For each of the following pairs of matrices
Question: For each of the following pairs of matrices $A$ and $B$, verify that $(A B)^{\prime}=\left(B^{\prime} A^{\prime}\right)$ : $A=\left[\begin{array}{ll}3 -1 \\ 2 -2\end{array}\right]$ and $B=\left[\begin{array}{cc}1 -3 \\ 2 -1\end{array}\right]$ Solution:...
Read More →For each of the following pairs of matrices
Question: For each of the following pairs of matrices $A$ and $B$, verify that $(A B)^{\prime}=\left(B^{\prime} A^{\prime}\right)$ : $A=\left[\begin{array}{ll}1 3 \\ 2 4\end{array}\right]$ and $B=\left[\begin{array}{ll}1 4 \\ 2 5\end{array}\right]$ Solution:...
Read More →Solve this
Question: Express the matrix $A=\left[\begin{array}{lll}3 2 5 \\ 4 1 3 \\ 0 6 7\end{array}\right]$ as sum of two matrices such that one is symmetric and the other is skew-symmetric. Solution: $\Rightarrow \frac{1}{2}\left[\begin{array}{ccc}6 6 5 \\ 6 2 9 \\ 5 9 14\end{array}\right]$...
Read More →Ten coins are tossed.
Question: Ten coins are tossed. What is the probability of getting at least 8 heads? Solution: Here we have, n = 10, p = and q = 1 = P(X 8) = P(x = 8) + P(x = 9) + P(x = 10) Therefore, the required probability is 7/128....
Read More →A die is thrown 5 times.
Question: A die is thrown 5 times. Find the probability that an odd number will come up exactly three times. Solution: Here, p = 1/6 + 1/6 + 1/6 = ⇒ q = 1 = and n = 5 Now, P(x = r) =nCrprqn r=5C3(1/2)3(1/2)5-3 = [5!/(3!2!)].(1/2)3(1/2)2= 10.1/8.1/4 = 5/16 Therefore, the required probability is 5/16....
Read More →Express the matrix A as the sum of a symmetric and a skew-symmetric matrix,
Question: Express the matrix A as the sum of a symmetric and a skew-symmetric matrix, where $A=\left[\begin{array}{ccc}3 -1 0 \\ 2 0 3 \\ 1 -1 2\end{array}\right]$. Solution: $\Rightarrow \frac{1}{2}\left[\begin{array}{ccc}0 -3 -1 \\ 3 0 4 \\ 1 -4 0\end{array}\right]$...
Read More →Four cards are successively drawn without replacement from
Question: Four cards are successively drawn without replacement from a deck of 52 playing cards. What is the probability that all the four cards are kings? Solution: Let E1, E2, E3and E4be the events that first, second, third and fourth card is King respectively. Therefore, the required probability is 1/27075....
Read More →A box has 5 blue and 4 red balls.
Question: A box has 5 blue and 4 red balls. One ball is drawn at random and not replaced. Its colour is also not noted. Then another ball is drawn at random. What is the probability of second ball being blue? Solution: Given that the box has 5 blue and 4 red balls. Lets consider E1be the event that first ball drawn is blue and E2be the event that first ball drawn is red. And, E is the event that second ball drawn is blue. Now, the probability of E is: Therefore, the required probability is 5/9....
Read More →Bag I contains 3 black and 2 white balls,
Question: Bag I contains 3 black and 2 white balls, Bag II contains 2 black and 4 white balls. A bag and a ball is selected at random. Determine the probability of selecting a black ball. Solution: Given that: Bag 1 has 3B, 2W balls and Bag 2 has 2B, 4W balls. Let E1= The event that bag 1 is selected E2= The event that bag 2 is selected And, E = The event that a black ball is selected Now, the probabilities are: Therefore, the required probability is 7/15....
Read More →A bag contains 4 white and 5 black balls.
Question: A bag contains 4 white and 5 black balls. Another bag contains 9 white and 7 black balls. A ball is transferred from the first bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white. Solution: Lets take W1and W2to be two bags containing (4W, 5B) and (9W, 7B) balls respectively. Let E1be the event that the transferred ball from the bag W1to W2is white and E2the event that the transferred ball is black. And, E be the ev...
Read More →Suppose 10,000 tickets are sold in a lottery each for Re 1.
Question: Suppose 10,000 tickets are sold in a lottery each for Re 1. First prize is of Rs 3000 and the second prize is of Rs. 2000. There are three third prizes of Rs. 500 each. If you buy one ticket, what is your expectation. Solution: Lets take X to be the random variable where X = 0, 500, 2000 and 3000...
Read More →Solve this following
Question: Express the matrix $A=\left[\begin{array}{ccc}-1 5 1 \\ 2 3 4 \\ 7 0 9\end{array}\right]$ as the sum of a symmetric and a skew-symmetric matrix. Solution: $A=\left[\begin{array}{ccc}-1 5 1 \\ 2 3 4 \\ 7 0 9\end{array}\right]$...
Read More →Three dice are thrown at the same time.
Question: Three dice are thrown at the same time. Find the probability of getting three twos, if it is known that the sum of the numbers on the dice was six. Solution: Given that the dice is thrown three times So, the sample space n(S) = 63= 216 Let E1be the event when the sum of number on the dice was 6 and E2be the event when three 2s occur. E1= {(1, 1, 4), (1, 2, 3), (1, 4, 1), (2, 1, 3), (2, 2, 2), (2, 3, 1), (3, 1, 2), (3, 2, 1), (4, 1, 1)} n(E1) =10 and n(E2) = 1 [Since, E2= (2, 2, 2)] Thu...
