1. If 3 books are picked at random from a shelf containing 5 novels, 3 books of poems, and a
dictionary, what is the probability that
a) The dictionary is selected?
b) 2 novels and 1 book of poems are selected?
2. A lot consists of 20 defective and 80 non-defective items from which two items are chosen
without replacement. Events A & B are defined as A = the first item chosen is defective,
B = the second item chosen is defective
a) What is the probability that both items are defective?
b) What is the probability that the second item is defective?
3. In how many ways can a committee of three be chosen from 10 female and 20 male? if
a) All are equally eligible?
b) One female?
c) All are female?
d) Two female?
e) What is probability of selecting one female in a committee?
f) What is probability of selecting two female in a committee?
g) What is probability of selecting three female in a committee?
4. The joint probability function of two discrete random variables X and Y is given by
f(x, y) = cxy for x = 1, 2, 3 and y = 1, 2, 3, and zero otherwise. Then Find,
(a) The constant c? (b) P(X ≤ 2, Y ≤ 3) (c) P(1 ≤ X ≤ 2, Y≤ 2)
(d) P(X ≥ 2) (e) P(Y < 2) (f) P(X = 1)
(g) P(Y = 3) (h) Find the marginal probability functions of X & Y (i)
Find the conditional probability function of X given Y and Y given X?
5. If the random variables X and Y have joint density function
otherwise
x y
xy
f x y
0
0 4, 1 5
( , ) 96
a) The marginal distribution of X and Y.
b) E(x), E(y) and E(xy)
c) Conditional distributions of X given Y
d) Conditional distributions of Y given X
e) P(0 ≤ 𝑥 ≤ 1 𝑎𝑛𝑑 1 ≤ 𝑦 ≤ 2
6. Let the two dimensional discrete random variable (X,Y) have the following joint
probability distribution.
X
Y
1 2 3
0 2p 0.1 0.25
1 0.08 0.1 0.08
2 0.05 p 0.13
a. Find the value of p that makes the distribution a legitimate distribution
b. Find the marginal probability distributions of X and Y
c. Are X and Y are independent? Why?
a. Find the marginal distribution of X for a given Y=2 i.e𝑓(𝑋|𝑌 = 2)�
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.