1. (a) P(Positive results) = 0.2 * 0.95 + 0.8 * 0.1 = 0.27P(both citizens have positive results)= P(positive results) * P(positive results)= 0.27 * 0.27= 0.0729

(b) P(negative results) = 0.2 * 0.05 + 0.8 * 0.9 = 0.73P(both citizens have negative results)= P(negative results) * P(negative results)= 0.73 * 0.73= 0.5329

2. (a) P(all fail) = 1/5 * 2/7 * 1/3 = 2/105

P(at least one pass) = 1 - 2/105 = 103/105

(b) P(only Alif will pass the art examination) = 4/5 * 2/7 * 1/3 = 8/105

3. Let A = Approve, D = disapprove

(a) P(all of them have opinions) = 0.85 * 0.85 * 0.85 = 4913/8000

(b) Let X = not disapproveP(X) = 1 - 0.4 = 0.6

P(at least 2 disapprove) = P(DDX) + P(DXD) + P(XDD) + P(DDD)= 3 (0.4 * 0.4 * 0.6) + 0.4 * 0.4 * 0.4=44/125

(c) Let NA = not approveP(NA) = 1 - 0.45 = 0.55

P(at most 1 approve) = P(all NA) + 3 P(A.NA.NA)= 0.55 * 0.55 * 0.55 + 3 (0.45 * 0.55 * 0.55)= 2299/4000

4. X = { x : 5, 6, 7, 8, 9, 10}Y = { y : 3, 6, 9}Total number of outcomes = 6 numbers * 3 numbers = 18

(i) Perfect squares = 6² and 9²so, probability that product xy is a perfect square = 2/18 = 1/9

(ii) xy is an odd number means both x and y must be odd numbersNumber of odd numbers in x = 3Number of odd numbers in y = 3so, probability that product xy is an odd number = 3 * 3 / 18 = 1/2

(iii) Number of terms of xy < 20 : 5 * 3, 6 * 3===> only 2 terms.

So, 16 terms are greater than 20Probability of product xy > 20 = 16/18 = 8/9

5. In each bag, there are 3 red balls, 1 white and 87 blue.

(i) P(ball is not blue) = 4/91

(ii) P(ball is blue) = 87/91

(iii)(a) P(both balls are blue) = 87/91 * 87/91= 7569/8281

(b) P(both balls same colour)= P(both red) + P(both white) + P(both blue)= 3/91 * 3/91 + 1/91 * 1/91 + 7569/91= 7579/8281

P(both balls are different colour) = 1 - P(both balls same colour)= 1 - 7579/8281= 702/8281

6. (i) 13/30 (13 boys out of 30 boys)

(ii) P(both boys support City)= P(1st boy support City) * P(2nd boy support City)= 13/30 * 12/29= 26/145

(iii) P(both boys support the same team)= P(both boys support City) + P(both boys support Rovers) + P(both boys support United)= 26/145 + 10/30 * 9/29 + 7/30 * 6/29= 26/145 + 3/29 + 7/145= 48/145

(iv) P(two boys chosen will support different teams)= 1 - P(both boys support the same team)= 1 - 48/145= 97/145

7. (a) P(a box of matches chosen at random will not contain exactly 50 matches)= 1 - 5/8= 3/8

(b)The probability that a box chosen at random will contain more than 50 matches is twice the probability that it will contain less than 50 matches====> P(a box chosen at random will contain more than 50 matches) = 2/8====> P(a box chosen at random will contain less than 50 matches) = 1/8

(i) P(one box chosen at random will contain at least 50 matches)= P(a box chosen at random will contain more than 50 matches) + P((a box chosen at random will contain exactly 50 matches)= 2/8 + 5/8= 7/8

(ii) P(at least one will contain less than 50 matches)= 1 - P(both boxes contain at least 50 matches)= 1 - 7/8 * 7/8= 15/64

8. (i) P(Disc is blue)= 2/(2 + 8)= 2/10=1/5

(ii) P(both discs are blue)= P(disc from B is blue and disc from C is blue)= 4/(6+4) * 5/(3+5)= 4/10 * 5/8= 1/4

(iii) P(at least one is blue)= P(B,R) + P(R,B) + P(B,B)= 7/8 * 1/7 + 1/8 * 7/7 + 7/8 * 6/7= 1/8 + 1/8 + 6/8= 1

9. (a) Prime numbers are 2, 3, 5Hence, probability that it is a prime number = 3/6 = 1/2

(i) P(sum of numbers is 10)= P(1,9) + P(2,8) + P(3,7) + P(4,6) + P(5,5) + P(6,4) => Summing all the possible combinations= 1/6 * 1/8 * 6 times= 1/8

(ii) P(two numbers are not equal)= 1 - P(two numbers are equal)= 1 - {P(3,3) + P(4,4) + P(5,5) + P(6,6)}= 1 - {1/6 * 1/8 * 4 times}= 1 - 1/12= 11/12

(b)(i) Probability that all 3 go off= 7/10 * 8/10 * 9/10= 63/125

(ii) P(Student is awakened)= P(at least 2 alarms goes off)= P(1,2, 3 don't go off) + P(2,3, 1 don't go off) + P(1,3, 2 don't go off) + P(all 3 go off) ==> P(which alarm goes off)= 0.7 * 0.8 * 0.1 + 0.8 * 0.9 * 0.3 + 0.7 * 0.9 * 0.2 + 63/125= 0.398 + 63/125= 451/500

10. (i) P(both numbers thrown are not 6)= P(1st dice not 6 and 2nd dice not 6)= 15/16 * 5/6= 25/32

(ii) P(at least one of the two numbers thrown is a 6)= 1 - P(both numbers thrown are not 6)= 1 - 25/32= 7/32

(iii) For the biased dice, the probability of getting 1, 2, 3, 4 or 5 is (1 - 1/16)/5 = 3/16

P(exactly one of the numbers thrown is a 4)= P(1st dice is 4, and 2nd dice is not 4) + P(1st dice is not 4, and 2nd dice is 4)= 3/16 * 5/6 + 13/16 * 1/6= 5/32 + 13/96= 7/24

11. (a) (i) P(both shows same number) = 1/6 * 1/6 * 6 (6 different numbers) = 1/6

(ii) P(both shows different number) = 1 - both show same number = 5/6

(b) (i) P(all different numbers) = P(first 2 dice are different and 3rd dice is also different number)

= P(first 2 dice different) * P(3rd dice different number from first 2 dice)

= 5/6 * 4/6 (4 numbers out of 6 left since the first 2 dice have already taken/used up 2 numbers)

= 5/9

(ii) P (at least two show same number) = 1 - P(all different numbers) = 4/9

12. (a) Probability that both contain hazelnuts= probability of first chocolate containing hazelnuts and probability of second chocolate containing hazelnuts= 3/12 * 2/11= 1/22

(b) Probability that one contains almonds and one contains hazelnuts= probability of first chocolate containing almonds and probability of second chocolate containing hazelnuts OR probability of first chocolate containing hazelnuts and probability of second chocolate containing almonds= 7/12 * 3/11 + 3/12 * 7/11= 7/44 + 7/44= 7/22

(c) Probability that both are without sultanas= probability of first chocolate NOT containing sultanas and probability of second chocolate NOT containing sultanas= 10/12 * 9/11= 15/22