Problem 1 :
Two dice are numbered 1,2,3,4,5,6 and 1,1,2,2,3,3 respectively. They are rolled and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.
Solution :
Sample space = {(1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) }
n(S) = 36
Let A, B, C, D, E, F, G, H be the event of getting the sum 2, 3, 4, 5, 6, 7, 8 and 9 respectively.
(i) Sum 2
A = {(1, 1) (1, 1)}
n(A) = 2
P(A) = 2/36
(ii) Sum 3
B = { (2, 1) (2, 1) (1, 2) (1,2) }
n(B) = 4
P(B) = 4/36
(iii) Sum 4
C = { (3,1) (3,1) (2,2) (2,2) (1,3) (1,3) }
n(C) = 6
P(C) = 6/36
(iv) Sum 5
D = {(4,1) (4,1) (3,2)(3,2) (2,3) (2,3)}
n(C) = 6
P(C) = 6/36
(v) Sum 6
E = {(5, 1) (5, 1) (4, 2) (4, 2) (3, 3) (3, 3) }
n(E) = 6
P(E) = 6/36
(vi) Sum 7
F = {(6, 1) (6, 1) (5, 2) (5, 2) (4, 3) (4, 3) }
n(F) = 6
P(F) = 6/36
(vii) Sum 8
G = {(6,2) (6,2) (5,3) (5,3)}
n(G) = 4
P(G) = 4/36
(viii) Sum 9
H = { (6,3)(6,3)}
n(H) = 2
P(H) = 2/36
Problem 2 :
A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is (i) white (ii) black or red (iii) not white (iv) neither white nor black
Solution :
Total number of balls = 5 red + 6 white + 7 green + 8 black
n(S) = 26
(i) Let "A" be the event of getting white ball
n(A) = 6
P(A) = n(A)/n(S)
P(A) = 6/26
P(A) = 3/13
(ii) black or red
Let "B" be the event of getting black or red ball
n(B) = 5 + 8
n(B) = 13
P(B) = 13/36
P(B) = 1/2
(iii) not white
P(A bar) = 1 - P(A)
= 1 - (3/13)
= (13 - 3)/13
P(A bar) = 10/13
(iv) neither white nor black
Let "C" be the event of getting neither white nor black
C = 5 red + 7 green
n(C) = 12
P(C) = 12/26
P(C) = 6/13
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