(a)

(b)(i)

\begin{align*} \text{P(Black marble from first bag)} & = {5 \over 9} \end{align*}

(b)(ii)

\begin{align*} \text{P(Red marble from second bag given that black marble is drawn from first bag)} & = {4 \over 9} \end{align*}

(b)(iii)

\begin{align*} \text{P(B, R)} & = {5 \over 9} \times {4 \over 9} \\ & = {20 \over 81} \end{align*}

(b)(iv)

\begin{align*} \text{P(Red marble from second bag)} & = \text{P(B, R)} + \text{P(R, R)} \phantom{000} [\text{2 cases}] \\ & = {20 \over 81} + {4 \over 9} \times {4 \over 9} \\ & = {4 \over 9} \end{align*}

(a)

(b)(i)

\begin{align*} \text{P(R, R)} & = {3 \over 5} \times {3 \over 5} \\ & = {9 \over 25} \end{align*}

(b)(ii)

\begin{align*} \text{P(one ball of each colour)} & = \text{P(R, Y)} + \text{P(Y, R)} \phantom{00000} [\text{2 cases}] \\ & = {3 \over 5} \times {2 \over 5} + {2 \over 5} \times {3 \over 5} \\ & = {12 \over 25} \end{align*}

(b)(iii)

\begin{align*} \text{P(Yellow ball on second draw)} & = \text{P(R, Y)} + \text{P(Y, Y)} \phantom{00000} [\text{2 cases}] \\ & = {3 \over 5} \times {2 \over 5} + {2 \over 5} \times {2 \over 5} \\ & = {2 \over 5} \end{align*}

(a)

(b)(i)

\begin{align*} \text{P(First number less than or equals to second number)} & = \text{P(10, 30)} + \text{P(20, 30)} + \text{P(30, 30)} \phantom{00000} [\text{3 cases}] \\ & = {1 \over 4} \times {3 \over 4} + {1 \over 2} \times {3 \over 4} + {1 \over 4} \times {3 \over 4} \\ & = {3 \over 4} \end{align*}

(b)(ii)

\begin{align*} \text{P(Second number is 0)} & = \text{P(10, 0)} + \text{P(20, 0)} + \text{P(30, 0)} \phantom{00000} [\text{3 cases}] \\ & = {1 \over 4} \times {1 \over 4} + {1 \over 2} \times {1 \over 4} + {1 \over 4} \times {1 \over 4} \\ & = {1 \over 4} \end{align*}

(c)(i)

\begin{align*} \text{P(Receive \$2)} & = \text{P(10, 30)} + \text{P(20, 0)} + \text{P(30, 0)} \\ & = {1 \over 4} \times {3 \over 4} + {1 \over 2} \times {1 \over 4} + {1 \over 4} \times {1 \over 4} \\ & = {3 \over 8} \end{align*}

(c)(ii)

\begin{align*} \text{P(Receive \$5)} & = \text{P(20, 30)} + \text{P(30, 30)} \\ & = {1 \over 2} \times {3 \over 4} \times {1 \over 4} \times {3 \over 4} \\ & = {9 \over 16} \end{align*}

(c)(iii) It is not possible 2 dollars and 5 dollars at the same time (mutually exclusive event)

\begin{align*} \text{P(Receive \$2 or \$5)} & = \text{P(Receive \$2)} + \text{P(Receive \$5)} \\ & = {3 \over 8} + {9 \over 16} \\ & = {15 \over 16} \end{align*}

(c)(iv)

\begin{align*} \text{P(Receive nothing)} & = \text{P(10, 0)} \\ & = {1 \over 4} \times {1 \over 4} \\ & = {1 \over 16} \end{align*}

(i)

\begin{align*} \text{P(Both buses punctual)} & = {2 \over 3} \times {7 \over 8} \\ & = {7 \over 12} \end{align*}

