisfying the following constraints: $P(E_1) = P(E_2)$, $P(E_1 \cup E_2) = 1$.  If $E_1$ and $E_2$ are independent then $P(E_1) = $  (A) 0,     (B) $\dfrac{1}{4}$,     (C) $\dfrac{1}{2}$,     (D) 1  (GATE 2000)  Answer: (D) 1   Explanation:  Given, $P(E_1) = P(E_2)$, $P(E_1 \cup E_2) = 1$, and $E_1$ and $E_2$ are independent events.  Hence, $P(E_1 \cap E_2) = P(E_1) \times P(E_2)$.  Now, $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1) \times P(E_2)$  $\Rightarrow  1 = P(E_2) + P(E_2) - P(E_2) \times P(E_2)$  $\Rightarrow  \{P(E_2)\}^2 - 2 \times P(E_2) + 1 = 0$  $\Rightarrow  (P(E_2) - 1)^2 = 0$  $\Rightarrow  P(E_2) = 1$. 

The probability that two friends share the same birth-month is (A) $\dfrac{1}{6}$,      (B) $\dfrac{1}{12}$,     (C) $\dfrac{1}{144}$,     (D) $\dfrac{1}{24}$  (GATE 1998)  Answer:  (B) $\dfrac{1}{12}$  Explanation:  Sample space, $S = $  $\{(Jan, Jan), (Jan, Feb), \ldots, (Jan, Dec), $ $\cdots$     $\cdots$      $\cdots$    $\cdots$      $\cdots$    $\cdots$      $\cdots, $ $(Dec, Jan), (Dec, Feb), \ldots, (Dec, Dec)\}$.  No. of sample points, $n(S) = 12 \times 12 = 144$.  Favorable events, $E = \{(Jan, Jan), (Feb, Feb), \ldots, (Dec, Dec)\}$.  No. of event points, $n(E) = 12$.  Therefore, the required probability that the two friends share the same birth-month is  $= \dfrac{n(E)}{n(S)} = \dfrac{12}{144} = \dfrac{1}{12}$. 

A die is rolled three times. The probability that exactly one odd number turns up among the three outcomes is  (A) $\dfrac{1}{6}$, (B) $\dfrac{3}{8}$, (C) $\dfrac{10}{8}$, (D) $\dfrac{1}{2}$.  (GATE 1998)  Answer: (B) $\dfrac{3}{8}$  Explanation:  Probability of getting an odd number when a die is rolled = $\dfrac{3}{6} = \dfrac{1}{2}$.  Hence, the probability of 'success', $p = \dfrac{1}{2}$.   No. of trials, $n = 3$.  Therefore, the required probability is  $= ^3C_1 \times \dfrac{1}{2} \times \left(\dfrac{1}{2}\right)^{3-1}$, from the Binomial distribution with parameters $n = 3$ and $p = \dfrac{1}{2}$  $= \dfrac{3}{8}$. 

The probability that it will rain today is 0.5. The probability that it will rain tomorrow is 0.6. The probability that it will rain either today or tomorrow is 0.7. What is the probability that it will rain today and tomorrow?  (A) 0.3,      (B) 0.25,     (C) 0.35,     (D) 0.4  (GATE 1997)  Answer:  (D) 0.4  Explanation:  Let $E_1$ be the event of raining today, $E_2$ be the event of raining tomorrow.  According to the given problem, $P(E_1) = 0.5$, $P(E_2) = 0.6$, and $P(E_1 \cup E_2) = 0.7$.  Therefore, the required probability is  $= P(E_1 \cap E_2) $ $= P(E_1) + P(E_2) - P(E_1 \cup E_2)$ $= 0.5 + 0.6 - 0.7 = 0.4$.  

The probability that a number selected at random between100 and 999 (both inclusive) will not contain the digit 7 is  (A) $\frac{16}{9}$,  (B) $\left(\frac{9}{10}\right)^3$,  (C) $\frac{27}{75}$,  (D) $\frac{18}{25}$.  (GATE 1995) Answer: (D) $\frac{18}{25}$.  Explanation:  All the numbers between 100 and 999 (both inclusive) have three digits.  No. of ways of choosing the left-most digit from 1 to 9 excluding 7 = 8.  No. of ways of choosing the middle digit from 0 to 9 excluding 7 = 9.  No. of ways of choosing the right-most digit from 0 to 9 excluding 7 = 9.  So, the total number of ways of selecting a number between 100 and 999 (both inclusive) that will not contain the digit 7 is = $8 \times 9 \times 9$.  Again, the total number of ways of selecting any number between 100 and 999 (both inclusive) is = $^{900}C_1 = 900$.  Therefore, the required probability is = $\frac{8 \times 9 \times 9}{900} = \frac{18}{25}...