(i)

P(both G and H wins) = P(G wins) and P(H wins)

Note: and means mutliplication

P(G wins) = $\frac{1}{5}$

P(H wins) = $\frac{3}{8}$

P(both G and H wins) = $\frac{1}{5}$ x $\frac{3}{8}$

P(both G and H wins) = $\frac{\mathrm{1\; x\; 3}}{\mathrm{5\; x\; 8}}$ = $\frac{3}{40}$

(ii)

P(Only one contract wins) = P(G wins and H loses) or P(G loses and H wins)

Note: and means multiplication and or means addition

P(Not) = 1 - P(Is)

P(G loses) = 1 - P(G wins)

P(G loses) = 1 - $\frac{1}{5}$

Every number is being divided by 1

1 = $\frac{1}{1}$

P(G loses) = $\frac{1}{1}$ - $\frac{1}{5}$

P(G loses) = $\frac{\mathrm{5\; x\; 1\; -\; 1\; x\; 1}}{5}$

P(G loses) = $\frac{\mathrm{5\; -\; 1}}{5}$

P(G loses) = $\frac{4}{5}$

P(H loses) = 1 - P(H wins)

P(H loses) = 1 - $\frac{3}{8}$

P(H loses) = $\frac{1}{1}$ - $\frac{3}{8}$

P(H loses) = $\frac{\mathrm{8\; x\; 1\; -\; 1\; x\; 3}}{8}$

P(H loses) = $\frac{\mathrm{8\; -\; 3}}{8}$

P(H loses) = $\frac{5}{8}$

P(Only one contract wins) = P(G wins and H loses) or P(G loses and H wins)

P(G wins) = $\frac{1}{5}$

P(H loses) = $\frac{5}{8}$

P(H wins) = $\frac{3}{8}$

P(G loses) = $\frac{4}{5}$

P(Only one contract wins) = $\frac{1}{5}$ x $\frac{5}{8}$ + $\frac{4}{5}$ x $\frac{3}{8}$

P(Only one contract wins) = $\frac{\mathrm{1\; x\; 5}}{\mathrm{5\; x\; 8}}$ + $\frac{\mathrm{4\; x\; 3}}{\mathrm{5\; x\; 8}}$

P(Only one contract wins) = $\frac{5}{40}$ + $\frac{12}{40}$

P(Only one contract wins) = $\frac{\mathrm{5\; +\; 12}}{40}$

P(Only one contract wins) = $\frac{17}{40}$

(i)

J = 3 + $\frac{1}{2}$ x L

Note:

1. half = $\frac{1}{2}$

2. Of means multiplication (x)

3. one-third = $\frac{1}{3}$

K = $\frac{1}{3}$ x L

Total number of books = J + K + L

Total number of books = 25

J + K + L = 25

Substitute the equation of J and K to solve for L

3 + $\frac{1}{2}$ x L + $\frac{1}{3}$ x L + L = 25

Get rid of the fraction by multiplying each sides of the equation by the L.C.M of the denominators. The L.C.M of 2 and 3 is 6.

6 x 3 + 6 x $\frac{1}{2}$ x L + 6 x $\frac{1}{3}$ x L + 6 x L = 25 x 6

18 + 3L + 2L + 6L = 150
11L = 150 - 18
11L = 132

Divide both sides by 11

L = $\frac{132}{11}$

L = 12

J = 3 + $\frac{1}{2}$ x L

But L = 12

J = 3 + $\frac{1}{2}$ x 12

Note: 2 divides itself 1 and 12, 6 times.

J = 3 + 6
J = 9

K = $\frac{1}{3}$ x L

K = $\frac{1}{3}$ x 12

Note: 3 divides itself 1 and 12, 4 times.

K = 4

(ii)

Probability = $\frac{\mathrm{Number\; of\; possible\; selection}}{\mathrm{Number\; of\; samples}}$

Total number of books = 25
K = 4
L = 12

P(K) = $\frac{4}{25}$

P(L) = $\frac{12}{25}$

P(K or L) = P(K) + P(L)

P(K or L) = $\frac{4}{25}$ + $\frac{12}{25}$

P(K or L) = $\frac{\mathrm{4\; +\; 12}}{25}$

P(K or L) = $\frac{16}{25}$

(i)

Prime numbers are numbers with only two factors (1 and itself). Thus only 1 and the number itself can divide the number without a remainder.

Note: 1 is not a prime number.