Read More →In a dice game,
Question: In a dice game, a player pays a stake of Re1 for each throw of a die. She receives Rs 5 if the die shows a 3, Rs 2 if the die shows a 1 or 6, and nothing otherwise. What is the players expected profit per throw over a long series of throws? Solution: Lets take X to be the random variable of profit per throw. As, she loses Rs 1 for giving any od 2, 4, 5. So, P(X = -1) = 1/6 + 1/6 + 1/6 = 3/6 = P(X = 1) = 1/6 + 1/6 = 2/6 = 1/3 [Since, die showing 1 or 6] P(X = 4) = 1/6 [Since, die shows ...
Read More →Express the matrix
Question: Express the matrix $A=\left[\begin{array}{rr}3 -4 \\ 1 -1\end{array}\right]$ as the sum of a symmetric matrix and a skew-symmetric matrix. Solution:...
Read More →If X is the number of tails in three tosses of a coin,
Question: If X is the number of tails in three tosses of a coin, determine the standard deviation of X. Solution: Given, X = 0, 1, 2, 3 P(X = r) =nCrprqn-r Where n = 3, p = , q = and r = 0, 1, 2, 3 P(X = 0) = x x = 1/8 P(X = 1) = 3 x x x = 3/8 P(X = 2) = 3 x x x = 3/8 P(X = 3) = x x = 3/8 The probability distribution table is: Now, E(X) = 0 + 1 x 3/8 + 2 x 3/8 + 3 x 1/8 = 3/8 + 6/8 + 3/8 = 12/8 = 3/2 E(X2) = 0 + 1 x 3/8 + 4 x 3/8 + 9 x 1/8 = 3/8 + 12/8 + 9/8 = 24/8 = 3 W.k.t, Var(X) = E(X2) [E(X...
Read More →Prove that (i) P(A) = P(AÇ B) + P(AÇ B )
Question: Prove that (i) P(A) = P(AÇ B) + P(AÇB) (ii) P(AÈ B) = P(AÇ B) + P(AÇB) + P(AÇ B) Solution:...
Read More →Express the matrix
Question: Express the matrix $A=\left[\begin{array}{cc}2 3 \\ -1 4\end{array}\right]$ as the sum of a symmetric matrix and a skew-symmetric matrix. Solution: $\Rightarrow\left[\begin{array}{cc}2 3 \\ -1 4\end{array}\right]$...
Read More →A discrete random variable X has the probability distribution
Question: A discrete random variable X has the probability distribution given as below: (i) Find the value ofk (ii) Determine the mean of the distribution. Solution: For a probability distribution, we know that if Pi 0...
Read More →Solve this following
Question: Show that the matrix $A=\left[\begin{array}{ccc}0 \mathrm{a} \mathrm{b} \\ -\mathrm{a} 0 \mathrm{c} \\ -\mathrm{b} -\mathrm{c} 0\end{array}\right]$ is skew-symmetric. HINT: Show that $A^{\prime}=-A$. Solution:...
Read More →Let E1 and E2 be two independent events such that p(E1) = p1 and P(E2) = p2.
Question: Let E1and E2be two independent events such thatp(E1) =p1and P(E2) = p2. Describe in words of the events whose probabilities are: (i)p1p2 (ii) (1p1)p2 (iii) 1 (1 p1)(1 p2) (iv)p1+p2 2p1p2 Solution: Here, P(E1) = p1and P(E2) = p2 Now, its clearly seen that either E1or E2occurs but not both....
Read More →Solve this following
Question: If $\mathrm{A}=\left[\begin{array}{ll}3 -4 \\ 1 -1\end{array}\right]$, show that $\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is skew-symmetric. Solution: $\Rightarrow \mathrm{P}^{\prime}=\mathrm{P}$ Hence $A-A^{\prime}$ is a skew symmetric matrix....
Read More →Three events A, B and C have probabilities 2/5,
Question: Three events A, B and C have probabilities 2/5, 1/3 and respectively. Given that P(AÇ C) = 1/5 and P(B Ç C) = , find the values of P(C | B) and P(AÇ C). Solution: Given, P(A) = 2/5, P(B) = 1/3 and P(C) = P(A Ç C) = 1/5 and P(B Ç C) = So, P(C/B) = P(B Ç C)/ P(B) = ()/ (1/3) = P(A Ç C) = 1 P(A ⋃ C) = 1 [P(A) + P(C) P(A Ç C)] = 1 [2/5 + 1/5] = 1 7/10 = 3/10 Therefore, the required probabilities are and 3/10....
Read More →Solve this following
Question: If $\mathrm{A}=\left[\begin{array}{cc}4 1 \\ 5 8\end{array}\right]$, show that $\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is symmetric. Solution:...
Read More →A and B are two events such that P(A) = 1/2,
Question: A and B are two events such that P(A) = 1/2, P(B) = 1/3 and P(A Ç B) = 1/4. Find: (i) P(A|B) (ii) P(B|A) (iii) P(A|B) (iv) P(A|B) Solution: Given, P(A) = 1/2, P(B) = 1/3 and P(A Ç B) = 1/4. P(A) = 1 1/2 = 1/2, P(B) = 1 1/3 = 2/3 Now,...
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