(ii)

\begin{align*} \text{P(Bus } A \text{ is late)} & = 1 - {2 \over 3} \\ & = {1 \over 3} \\ \\ \text{P(Bus } A \text{ is late and bus } B \text{ is punctual)} & = {1 \over 3} \times {7 \over 8} \\ & ={7 \over 24} \end{align*}

(iii)

\begin{align*} \text{P(Bus } B \text{ is late)} & = 1 - {7 \over 8} \\ & = {1 \over 8} \\ \\ \text{P(Only one bus is late)} & = \text{P(}A \text{ late but } B \text{ punctual)} + \text{P(} A \text{ punctual but } B \text{ late)} \phantom{00000} [\text{2 cases}] \\ & = {7 \over 24} + {2 \over 3} \times {1 \over 8} \\ & = {3 \over 8} \end{align*}

First part

\begin{align*} \text{P(Boy chosen is left-handed)} & = {3 \over 8} \end{align*}

(a)

\begin{align*} \text{P(Second boy chosen is left-handed given that first boy chosen is left-handed)} & = {2 \over 7} \end{align*}

(b)

(c)(i)

\begin{align*} \text{P(}R, L ) & = {5 \over 8} \times {3 \over 7} \\ & = {15 \over 56} \end{align*}

(c)(ii)

\begin{align*} \text{P(}L, L) & = {3 \over 8} \times {2 \over 7} \\ & = {3 \over 28} \end{align*}

(c)(iii)

\begin{align*} \text{P(Second boy chosen is left-handed)} & = \text{P(}L, L) + \text{P(}R, L) \\ & = {3 \over 28} + {15 \over 56} \\ & = {3 \over 8} \end{align*}

(i)

\begin{align*} \text{P(First rep. is a girl)} & = {30 \over 45} \\ & = {2 \over 3} \end{align*}

(ii)

\begin{align*} \text{P(Second rep. is a girl given that first rep. is a boy)} & = {30 \over 45 - 1} \\ & = {15 \over 22} \end{align*}

(iii)

\begin{align*} \text{P(First rep. is a boy and second rep. is a girl)} & = {15 \over 45} \times {15 \over 22} \\ & = {5 \over 22} \end{align*}

(iv)

\begin{align*} \text{P(A boy and a girl are selected)} & = \text{P(B, G)} + \text{P(G, B)} \phantom{00000} [\text{2 cases}] \\ & = {15 \over 45} \times {30 \over 44} + {30 \over 45} \times {15 \over 44} \\ & = {5 \over 11} \end{align*}

(a)

\begin{align*} \text{P(Green card drawn)} & = {6 \over 10} \\ & = {3 \over 5} \end{align*}

(b)(i)

\begin{align*} \text{P(} G, G) & = {3 \over 5} \times {5 \over 9} \\ & = {1 \over 3} \end{align*}

(b)(ii)

\begin{align*} \text{P(One card of each colour)} & = \text{P(} G, B) + \text{P(} B, G) \\ & = {3 \over 5} \times {4 \over 9} + {2 \over 5} \times {2 \over 3} \\ & = {8 \over 15} \end{align*}

(b)(iii)

\begin{align*} \text{P(At least one blue card)} & = 1 - \text{P(No blue card)} \\ & = 1 - \text{P(}G, G) \\ & = 1 - {1 \over 3} \\ & = {2 \over 3} \end{align*}

(a)

(b)

\begin{align} & \text{Yes. The selection in bag A does not affect the selection in bag B.} \\ & \text{This implies that event '} RR \text{' are independent events.} \end{align}

(a)

(b)(i)

\begin{align*} \text{P(2, 5, 6)} & = {1 \over 3} \times {5 \over 6} \times {1 \over 6} \\ & = {5 \over 108} \end{align*}

(b)(ii)

\begin{align*} \text{P(3, 6, 6)} & = {1 \over 2} \times {1 \over 6} \times {1 \over 6} \\ & = {1 \over 72} \end{align*}