Prime numbers from sample = {2, 3, 5, 7, 11, 13}

Number of prime samples = 6

P(Prime) = $\frac{6}{15}$ = $\frac{2}{5}$

(ii)

Multiple of 3 from sample = {3, 6, 9, 12, 15}

Number of multiple of 3 samples = 5

P(Multiple of 3) = $\frac{5}{15}$ = $\frac{1}{3}$

(iii)

Divisible by 5 from sample = {5, 10, 15}

Number of divisible by 5 samples = 3

P(Divisible by 5) = $\frac{3}{15}$ = $\frac{1}{5}$

(a)

Obtaining an odd number on the first die and 6 on the second

Possible outcomes = {1, 6}, {3, 6}, {5, 6}

Number of possible outcomes = 3

P(odd then 6) = $\frac{3}{36}$ = $\frac{1}{12}$

(b)

Obtaining a number greater than 4 on each dice

Possible outcomes = {5, 5}, {5, 6}, {6, 5}, {6, 6}

Number of possible outcomes = 4

P(greater than 4 on each dice) = $\frac{4}{36}$ = $\frac{1}{9}$

(c)

Obtaining a total of 9 or 11

Possible outcomes = {3, 6}, {4, 5}, {5, 4}, {5, 6}, {6, 3}, {6, 5}

Number of possible outcomes = 6

P(a total of 9 or 11) = $\frac{6}{36}$ = $\frac{1}{6}$

(i)

Winning only one event = winning first event and losing second event or losing first event and winning second event

And is multiplication (x) and or is addition (+)

Probability of winning first event = $\frac{3}{5}$

Probability of losing first event = $\frac{2}{5}$

Probability of winning second event = $\frac{2}{9}$

Probability of losing second event = $\frac{7}{9}$

Winning only one event = $\frac{3}{5}$ x $\frac{7}{9}$ + $\frac{2}{5}$ x $\frac{2}{9}$

Winning only one event = $\frac{\mathrm{3\; x\; 7}}{\mathrm{5\; x\; 9}}$ + $\frac{\mathrm{2\; x\; 2}}{\mathrm{5\; x\; 9}}$

Winning only one event = $\frac{21}{45}$ + $\frac{4}{45}$

Winning only one event = $\frac{\mathrm{21\; +\; 4}}{45}$

Winning only one event = $\frac{25}{45}$ = $\frac{5}{9}$

(ii)

none of the events = losing first event and losing second event

And is multiplication (x)

Probability of losing first event = $\frac{2}{5}$

Probability of losing second event = $\frac{7}{9}$

Probability of winning none of the events = $\frac{2}{5}$ x $\frac{7}{9}$

Probability of winning none of the events = $\frac{\mathrm{2\; x\; 7}}{\mathrm{5\; x\; 9}}$

Probability of winning none of the events = $\frac{14}{45}$

(a)

Number of possible selection = 32 + 26 + 26 = 84

P(Boy) = $\frac{84}{192}$ = $\frac{7}{16}$

(b)

Number of possible selection = 44

P(A girl in JHS 2) = $\frac{44}{192}$ = $\frac{11}{48}$

(a)

Probability (James failing examination) = 1 - $\frac{3}{4}$

Every number is being divided by 1

1 = $\frac{1}{1}$

Probability (James failing examination) = $\frac{1}{1}$ - $\frac{3}{4}$ = $\frac{\mathrm{4\; x\; 1\; -\; 1\; x\; 3}}{4}$ = $\frac{\mathrm{4\; -\; 3}}{4}$ = $\frac{1}{4}$

Probability (Juliet failing examination) = 1 - $\frac{3}{5}$

Probability (Juliet failing examination) = $\frac{1}{1}$ - $\frac{3}{5}$ = $\frac{\mathrm{5\; x\; 1\; -\; 1\; x\; 3}}{5}$ = $\frac{\mathrm{5\; -\; 3}}{5}$ = $\frac{2}{5}$

Probability(Both failing) = Probability(James fails) and Prability(Juliet fails)

Note: And means multiplication (x)

Probability(Both failing) = Probability(James fails) x Prability(Juliet fails)

Probability(Both failing) = $\frac{1}{4}$ x $\frac{2}{5}$ = $\frac{\mathrm{1\; x\; 2}}{\mathrm{4\; x\; 5}}$ = $\frac{1}{10}$ or 0.1

(b)

Total number of balls = 8 + 2 + 4 + 6 = 20

Number of new green balls = 8

Number of old blue balls = 6

Probability (New green ball) = $\frac{8}{20}$ = $\frac{2}{5}$

Probability (Old blue ball) = $\frac{6}{20}$ = $\frac{3}{10}$

Probability(2 new green or 2 old blue balls) = Probability(First being new green ball and second being new green ball) or Probability(First being old blue ball and second being old blue ball)

Notes:

1. And means multiplication (x)

2. Or means addition (+)

Probability(2 new green or 2 old blue balls) = $\frac{2}{5}$ x $\frac{2}{5}$ + $\frac{3}{10}$ x $\frac{3}{10}$

Probability(2 new green or 2 old blue balls) = $\frac{4}{25}$ + $\frac{9}{100}$

Probability(2 new green or 2 old blue balls) = $\frac{\mathrm{4\; x\; 4\; +\; 1\; x\; 9}}{100}$

Probability(2 new green or 2 old blue balls) = $\frac{\mathrm{16\; +\; 9}}{100}$

Probability(2 new green or 2 old blue balls) = $\frac{25}{100}$

Probability(2 new green or 2 old blue balls) = $\frac{1}{4}$ or 0.25