Solve this following
Question: Construct a $3 \times 2$ matrix whose elements are given by $a_{i j}=\frac{1}{2}(i-2 j)^{2}$ Solution:...
Read More →The probability distribution of a random variable x is given as under:
Question: The probability distribution of a random variablexis given as under: $\mathrm{P}(\mathrm{X}=x)=\left\{\begin{array}{l}k x^{2} \text { for } x=1,2,3 \\ 2 k x \text { for } x=4,5,6 \\ 0 \quad \text { otherwise }\end{array}\right.$ wherekis a constant. Calculate (i) E(X) (ii) E (3X2) (iii) P(X 4) Solution: The probability distribution of random variable X is given by: (i) E(X) = = 1 x k + 2 x 4k + 3 x 9k + 4 x 8k + 5 x 10k + 6 x 12k = k + 8k + 27k + 32k + 50k + 72k = 190k = 190 x 1/44 = 9...
Read More →Using elementary row transformations, find the inverse of each of the following matrices:
Question: Using elementary row transformations, find the inverse of each of the following matrices: $\left[\begin{array}{ccc}-1 1 2 \\ 1 2 3 \\ 3 1 1\end{array}\right]$ Solution:...
Read More →The probability distribution of a discrete random
Question: The probability distribution of a discrete random variable X is given as under: Calculate: (i) The value of A if E(X) = 2.94 (ii) Variance of X. Solution: (i) We know that:...
Read More →Using elementary row transformations, find the inverse of each of the following matrices:
Question: Using elementary row transformations, find the inverse of each of the following matrices: $\left[\begin{array}{ccc}3 0 -1 \\ 2 3 0 \\ 0 4 1\end{array}\right]$ Solution:...
Read More →Let X be a discrete random variable
Question: Let X be a discrete random variable whose probability distribution is defined as follows: $P(\mathrm{X}=x)= \begin{cases}k(x+1) \text { for } x=1,2,3,4 \\ 2 k x \text { for } x=5,6,7 \\ 0 \text { otherwise }\end{cases}$ wherekis a constant. Calculate (i) the value ofk (ii) E (X) (iii) Standard deviation of X. Solution: (i) Given, P(X = x) = k(x + 1) for x = 1, 2, 3, 4 So, P(X = 1) = k(1 + 1) = 2k P(X = 2) = k(2 + 1) = 3k P(X = 3) = k(3 + 1) = 4k P(X = 4) = k(4 + 1) = 5k Also, P(X = x) ...
Read More →Using elementary row transformations, find the inverse of each of the following matrices:
Question: Using elementary row transformations, find the inverse of each of the following matrices: $\left[\begin{array}{ccc}1 2 3 \\ 2 5 7 \\ -2 -4 -5\end{array}\right]$ Solution:...
Read More →An item is manufactured by three machines A, B and C.
Question: An item is manufactured by three machines A, B and C. Out of the total number of items manufactured during a specified period, 50% are manufactured on A, 30% on B and 20% on C. 2% of the items produced on A and 2% of items produced on B are defective, and 3% of these produced on C are defective. All the items are stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A? Solution: Lets consider: E1= The...
Read More →Using elementary row transformations, find the inverse of each of the following matrices:
Question: Using elementary row transformations, find the inverse of each of the following matrices:. $\left[\begin{array}{ccc}1 3 -2 \\ -3 0 -1 \\ 2 1 0\end{array}\right]$ Solution:...
Read More →By examining the chest X ray,
Question: By examining the chest X ray, the probability that TB is detected when a person is actually suffering is 0.99. The probability of an healthy person diagnosed to have TB is 0.001. In a certain city, 1 in 1000 people suffers from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB? Solution: Let E1= Event that a person has TB E2= Event that a person does not have TB And H = Event that the person is diagnosed to have TB. So, P(E1...
Read More →Using elementary row transformations, find the inverse of each of the following matrices:
Question: Using elementary row transformations, find the inverse of each of the following matrices: $\left[\begin{array}{ccc}3 -1 -2 \\ 2 0 -1 \\ 3 -5 0\end{array}\right]$ Solution:...
Read More →There are three urns containing 2 white and 3 black balls,
Question: There are three urns containing 2 white and 3 black balls, 3 white and 2 black balls, and 4 white and 1 black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn. Solution: Given, we have 3 urns: Urn 1 = 2 white and 3 black balls Urn 2 = 3 white and 2 black balls Urn 3 = 4 white and 1 black balls Now, the probabilities of c...
Read More →There are two bags, one of which contains 3 black and 4
Question: There are two bags, one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls. A die is thrown. If it shows up 1 or 3, a ball is taken from the Ist bag; but it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball. Solution: Let E1be the event of selecting Bag 1 and E2be the event of selecting Bag 2. Also, let E3be the event that black ball is selected Now, P(E1) = 2/6 = 1/3 and P(E2) = 1 ...