(b)(iii)

\begin{align*} \text{P(Exactly two sixes)} & = \text{P(1, 6, 6)} + \text{P(2, 6, 6)} + \text{P(3, 6 , 6)} \\ & = {1 \over 6} \times {1 \over 6} \times {1 \over 6} + {1 \over 3} \times {1 \over 6} \times {1 \over 6} + {1 \over 2} \times {1 \over 6} \times {1 \over 6} \\ & = {1 \over 36} \end{align*}

(b)(iv)

\begin{align*} \text{P(Sum of 12)} & = \text{P(1, 5, 6)} + \text{P(1, 6, 5)} + \text{P(2, 5, 5)} \\ & = {1 \over 6} \times {5 \over 6} \times {1 \over 6} + {1 \over 6} \times {1 \over 6} \times {5 \over 6} + {1 \over 3} \times {5 \over 6} \times {5 \over 6} \\ & = {5 \over 18} \end{align*}

(b)(v)

\begin{align*} \text{P(Sum divisible by 3)} & = \text{P(Sum of 12)} + \text{P(Sum of 15)} \\ & = {5 \over 18} + {1 \over 2} \times {1 \over 6} \times {1 \over 6} \\ & = {7 \over 24} \end{align*}

(i)

\begin{align*} \text{P(She will not buy in a particular month)} & = 1 - {4 \over 9} \\ & = {5 \over 9} \end{align*}

(ii)

\begin{align*} \text{P(She will not buy in two particular consecutive months)} & = {5 \over 9} \times {5 \over 9} \\ & = {25 \over 81} \end{align*}

(iii)

\begin{align*} \text{P(She buys in just one of two particular months)} & = \text{P(Buy, don't buy)} + \text{P(Don't buy, buy)} \\ & = {4 \over 9} \times {5 \over 9} + {5 \over 9} \times {4 \over 9} \\ & = {40 \over 81} \end{align*}

(i)

\begin{align*} \text{P(F, F, F)} & = {36 \over 40 + 36} \times {35 \over 55 + 35} \times {52 \over 38 + 52} \\ & = {91 \over 855} \end{align*}

(ii)

\begin{align*} \text{P(M, F, F)} & = {40 \over 40 + 36} \times {35 \over 55 + 35} \times {52 \over 38 + 52} \\ & = {182 \over 1539} \end{align*}

(iii)

\begin{align*} \text{P(Exactly one representative is male)} & = \text{P(M, F, F)} + \text{P(F, M, F)} + \text{P(F, F, M)} \\ & = {182 \over 1539} + {36 \over 40 + 36} \times {55 \over 55 + 35} \times {52 \over 38 + 52} + {36 \over 40 + 36} \times {35 \over 55 + 35} \times {38 \over 38 + 52} \\ & = {5591 \over 15390} \end{align*}

(a)(i)

\begin{align*} \text{P(B, B)} & = {8 \over 16} \times {7 \over 15} \\ & = {7 \over 30} \end{align*}

(a)(ii)

\begin{align*} \text{P(One black and one white)} & = \text{P(B, W)} + \text{P(W, B)} \phantom{00000} [\text{2 cases}] \\ & = {8 \over 16} \times {6 \over 15} + {6 \over 16} \times {8 \over 15} \\ & = {2 \over 5} \end{align*}

(a)(iii)

\begin{align*} \text{P(Two shirts of the same colour)} & = \text{P(B, B)} + \text{P(W, W)} + \text{P(Blue, Blue)} \phantom{00000} [\text{3 cases}] \\ & = {7 \over 30} + {6 \over 16} \times {5 \over 15} + {2 \over 16} \times {1 \over 15} \\ & = {11 \over 30} \end{align*}

(b)

\begin{align*} \text{P(B, B, B)} & = {8 \over 16} \times {7 \over 15} \times {6 \over 14} \\ & = {1 \over 10} \end{align*}

