(i)

\begin{align*} \text{P('3', Head)} & = {1 \over 6} \times {1 \over 2} \\ & = {1 \over 12} \end{align*}

(ii)

\begin{align*} \text{P(Even number, tail)} & = {3 \over 6} \times {1 \over 2} \\ & = {1 \over 4} \end{align*}

(i)

\begin{align*} \text{P(Same number)} & = \text{P(1, 1)} + \text{P(2, 2)} + \text{P(3, 3)} + \text{P(4, 4)} + \text{P(5, 5)} + \text{P(6, 6)} \\ & = {1 \over 6} \times {1 \over 6} + {1 \over 6} \times {1 \over 6} + {1 \over 6} \times {1 \over 6} + {1 \over 6} \times {1 \over 6} + {1 \over 6} \times {1 \over 6} + {1 \over 6} \times {1 \over 6} \\ & = {1 \over 6} \end{align*}

(ii)

\begin{align*} \text{P(Two even numbers)} & = {3 \over 6} \times {3 \over 6} \\ & = {1 \over 4} \end{align*}

(iii)

\begin{align*} \text{P(Two odd numbers)} & = {3 \over 6} \times {3 \over 6} \\ & = {1 \over 4} \end{align*}

(iv)

\begin{align*} \text{P(One odd and one even number)} & = \text{P(1st dice odd, 2nd dice even)} + \text{P(1st dice even, 2nd dice odd)} \\ & = {3 \over 6} \times {3 \over 6} + {3 \over 6} \times {3 \over 6} \\ & = {1 \over 2} \end{align*}

(i)

\begin{align*} \text{P(Number greater than 28)} & = {50 - 28 \over 50} \\ & = {11 \over 25} \end{align*}

(ii)

\begin{align*} \text{P(Number includes digit '3')} & = \text{P('3', '13', '23', '43')} + \text{P(All numbers between '30' and '39' inclusive)} \\ & = {4 \over 50} + {10 \over 50} \\ & = {7 \over 25} \end{align*}

(iii) A prime number has only two factors, 1 and itself.

\begin{align*} \text{P(Number is prime)} & = \text{P('2', '3', '5', '7', '11', '13', '17', '19', '23', '29', '31', '37', '41', '43', '47')} \\ & = {15 \over 50} \\ & = {3 \over 10} \end{align*}

(iv)

\begin{align*} \text{P(Number is divisible by 4)} & = \text{P('4', '8', '12', '16', '20', '24', '28', '32', '36', '40', '44', '48')} \\ & = {12 \over 50} \\ & = {6 \over 25} \end{align*}

(i)

\begin{align*} \text{P(Two people born in same month)} & = \text{P(Jan, Jan)} + \text{P(Feb, Feb)} + ... + \text{P(Dec, Dec)} \phantom{0000} [\text{12 possible months}] \\ & = 12 \left( {1 \over 12} \times {1 \over 12} \right) \\ & = {1 \over 12} \end{align*}

(ii)

\begin{align*} \text{P(Three people selected not born in the same month)} & = 1 - \text{P(Three people selected born in the same month)} \\ & = 1 - 12 \left( {1 \over 12} \times {1 \over 12} \times {1 \over 12} \right) \\ & = {143 \over 144} \end{align*}

(iii)

\begin{align*} \text{P(Four people selected born in the same month)} & = 12 \left({1 \over 12} \times {1 \over 12} \times {1 \over 12} \times {1 \over 12} \right) \\ & = {1 \over 1728} \end{align*}

(i)

\begin{align*} \text{P(Huixian catch the bus)} & = 1 - {1 \over 7} \\ & = {6 \over 7} \end{align*}

(ii)

\begin{align*} \text{P(Miss on two particular consecutive days)} & = {1 \over 7} \times {1 \over 7} \\ & = {1 \over 49} \end{align*}