Read More →A letter is known to have come either from TATA NAGAR
Question: A letter is known to have come either from TATA NAGAR or from CALCUTTA. On the envelope, just two consecutive letter TA are visible. What is the probability that the letter came from TATA NAGAR. Solution: Let E1 be the event that the letter comes from TATA NAGAR, E2 be the event that the letter comes from CALCUTTA And, E3 be the event that on the letter, two consecutive letters TA are visible Now, P(E1) = , P(E2) = , P(E3/E1) = 2/8 and P(E3/E2) = 1/7 For TATA NAGAR, the two consecutive...
Read More →A shopkeeper sells three types of flower seeds A1, A2 and A3.
Question: A shopkeeper sells three types of flower seeds A1, A2and A3. They are sold as a mixture where the proportions are 4:4:2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35%. Calculate the probability (i) of a randomly chosen seed to germinate (ii) that it will not germinate given that the seed is of type A3, (iii) that it is of the type A2 given that a randomly chosen seed does not germinate. Solution: Given that: A1: A2: A3= 4: 4: 2 So, the probabilitie...
Read More →Using elementary row transformations, find the inverse of each of the following matrices:
Question: Using elementary row transformations, find the inverse of each of the following matrices: $\left[\begin{array}{ccc}1 2 -3 \\ 2 3 2 \\ 3 -3 -4\end{array}\right]$ Solution:...
Read More →Refer to Question 41 above.
Question: Refer to Question 41 above. If a white ball is selected, what is the probability that it came from (i) Bag 2 (ii) Bag 3 Solution: Referring the Question 41, here we will use Bayes Theorem Therefore, the required probabilities are 2/11 and 9/11....
Read More →Three bags contain a number of red and white balls as follows:
Question: Three bags contain a number of red and white balls as follows: Bag 1 : 3 red balls, Bag 2 : 2 red balls and 1 white ball Bag 3 : 3 white balls. The probability that bagiwill be chosen and a ball is selected from it is i/6, i = 1, 2, 3. What is the probability that (i) a red ball will be selected? (ii) a white ball is selected? Solution: Given: Bag 1 : 3 red balls, Bag 2 : 2 red balls and 1 white ball Bag 3 : 3 white balls Now, let E1, E2and E3be the events of choosing Bag 1, Bag 2 and ...
Read More →Using elementary row transformations,
Question: Using elementary row transformations, find the inverse of each of the following matrices: $\left[\begin{array}{lll}3 0 2 \\ 1 5 9 \\ 6 4 7\end{array}\right]$ Solution:...
Read More →An urn contains m white and n black balls.
Question: An urn containsmwhite andnblack balls. A ball is drawn at random and is put back into the urn along withkadditional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend onk. Solution: Lets consider A to be the event of having m white and n black balls E1= First ball drawn of white colour E2= First ball drawn of black colour E3= Second ball drawn of white colour Therefore, the probabili...
Read More →Two dice are tossed.
Question: Two dice are tossed. Find whether the following two events A and B are independent: A = {(x,y) :x+y= 11} B = {(x,y) :x 5} where (x,y) denotes a typical sample point. Solution: Given, two events A and B are independent such that A = {(x,y) :x+y= 11} B = {(x,y) :x 5} Now, A = {(5, 6), (6, 5)} B = {(1, 1), (1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2), (2, 4), (2, 5), (2, 6), (3, 1), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (...
Read More →Using elementary row transformations, find the inverse of each of the following matrices:
Question: Using elementary row transformations, find the inverse of each of the following matrices: $\left[\begin{array}{ccc}2 -3 3 \\ 2 2 3 \\ 3 -2 2\end{array}\right]$ Solution: Let, $A=\left[\begin{array}{ccc}2 -3 3 \\ 2 2 3 \\ 3 -2 2\end{array}\right]$ Now we are going to write the Augmented Matrix followed by matrix A and the Identity matrix I, i.e.,...
Read More →A and B throw a pair of dice alternately.
Question: A and B throw a pair of dice alternately. A wins the game if he gets a total of 6 and B wins if she gets a total of 7. It A starts the game, find the probability of winning the game by A in third throw of the pair of dice. Solution: Lets take A1 to be the event of getting a total of 6 A1= {(2, 4), (4, 2), (1, 5), (5, 1), (3, 3)} And, B1be the event of getting a total of 7 B1= {(2, 5), (5, 2), (1, 6), (6, 1), (3, 4), (4, 3)} Let P(A1) is the probability, if A wins in a throw = 5/36 And ...
Read More →Find the variance of the distribution:
Question: Find the variance of the distribution: Solution: We know that, Variance(X) = E(X2) [E(X)]2...
Read More →Using elementary row transformations, find the inverse of each of the following matrices:
Question: Using elementary row transformations, find the inverse of each of the following matrices: $\left[\begin{array}{lll}0 1 2 \\ 1 2 3 \\ 3 1 1\end{array}\right]$ Solution:...
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