(c)

$$ \text{No. The three events are dependent events} $$

Letter count:

(i)

\begin{align*} \text{P(First card bears letter 'O')} & = {3 \over 10} \end{align*}

(ii)

\begin{align*} \text{P('P', 'O')} & = {2 \over 10} \times {3 \over 9} \\ & = {1 \over 15} \end{align*}

(iii)

\begin{align*} \text{P(Two cards with letters 'P' and 'O')} & = \text{P('P', 'O')} + \text{P('O', 'P')} \\ & = {1 \over 15} + {3 \over 10} \times {2 \over 9} \\ & = {2 \over 15} \end{align*}

(iv)

\begin{align*} \text{P(Two cards with same letters)} & = \text{P('P', 'P')} + \text{P('O', 'O')} + \text{P('R', 'R')} \\ & = {2 \over 10} \times {1 \over 9} + {3 \over 10} \times {2 \over 9} + {2 \over 10} \times {1 \over 9} \\ & = {1 \over 9} \end{align*}

(a)

\begin{align*} \text{P(Ball is numbered '8')} & = {1 \over 5} \end{align*}

(b)(i)

\begin{align*} \text{P(Number on each ball is even)} & = {2 \over 5} \times {1 \over 4} \\ & = {1 \over 10} \end{align*}

(b)(ii)

\begin{align*} \text{P(Sum is more than 10)} & = \text{P('2', '9')} + \text{P('5', '8' or '9')} + \text{P('8', '5' or '9')} + \text{P('9', '2' or '5' or '8')} \\ & = {1 \over 5} \times {1 \over 4} + {1 \over 5} \times {2 \over 4} + {1 \over 5} \times {2 \over 4} + {1 \over 5} \times {3 \over 4} \\ & = {2 \over 5} \end{align*}

(b)(iii) A prime number has only two factors, 1 and itself. The non-prime numbers are 1, 8 and 9.

\begin{align*} \text{P(Each number is not a prime number)} & = {3 \over 5} \times {2 \over 4} \\ & = {3 \over 10} \end{align*}

(b)(iv)

\begin{align*} \text{P(Only one ball bears an odd number)} & = \text{P(Even, odd)} + \text{P(Odd, even)} \\ & = {2 \over 5} \times {3 \over 4} + {3 \over 5} \times {2 \over 4} \\ & = {3 \over 5} \end{align*}

(a)

(b)(i)

\begin{align*} \text{P(More yellow balls than blue balls)} & = 0 \end{align*}

(b)(ii)

\begin{align*} \text{P(7 B, 5 Y)} & = {7 \over 12} \times {4 \over 11} + {5 \over 12} \times {8 \over 11} \\ & = {17 \over 33} \end{align*}

(b)(iii)

\begin{align*} \text{P(Twice as many blue balls as yellow balls)} & = \text{P(8B, 4Y)} \\ & = {5 \over 12} \times {3 \over 11} \\ & = {5 \over 44} \end{align*}

(i)

\begin{align*} \text{P(Student initially from Class A selected)} & = {1 \over 14 + 22 + 1} \\ & = {1 \over 37} \end{align*}

(ii)

\begin{align*} \text{P(Student selected is a boy)} & = \text{P(Boy transferred, then boy selected from B)} + \text{P(Girl transferred, then boy selected from B)} \\ & = {18 \over 35} \times {14 + 1 \over 14 + 22 + 1} + {17 \over 35} \times {14 \over 14 + 22 + 1} \\ & = {508 \over 1295} \end{align*}

(i)

\begin{align*} \text{P(Volcanic eruptions occur in A, B and C)} & = 0.03 \times 0.12 \times 0.3 \\ & = 0.001 \phantom{.} 08 \end{align*}

(ii)

\begin{align*} \text{P(No volcanic eruptions occur in A, B and C)} & = (1 - 0.03) \times (1 - 0.12) \times (1 - 0.3) \\ & = 0.597 \phantom{.} 52 \end{align*}