(iii)

\begin{align*} \text{P(Miss on just one of two particular consecutive days)} & = \text{P(Miss, catch)} + \text{P(Catch, miss)} \\ & = {1 \over 7} \times {6 \over 7} + {6 \over 7} \times {1 \over 7} \\ & = {12 \over 49} \end{align*}

(iv)

\begin{align*} \text{P(Catch her bus on three particular consecutive days)} & = {6 \over 7} \times {6 \over 7} \times {6 \over 7} \\ & = {216 \over 343} \end{align*}

(i) Note: There can only be 1 gold medal winner!

\begin{align*} \text{P(One of them wins the gold medal)} & = \text{P(Rui Feng win)} + \text{P(Michael win)} + \text{P(Khairul win)} \\ & = {1 \over 2} + {1 \over 6} + {1 \over 8} \\ & = {19 \over 24} \end{align*}

(ii)

\begin{align*} \text{P(None of them wins the gold medal)} & = 1 - \text{P(One of them wins the gold medal)} \\ & = 1 - {19 \over 24} \\ & = {5 \over 24} \end{align*}

(iii)

\begin{align*} \text{P(Rui Feng fails to win the gold medal)} & = 1 - \text{P(Rui Feng wins the gold medal)} \\ & = 1 - {1 \over 2} \\ & = {1 \over 2} \end{align*}

(i)

(ii)

\begin{align*} \text{P(Number is divisible by 5)} & = {5 \over 30} \\ & = {1 \over 6} \end{align*}

(iii) A prime number has only two factors, 1 and itself. Thus the relevant prime numbers are 13, 23, 31, 41, 43, 53 and 61.

\begin{align*} \text{P(Number is a prime number)} & = {7 \over 30} \end{align*}

(iv) The perfect squares are 16, 25, 36 and 64.

\begin{align*} \text{P(Number is a perfect square)} & = {4 \over 30} \\ & = {2 \over 15} \end{align*}

(i)

\begin{align*} \text{P('O', 'O')} & = {2 \over 6} \times {1 \over 5} \\ & = {1 \over 15} \end{align*}

(ii) Since there's only 1 card with F, for the second card to be labelled with F, the first card must not be labelled with F.

\begin{align*} \text{P(Not 'F', 'F)} & = {5 \over 6} \times {1 \over 5} \\ & = {1 \over 6} \end{align*}

(iii)

\begin{align*} \text{P('FOLLOW' obtained)} & = {1 \over 6} \times {2 \over 5} \times {2 \over 4} \times {1 \over 3} \times {1 \over 2} \times {1 \over 1} \\ & = {1 \over 180} \end{align*}

(i)

\begin{align*} \text{P(Khairul selects a dark chocolate)} & = {y \over x + y} \end{align*}

(ii)

\begin{align*} \text{P(Khairul selects a white chocolate, Priya selects a dark chocolate)} & = {x \over x + y} \times {y \over x + y - 1} \\ & = {xy \over (x + y)(x + y - 1)} \end{align*}

(iii)

\begin{align*} \text{P(Khairul and Priya select different types)} & = \text{P(Khairul selects dark, Priya selects white)} + \text{P(Khairul selects white, Priya selects dark)} \\ & = {xy \over (x + y)(x + y - 1)} + {y \over x + y} \times {x \over x + y - 1} \\ & = {xy \over (x + y)(x + y - 1)} + {xy \over (x + y)(x + y - 1)} \\ & = {2xy \over (x + y)(x + y - 1)} \end{align*}

\begin{align*} \text{P(Traffic jam)} & = \text{P(Rain and traffic jam)} + \text{P(No rain and traffic jam)} \\ & = {1 \over 4} \times {2 \over 5} + \left(1 - {1 \over 4}\right) \times {1 \over 5} \\ & = {1 \over 4} \end{align*}

