The probability of any given number is 1/8 = 0.125 . All the answers are a various multiples of that probability

1. 3 - 1 out of 8
2. 1, 2, 3, 4, 5, 6 and 7 - 7 out of 8
3. Primes are 2, 3, 5 and 7 - 4 out of 8 (careful not to include 1 and not toforget 2)
4. 1, 2 and 3 - 3 out of 8 (4 is not less than 4)
5. 7 and 8 - 2 out of 8
6. 3, 5 and 7 - 3 out of 8 (see answer to part 4 to help)
7. 1, 2, 3, 5 and 7 - 5 out of 8
8. None - 0 (There are no mulitples of 9 available)

You might need to revisit some of the work on set notation to help with this question.

Part 1 - 3 divided by 500 = 0.006 which is 6 x 10^-3 in standard index form

Part 2 - 15600 x 0.006 = 93.6 which rounds to 94

A sample space diagram should help to answer this question.

1. There are 4 possible colours for each of the 10 numbers of the dice
2. 5 out of 40
3. 7 out of 40
4. 14 out of 40, 20 out of 40 are Green OR Red, of those, 14 are bigger than 3

1. a = 1 - 0.1 = 0.9, b = 0.12 (given in the question), c = 1 - 0.12 = 0.88
2. 0.1 x 0.12 = 0.012
3. 0.1 x 0.88 = 0.088
4. From part 3, Ellie late but not Manny is 0.88. Ellie on time but Manny late = 0.9 x 0.12 = 0.108. Together we get 0.088 + 0.108 = 0.196

1. a = 2, if the 1st was purple there are only 2 left, b = 9, there are only 9 left all together, c = 7 - there are still 7 red ones left, d = 3 - there are still 3 red ones left, e = 6 - the ifrst was red so there is only six red ones left.

2. \(\frac { 7 }{ 10 } \times \frac { 2 }{ 9 } =\frac { 14 }{ 90 } =0.156\quad (3sf)\)

3. 1 subtract the probability they are both purple, \(1\quad -\quad \frac { 3 }{ 10 } \times \frac { 2 }{ 9 } =\quad 0.933\quad (3sf)\)

4. \(\frac { 3 }{ 10 } \times \frac { 7 }{ 9 } \quad +\quad \frac { 7 }{ 10 } \times \frac { 3 }{ 9 } \quad =\quad 0.467\quad (3sf)\)

1. a = 1 - 0.75, b = 0.4, given, c = 1 - 0.4, d = 0.8, given, e = 1 - 0.8

2. 0.25 x 0.2 = 0.05

3. 0.75 x 0.4 + 0.25 x 0.8 = 0.5

4. \(\frac { Probability\quad on\quad time\quad and\quad stuck }{ Probability\quad stuck } =\quad \frac { 0.75\times 0.4 }{ 0.5 } =\frac { 0.3 }{ 0.5 } =0.6\)

All answers rounded to 3sf.

1. 95/215 = 0.442

2. 80/215 = 0.372

3. 22/215 = 0.102 (Only 22 people are both)

4. 136/215 = 0.633 (120 females + 16 males who like horror)

5. 35/215 = 0.149 (Only 35 people who are both)

6. 107/215 = 0.498 (34 people who prefer comedy + 25 + 32 + 16 males who prefer other genres)

7. 32/95 = 0.337 (Only considering the 95 males of which 32 prefer drama)

8. 35/60 = 0.583 (Only considering the 60 people who prefer action of which 35 are female)

All answers rounded to 3sf.

1. 31/53 = 0.585 (31 elements in set A out of a total of 53 elements all together)

2.43/53 = 0.811 (The union includes everything in A OR B - careful not to count the 6 twice)

3. 6/53 = 0.113 (The intersection, in set A AND B)

4. 6/18 = 0.333 (Of the 18 elements in set B, 6 also belong to set A)

5. 35/53 = 0.660 (35 of the 53 elements are NOT in set B)

6. 10/53 = 0.189 (10 out of the 53 elements are NOT in the union of A and B)

7. 25/53 = 0.472 (25 elements are in the set A AND NOT B)

8. 12/22 = 0.545 (of the 22 elements that are NOT in set A, 12 are in set B)

9. N - it is possible to be in both sets

10. N - \(P(A)\) = 0.585 and \(P(A|B)\) = 0.333 Likelihood of A depends on membership of B

Part 1 - For combined events

\(P(A\cup B)\quad =\quad P(A)\quad +\quad P(B)\quad -\quad P(A\cap B)\\ So,\quad 0.58\quad =\quad 0.3\quad +\quad 0.4\quad -\quad P(A\cap B)\\ So\quad P(A\cap B)\quad =\quad 0.12\)

i) No - For mutually exclusive events \(P(A\cap B)\) = 0

ii) Yes - For independent events....

\(\\ P(A\cap B)\quad =\quad P(A)\quad \times \quad P(B)\\ in\quad this\quad case\\ 0.12\quad =\quad 0.3\quad \times \quad 0.4\)

NOTE - you would be expected to give this reasoning to get these marks.