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Two fair dice A and B, each with faces numbered 1 to 6, are thrown together. a) Construct a table showing all outcomes. b) List pairs summing to more than 10, and at least 10. c) Find probability sum

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4 parts
b(i)
(5,6), (6,5), (6,6)
b(ii)
(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)
c(i)

112\frac{1}{12}

c(ii)

16\frac{1}{6}

Step-by-step solution

6 steps

12) Two fair dice A and B, each with faces numbered 1 to 6, are thrown together.

Part a

Construct a table showing all the equally likely outcomes.

The table below shows the outcomes as (Die A, Die B) and their sum.

Die A  Die B1234561(1,1)=2(1,2)=3(1,3)=4(1,4)=5(1,5)=6(1,6)=72(2,1)=3(2,2)=4(2,3)=5(2,4)=6(2,5)=7(2,6)=83(3,1)=4(3,2)=5(3,3)=6(3,4)=7(3,5)=8(3,6)=94(4,1)=5(4,2)=6(4,3)=7(4,4)=8(4,5)=9(4,6)=105(5,1)=6(5,2)=7(5,3)=8(5,4)=9(5,5)=10(5,6)=116(6,1)=7(6,2)=8(6,3)=9(6,4)=10(6,5)=11(6,6)=12\begin{array}{|c|c|c|c|c|c|c|} \hline \textbf{Die A \ Die B} & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6} \\ \hline \textbf{1} & (1,1)=2 & (1,2)=3 & (1,3)=4 & (1,4)=5 & (1,5)=6 & (1,6)=7 \\ \hline \textbf{2} & (2,1)=3 & (2,2)=4 & (2,3)=5 & (2,4)=6 & (2,5)=7 & (2,6)=8 \\ \hline \textbf{3} & (3,1)=4 & (3,2)=5 & (3,3)=6 & (3,4)=7 & (3,5)=8 & (3,6)=9 \\ \hline \textbf{4} & (4,1)=5 & (4,2)=6 & (4,3)=7 & (4,4)=8 & (4,5)=9 & (4,6)=10 \\ \hline \textbf{5} & (5,1)=6 & (5,2)=7 & (5,3)=8 & (5,4)=9 & (5,5)=10 & (5,6)=11 \\ \hline \textbf{6} & (6,1)=7 & (6,2)=8 & (6,3)=9 & (6,4)=10 & (6,5)=11 & (6,6)=12 \\ \hline \end{array}

Part b(i)(5,6), (6,5), (6,6)

More than 10

From your table list the pair of numbers on the two dice for which the sum is

The sums greater than 10 are 11 and 12.
The pairs are:

(5,6),(6,5),(6,6)\boxed{(5,6), (6,5), (6,6)}

Part b(ii)(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)

At least 10

The sums at least 10 are 10, 11, and 12.
The pairs are:

(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)\boxed{(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)}

Part c(i)

112\frac{1}{12}

More than 10

Find the probability that the sum of the numbers on the two dice is

  1. Identify the number of outcomes where the sum is more than 10.

    From part b) i), there are 3 such outcomes: (5,6), (6,5), (6,6).

  2. The total number of possible outcomes is 6×6=366 \times 6 = 36.

  3. Calculate the probability.

    P(sum>10)=Number of outcomes with sum>10Total number of outcomes=336=112P(\text{sum} > 10) = \frac{\text{Number of outcomes with sum} > 10}{\text{Total number of outcomes}} = \frac{3}{36} = \frac{1}{12}

    The probability is 112\boxed{\frac{1}{12}}.

Part c(ii)

16\frac{1}{6}

At least 10

  1. Identify the number of outcomes where the sum is at least 10.

    From part b) ii), there are 6 such outcomes: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6).

  2. The total number of possible outcomes is 36.

  3. Calculate the probability.

    P(sum≥10)=Number of outcomes with sum≥10Total number of outcomes=636=16P(\text{sum} \ge 10) = \frac{\text{Number of outcomes with sum} \ge 10}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}

    The probability is 16\boxed{\frac{1}{6}}.

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