(a)(i)

\begin{align*} \text{P(One girl and one boy)} & = \text{P(G, B)} + \text{P(B, G)} \\ & = {10 \over 30} \times {20 \over 29} + {20 \over 30} \times {10 \over 29} \\ & = {40 \over 87} \end{align*}

(a)(ii)

\begin{align*} \text{P(No girls selected)} & = \text{P(B, B)} \\ & = {20 \over 30} \times {19 \over 29} \\ & = {38 \over 87} \end{align*}

(b)(i)

\begin{align*} \text{P(Both travel to school by bus)} & = {6 \over 10} \times {5 \over 9} \\ & = {1 \over 3} \end{align*}

(b)(ii)

\begin{align*} \text{P(Both travel to school by differnt means)} & = \text{P(Bus, car)} + \text{P(Car, bus)} \\ & = {6 \over 10} \times {4 \over 9} + {4 \over 10} \times {6 \over 9} \\ & = {8 \over 15} \end{align*}

(b)(iii)

\begin{align*} \text{P(At least one travels to school by bus)} & = 1 - \text{P(None travel to school by bus)} \\ & = 1 - \text{P(Both travel to school by car)} \\ & = 1 - {4 \over 10} \times {3 \over 9} \\ & = {13 \over 15} \end{align*}

(i)

\begin{align*} \text{P(Wet for next two days)} & = 0.6 \times 0.6 \\ & = 0.36 \end{align*}

(ii)

\begin{align*} \text{P(Tue wet, Wed fine)} & = 0.6 \times (1 - 0.6) \\ & = 0.24 \end{align*}

(iii)

\begin{align*} \text{P(One fine, one wet)} & = \text{P(Tue wet, Wed fine)} + \text{P(Tue fine, Wed wet)} \\ & = 0.24 + (1 - 0.6) (1 - 0.8) \\ & = 0.32 \end{align*}

(iv)

\begin{align*} \text{P(Two of next three days wet)} & = \text{P(Wet, wet, fine)} + \text{P(Wet, fine, wet)} + \text{P(Fine, wet, wet)} \\ & = 0.36(1 - 0.6) + 0.24(1 - 0.8) + (1 - 0.6)(1 - 0.8)(0.6) \\ & = 0.24 \end{align*}

(i)

\begin{align*} \text{P(First two sweets are different)} & = 1 - \text{P(First two sweets are same)} \\ & = 1 - [ \text{P(F, F)} + \text{P(M, M)} + \text{P(T, T)} ] \\ & = 1 - \left[ {2 \over 10} \times {1 \over 9} + {3 \over 10} \times {2 \over 9} + {5 \over 10} \times {4 \over 9} \right] \\ & = {31 \over 45} \end{align*}

(ii) Since there are only 2 fruit gums, it is not possible to choose 3 fruit gums in a row. In other words, it is only possible to pick 3 mints in a row or 3 toffees in a row.

\begin{align*} \text{P(First three sweets are same)} & = \text{P(M, M, M)} + \text{P(T, T, T)} \\ & = {3 \over 10} \times {2 \over 9} \times {1 \over 8} + {5 \over 10} \times {4 \over 9} \times {3 \over 8} \\ & = {11 \over 120} \end{align*}

(iii)

\begin{align*} \text{P(First two sweets are same, third will be a toffee)} & = \text{P(F, F, T)} + \text{P(M, M, T)} + \text{P(T, T, T)} \\ & = {2 \over 10} \times {1 \over 9} \times {5 \over 8} + {3 \over 10} \times {2 \over 9} \times {5 \over 8} + {5 \over 10} \times {4 \over 9} \times {3 \over 8} \\ & = {5 \over 36} \end{align*}

(i)

\begin{align*} \text{P(All three at Terminal 2)} & = {1 \over 4} \times {1 \over 3} \times {1 \over 6} \\ & = {1 \over 72} \end{align*}

