1.
a) The probability of selecting the dictionary is 1/9, since there is only 1 dictionary out of 9 total books.
b) The probability of selecting 2 novels and 1 book of poems can be calculated using the combination formula: C(5,2) * C(3,1) / C(9,3) = (10 * 3) / 84 = 30/84 = 5/14.
2.
a) The probability that both items are defective can be calculated as (20/100) * (19/99) = 380/9900 = 19/495.
b) The probability that the second item is defective can be calculated as (20/100) * (80/99) = 1600/9900 = 8/49.
3.
a) The number of ways to choose a committee of three from 30 people (10 female and 20 male) is C(30,3) = 4060.
b) The number of ways to choose a committee with one female and two males is C(10,1) * C(20,2) = 10 * 190 = 1900.
c) The number of ways to choose a committee with all female members is C(10,3) = 120.
d) The number of ways to choose a committee with two female and one male is C(10,2) * C(20,1) = 45 * 20 = 900.
e) The probability of selecting one female in a committee can be calculated as 1900/4060 = 95/203.
f) The probability of selecting two female in a committee can be calculated as 120/4060 = 3/101.
g) The probability of selecting three female in a committee can be calculated as 1/406.
4.
(a) The constant c can be found by summing up all the probabilities and setting it equal to 1. In this case, c * (1+2+3+2+4+6+3+6+9) = 1, so c = 1/56.
(b) P(X ≤ 2, Y ≤ 3) = P(X = 1, Y = 1) + P(X = 1, Y = 2) + P(X = 1, Y = 3) + P(X = 2, Y = 1) + P(X = 2, Y = 2) + P(X = 2, Y = 3) = 1/56 + 2/56 + 3/56 + 2/56 + 4/56 + 6/56 = 18/56.
(c) P(1 ≤ X ≤ 2, Y ≤ 2) = P(X = 1, Y = 1) + P(X = 1, Y = 2) + P(X = 2, Y = 1) + P(X = 2, Y = 2) = 1/56 + 2/56 + 2/56 + 4/56 = 9/56.
(d) P(X ≥ 2) = P(X = 2) + P(X = 3) = 4/56 + 9/56 = 13/56.
(e) P(Y < 2) = P(Y = 1) = 6/56.
(f) P(X = 1) = 1/56.
(g) P(Y = 3) = P(X = 1, Y = 3) + P(X = 2, Y = 3) + P(X = 3, Y = 3) = 3/56 + 6/56 + 9/56 = 18/56.
(h) The marginal probability function of X is obtained by summing up the probabilities of each value of X: P(X = 1) = 1/56, P(X = 2) = 2/56 + 2/56 + 2/56 = 6/56, P(X = 3) = 3/56 + 6/56 + 9/56 = 18/56. The marginal probability function of Y is obtained by summing up the probabilities of each value of Y: P(Y = 1) = 6/56, P(Y = 2) = 4/56 + 6/56 + 9/56 = 19/56, P(Y = 3) = 3/56 + 6/56 + 9/56 = 18/56.
(i) The conditional probability function of X given Y can be obtained by dividing the joint probability function by the marginal probability function of Y for each value of Y. The conditional probability function of Y given X can be obtained by dividing the joint probability function by the marginal probability function of X for each value of X.
5.
a) The marginal distribution of X can be obtained by integrating the joint density function over the range of Y: f(x) = ∫(f(x,y)dy) = ∫(x*y*f(x,y)dy) = x * ∫(y*f(x,y)dy) = x * ∫(y^2/96 dy) = x * (y^3/288) evaluated from 0 to 4, which is equal to (4x)/3.
Similarly, the marginal distribution of Y can be obtained by integrating the joint density function over the range of X: f(y) = ∫(f(x,y)dx) = ∫(x*y*f(x,y)dx) = y * ∫(x*f(x,y)dx) = y * ∫(x^2/120 dx) = y * (x^3/360) evaluated from 0 to 5, which is equal to (5y)/12.
b) E(x) = ∫(x*f(x,y)dxdy) = ∫(x^2*y/96 dxdy) = (x^3*y/288) evaluated from x=0 to 4, y=0 to 1, which is equal to 8/3.
E(y) = ∫(y*f(x,y)dxdy) = ∫(x*y^2/96 dxdy) = (x*y^3/288) evaluated from x=0 to 4, y=0 to 1, which is equal to 1/3.
E(xy) = ∫(x*y*f(x,y)dxdy) = ∫(x^2*y^2/96 dxdy) = (x^3*y^3/576) evaluated from x=0 to 4, y=0 to 1, which is equal to 16/27.
c) The conditional distribution of X given Y can be obtained by dividing the joint density function by the marginal density function of Y for each value of Y. In this case, f(x|y) = f(x,y) / f(y).
f(x|y) = (x*y) / [(5y)/12] = 12x/5.
d) The conditional distribution of Y given X can be obtained by dividing the joint density function by the marginal density function of X for each value of X. In this case, f(y|x) = f(x,y) / f(x).
f(y|x) = (x*y) / [(4x)/3] = 3y/4.
e) P(0 ≤ x ≤ 1 and 1 ≤ y ≤ 2) can be calculated by integrating the joint density function over the given range:
∫∫(f(x,y)dxdy) over (0 ≤ x ≤ 1 and 1 ≤ y ≤ 2) = ∫(∫(xy/96)dy)dx over (0 ≤ x ≤ 1) = ∫(x/96 * y^2/2) from 1 to 2 = ∫(x/96 * 4/2 - x/96)dx from 0 to 1 = ∫(x/48 - x/96)dx from 0 to 1 = [(x^2/96 - x^2/192)] from 0 to 1 = (1/96 - 1/192) = 1/192.
6.
a. To find the value of p that makes the distribution a legitimate distribution, we need to calculate the sum of all probabilities in the joint distribution and set it equal to 1: 2p + 0.1 + 0.25 + 0.08 + 0.1 + 0.08 + 0.05 + p + 0.13 = 1. Solving this equation, we get p = 0.16.
b. The marginal probability distributions of X and Y can be obtained by summing up the probabilities for each value of X and Y, respectively:
Marginal distribution of X: P(X = 0) = 2p + 0.1 + 0.25 = 2(0.16) + 0.1 + 0.25 = 0.57, P(X = 1) = 0.08 + 0.1 + 0.08 = 0.26, P(X = 2) = 0.05 + p + 0.13 = 0.05 + 0.16 + 0.13 = 0.34.
Marginal distribution of Y: P(Y = 0) = 2p + 0.08 + 0.05 = 2(0.16) + 0.08 + 0.05 = 0.51, P(Y = 1) = 0.1 + 0.1 + p = 0.1 + 0.1 + 0.16 = 0.36, P(Y = 2) = 0.25 + 0.08 + p = 0.25 + 0.08 + 0.16 = 0.49, P(Y = 3) = 0.1 + 0.08 + 0.13 = 0.31.
c. X and Y are independent if and only if the joint probability distribution can be expressed as the product of the marginals: f(x,y) = f(x) * f(y). However, in this case, the joint probability distribution does not satisfy this condition. Therefore, X and Y are not independent.