The random variable X can take only the values 0, 1, 2.
Question: The random variable X can take only the values 0, 1, 2. Given that P(X = 0) = P (X = 1) =pand that E(X2) = E[X], find the value ofp. Solution: Given, X = 0, 1, 2 and P(X = 0) = P (X = 1) =p Let P(X) at X = 2 is x ⇒ p + p + x = 1 ⇒ x = 1 2p Now, we have the following distribution: So, E(X) = 0.p + 1.p + 2(1 2p) = p + 2 4p = 2 3p And, E(X2) = 0.p + 1.p + 4(1 2p) = p + 4 8p = 4 7p Also, given E(X) = E(X2) 2 3p = 4 7p 4p = 2 p = Therefore, the required value of p is ....
Read More →Find the probability distribution of the maximum
Question: Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice. Determine also the mean of the distribution. Solution: Let X be the random variable scores when a die is thrown twice. X = 1, 2, 3, 4, 5, 6 And S = {(1, 1), (1, 2), (2, 1), (2, 2) , (1, 3) , (2, 3) , (3, 1) , (3, 2) , (3, 3) , (3, 4) , (3, 5) , , (6, 6)} So, P(X = 1) = 1/6 . 1/6 = 1/36 P(X = 2) = 1/6 . 1/6 + 1/6 . 1/6 + 1/6 . 1/6 = 3/36 P(X = 3) = 1/6 . 1/6 + 1/6 . 1/6 + 1/6 . 1/6 + ...
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: $\left[\begin{array}{ll}6 7 \\ 8 9\end{array}\right]$ Solution:...
Read More →Two natural numbers r, s are drawn one at a time,
Question: Two natural numbersr,sare drawn one at a time, without replacement from the set S={1, 2, 3, .,n} . Find P[rp|sp] , wherepÎS. Solution: Given, S = {1, 2, 3, , n} So, P(r p/s p) = P(P ⋂ S)/ P(S) = p 1/n x n/(n 1) = (p 1)/(n 1) Therefore, the required probability is (p 1)/(n 1)....
Read More →Suppose that 6% of the people with blood group O are left-handed and 10%
Question: Suppose that 6% of the people with blood group O are left-handed and 10% of those with other blood groups are left-handed 30% of the people have blood group O. If a left-handed person is selected at random, what is the probability that he/she will have blood group O? Solution: Lets assume E1= The event that a person selected is of blood group O E2= The event that the people selected is of other group And H = The event that selected person is left handed Now, P(E1) = 0.30 and P(E2) = 0....
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: $\left[\begin{array}{ll}4 0 \\ 2 5\end{array}\right]$ Solution:...
Read More →Suppose you have two coins which appear identical
Question: Suppose you have two coins which appear identical in your pocket. You know that one is fair and one is 2-headed. If you take one out, toss it and get a head, what is the probability that it was a fair coin? Solution: Lets consider E1= Event that the coin is fair E2= Event that the coin is 2 headed And H = Event that the tossed coin gets head. Now, P(E1) = , P(E2) = , P(H/E1) = , P(H/E2) = 1 Using Bayes Theorem, we get Therefore, the required probability is 1/3....
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: $\left[\begin{array}{cc}2 -3 \\ 4 5\end{array}\right]$ Solution:...
Read More →A factory produces bulbs.
Question: A factory produces bulbs. The probability that any one bulb is defective is 1/50 and they are packed in boxes of 10. From a single box, find the probability that (i) none of the bulbs is defective (ii) exactly two bulbs are defective (iii) more than 8 bulbs work properly Solution: Lets assume X to be the random variable denoting a bulb to be defective. Here, n = 10, p = 1/50, q = 1 1/50 = 49/50 We know that, P(X = r) =nCrprqn r (i) None of the bulbs is defective, i.e., r = 0 P(x = 0) =...
Read More →Two probability distributions of the discrete random variable X and Y
Question: Two probability distributions of the discrete random variable X and Y are given below. Solution: The probability distribution of random variable X is...
Read More →Two biased dice are thrown together.
Question: Two biased dice are thrown together. For the first die P(6) = 1/2, the other scores being equally likely while for the second die, P(1) = 2/5 and the other scores are equally likely. Find the probability distribution of the number of ones seen. Solution:...
Read More →A die is thrown three times.
Question: A die is thrown three times. Let X be the number of twos seen. Find the expectation of X. Solution: Here, we have X = 0, 1, 2, 3 [Since, die is thrown 3 times] And p = 1/6, q = 5/6 Therefore, the required expression is ....