(ii)

\begin{align*} \text{P(Exactly two land at Terminal 1)} & = \text{P(T1, T1, T2)} + \text{P(T1, T2, T1)} + \text{P(T2, T1, T1)} \\ & = {3 \over 4} \times {2 \over 3} \times {1 \over 6} + {3 \over 4} \times {1 \over 3} \times {5 \over 6} + {1 \over 4} \times {2 \over 3} \times {5 \over 6} \\ & = {31 \over 72} \end{align*}

(iii)

\begin{align*} \text{P(Exactly one land at Terminal 1)} & = \text{P(T1, T2, T2)} + \text{P(T2, T2, T1)} + \text{P(T2, T1, T2)} \\ & = {3 \over 4} \times {1 \over 3} \times {1 \over 6} + {1 \over 4} \times {1 \over 3} \times {5 \over 6} + {1 \over 4} \times {2 \over 3} \times {1 \over 6} \\ & = {5 \over 36} \end{align*}

(i)

\begin{align*} \text{P(R, R)} & = {5 \over 13} \times {1 \over 3} \\ & = {5 \over 39} \end{align*}

(ii)

\begin{align*} \text{P(R, Y)} & = {5 \over 13} \times {7 \over 12} \\ & = {35 \over 156} \end{align*}

(iii) Since there's only 1 white disc in the bag, it is not possible to obtain two white discs.

\begin{align*} \text{P(W, W)} & = 0 \end{align*}

(iv)

\begin{align*} \text{P(Two discs of different colours)} & = 1 - \text{P(Two discs of same colour)} \\ & = 1 - [ \text{P(R, R)} + \text{P(Y, Y)} ] \\ & = 1 - \left[ {5 \over 39} + {7 \over 13} \times {1 \over 2} \right] \\ & = {47 \over 78} \end{align*}

(a)(i)

\begin{align*} \text{P(All three men hit the target)} & = {2 \over 3} \times {3 \over 5} \times {4 \over 7} \\ & = {8 \over 35} \end{align*}

(a)(ii)

\begin{align*} \text{P(All three men miss the target)} & = \left(1 - {2 \over 3}\right) \left(1 - {3 \over 5}\right) \left(1 - {4 \over 7}\right) \\ & = {2 \over 35} \end{align*}

(a)(iii)

\begin{align*} \text{P(Exactly two of them hit the target)} & = \text{P(R hit, RF hit, F miss)} + \text{P(R hit, RF miss, F hit)} + \text{P(R miss, RF hit, F hit)} \\ & = {2 \over 3} \times {3 \over 5} \times \left(1 - {4 \over 7}\right) + {2 \over 3} \times \left(1 - {3 \over 5}\right) \times {4 \over 7} + \left(1 - {2 \over 3}\right) \times {3 \over 5} \times {4 \over 7} \\ & = {46 \over 105} \end{align*}

(a)(iv)

\begin{align*} \text{P(At least one man hits the target)} & = 1 - \text{P(All three men miss the target)} \\ & = 1 - {2 \over 35} \\ & = {33 \over 35} \end{align*}

(b)(i)

\begin{align*} \text{P(Game ends after two shots)} & = \text{P(R miss, RF hit)} \\ & = \left(1 - {2 \over 3}\right) \times {3 \over 5} \\ & = {1 \over 5} \end{align*}

(b)(ii)

\begin{align*} \text{P(Game ends after three shots)} & = \text{P(R miss, RF miss, F hit)} \\ & = \left(1 - {2 \over 3}\right) \times \left(1 - {3 \over 5}\right) \times {4 \over 7} \\ & = {8 \over 105} \end{align*}

(b)(iii)

\begin{align*} \text{P(Game ends by third shot)} & = \text{P(Game ends in first shot or second shot or third shot)} \\ & = \text{P(R hit)} + \text{P(R miss, RF hit)} + \text{P(R miss, RF miss, F hit)} \\ & = {2 \over 3} + {1 \over 5} + {8 \over 105} \\ & = {33 \over 35} \end{align*}