(iii)

\begin{align*} \text{P(At least 1 volcanic eruptions)} & = 1 - \text{P(No volcanic eruptions occur in A, B and C)} \\ & = 1 - 0.597 \phantom{.} 52 \\ & = 0.402 \phantom{.} 48 \end{align*}

(iv)

\begin{align*} \text{P(Exactly two volcanic eruptions)} & = \text{P(} A \checkmark, B \checkmark, C \times) + \text{P(} A \checkmark, B \times, C \checkmark) + \text{P(} A \times, B \checkmark, C \checkmark) \\ & = (0.03)(0.12)(1-0.3) + (0.03)(1-0.12)(0.3) + (1-0.03)(0.12)(0.3) \\ & = 0.045 \phantom{.} 36 \end{align*}

(i)

\begin{align*} \text{P(R, B, B)} & = {10 \over 26} \times {9 \over 25} \times {8 \over 24} \\ & = {3 \over 65} \end{align*}

(ii)

\begin{align*} \text{P(R, Y, B)} & = {10 \over 26} \times {7 \over 25} \times {9 \over 24} \\ & = {21 \over 520} \end{align*}

(iii)

\begin{align*} \text{P(Three balls of different colours)} & = \text{P(R, Y, B)} + \text{P(R, B, Y)} + \text{P(B, R, Y)} + \text{P(B, Y, R)} + \text{P(Y, R, B)} + \text{P(Y, B, R)} \\ & = 6\left(21 \over 520\right) \\ & = {63 \over 260} \end{align*}

(a)(i)

\begin{align*} \text{P(Game ends on third roll)} & = \text{P(Not '6', not '6', '6')} \\ & = {5 \over 6} \times {5 \over 6} \times {1 \over 6} \\ & = {25 \over 216} \end{align*}

(a)(ii)

\begin{align*} \text{P(Game ends on fourth roll)} & = \text{P(Not '6', not '6', not '6', '6')} \\ & = {5 \over 6} \times {5 \over 6} \times {5 \over 6} \times {1 \over 6} \\ & = {125 \over 1296} \end{align*}

(a)(iii)

\begin{align*} \text{P(Game ends by fourth roll)} & = 1 - \text{P(Game ends after fourth roll)} \\ & = 1 - \left( {5 \over 6} \right)^4 \\ & = {671 \over 1296} \end{align*}

(a)(iii) Alternative solution

\begin{align*} \text{P(Game ends by fourth roll)} & = \text{P(Game ends in 1st roll or 2nd roll or 3rd roll or 4th roll)} \\ & = \text{P('6')} + \text{P(Not '6', '6')} + \text{P(Not '6', not '6', '6')} + \text{P(Not '6', not '6', not '6', '6')} \\ & = {1 \over 6} + {5 \over 6} \times {1 \over 6} + {25 \over 216} + {125 \over 1296} \\ & = {671 \over 1296} \end{align*}

(b)(i)

\begin{align*} \text{P(Game ends on third roll)} & = \text{P('6', not '6', '6')} + \text{P(Not '6', '6', '6')} \\ & = {1 \over 6} \times {5 \over 6} \times {1 \over 6} + {5 \over 6} \times {1 \over 6} \times {1 \over 6} \\ & = {5 \over 108} \end{align*}

(b)(ii) The sum of two '6's is 12 (an even number). Thus for the sum to be an odd number, the third score must be an odd number (1 or 3 or 5).

\begin{align*} \text{P(Game ends on third roll & sum is odd)} & = \text{P('6', odd, '6')} + \text{P(Odd, '6', '6')} \\ & = {1 \over 6} \times {3 \over 6} \times {1 \over 6} + {3 \over 6} \times {1 \over 6} \times {1 \over 6} \\ & = {1 \over 36} \end{align*}