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: $\left[\begin{array}{cc}2 5 \\ -3 1\end{array}\right]$ Solution:...
Read More →A biased die is such that P(4) = 1/10 and other scores being equally likely.
Question: A biased die is such that P(4) = 1/10 and other scores being equally likely. The die is tossed twice. If X is the number of fours seen, find the variance of the random variable X. Solution: Here, random variable X = 0, 1, 2 Therefore, the required variance = 0.18....
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: $\left[\begin{array}{cc}1 2 \\ 2 -1\end{array}\right]$ Solution:...
Read More →For the following probability distribution determine
Question: For the following probability distribution determine standard deviation of the random variable X. Solution: We know that: Standard deviation (S.D.) = Variance So, Var(X) = E(X2) [E(X)]2 E(X) = 2 x 0.2 + 3 x 0.5 + 4 x 0.3 = 0.4 + 1.5 + 1.2 = 3.1 E(X2) = 4 x 0.2 + 9 x 0.5 + 16 x 0.3 = 0.8 + 4.5 + 4.8 = 10.1 ⇒ V(X) = 10.1 (3.1)2= 10.1 9.61 = 0.49 Therefore, S.D. = 0.49 = 0.7...
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: $\left[\begin{array}{ll}1 2 \\ 3 7\end{array}\right]$ Solution:...
Read More →The probability distribution of a random variable X is given below:
Question: The probability distribution of a random variable X is given below: (i) Determine the value ofk. (ii) Determine P(X 2) and P(X 2) (iii) Find P(X 2) + P (X 2). Solution: (i) W.k.t P(0) + P(1) + P(2) + P(3) = 1 ⇒ k + k/2 + k/4 + k/8 = 1 (8k + 4k + 2k + k)/8 = 1 15k = 8 Hence, k = 8/15 (ii) P(X 2) = P(X = 0) + P(X = 1) + P(X = 2) = k + k/2 + k/4 = 7k/4 = 7/4 x 8/15 = 14/15 And P(X 2) = P(X = 3) = k/8 = 1/8 x 8/15 = 1/15 (iii) P(X 2) + P(X 2) = 14/15 + 1/15 = (14 + 1)/15 = 15/15 = 1...
Read More →Solve this following
Question: If matrix $A=\left[\begin{array}{lll}1 2 3\end{array}\right]$, write $A A^{\prime}$. Solution:...
Read More →Solve this
Question: If $A=\left[\begin{array}{cc}\cos \alpha \sin \alpha \\ -\sin \alpha \cos \alpha\end{array}\right]$, show that $A^{\prime} A=1$ Solution:...
Read More →Consider the probability distribution of a random variable X:
Question: Consider the probability distribution of a random variable X: Solution: Given:...
Read More →For each of the following pairs of matrices
Question: For each of the following pairs of matrices $A$ and $B$, verify that $(A B)^{\prime}=\left(B^{\prime} A^{\prime}\right)$ : $A=\left[\begin{array}{crc}-1 2 -3 \\ 4 -5 6\end{array}\right]$ and $B=\left[\begin{array}{cc}3 -4 \\ 2 1 \\ -1 0\end{array}\right]$ Solution: Let us take $C=A B$...
Read More →A lot of 100 watches is known to have 10 defective watches.
Question: A lot of 100 watches is known to have 10 defective watches. If 8 watches are selected (one by one with replacement) at random, what is the probability that there will be at least one defective watch? Solution: Given: Total number of watches = 100 and number of defective watches = 10 So, the probability of selecting a detective watch = 10/100 = 1/10 Now, n = 8, p = 1/10 and q = 1 1/10 = 9/10, r 1 P(X 1) = 1 P(x = 0) = 1 8C0(1/10)0(9/10)8 0= 1 (9/10)8 Therefore, the required probability ...
Read More →For each of the following pairs of matrices
Question: For each of the following pairs of matrices $A$ and $B$, verify that $(A B)^{\prime}=\left(B^{\prime} A^{\prime}\right)$ : $A=\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right]$ and $B=\left[\begin{array}{lll}-2 -1 -4\end{array}\right]$ Solution:...
Read More →The probability of a man hitting a target is 0.25. He shoots 7 times.
Question: The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice? Solution: Here, we have n = 7, p = 0.25 = 25/100 = and q = 1 = P(X 2) = 1 [P(X = 0) + P(X = 1)] Therefore, the required probability is 4547/8192....
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