Answer to Question #197453 in Statistics and Probability for Sherlock

Question

6. Two ordinary six-sided dice were tossed. Set up a sample space for this experiment and hence find the probability that

i. Sum of the points on the two dice is 7,

ii. Points on the first die are greater than the points on the second die,

iii. First die shows an even number,

iv. Points on the dice are the same, (i.e. throwing a double),

v. Difference in the outcomes of two dice is 2.

Answer 1

Probability – Sample space for two dice (outcomes):

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c :c :c:c } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & (1,1)& (1,2) & (1, 3) & (1, 4) & (1, 5) & (1, 6) \\ \hdashline2 & (2,1)& (2,2) & (2, 3) & (2, 4) & (2, 5) & (2, 6) \\ \hdashline3 & (3,1)& (3,2) & (3, 3) & (3, 4) & (3, 5) & (3, 6) \\ \hdashline4 & (4,1)& (4,2) & (4, 3) & (4, 4) & (4, 5) & (4, 6) \\ \hdashline5 & (5,1)& (5,2) & (5, 3) & (5, 4) & (5, 5) & (5, 6) \\ \hdashline 6 & (6,1)& (6,2) & (6, 3) & (6, 4) & (6, 5) & (6, 6) \\ \end{array}$$

i.

$$1+6=2+5=3+4=4+3=5+2=6+1=7$$

$$P(sum=7)=\dfrac{6}{36}=\dfrac{1}{6}$$

ii.

$$(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5,1), (5, 2),$$

$$(5, 3), (5, 4),(6,1), (6, 2), (6, 3), (6, 4), (6, 5)$$

$$P(first\ is\ greater)=\dfrac{15}{36}=\dfrac{5}{12}$$

iii.

$$(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),$$

$$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),$$

$$6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$$

$$P(first\ is\ even)=\dfrac{18}{36}=\dfrac{1}{2}$$

iv.

$$(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$$

$$P(first\ and\ second\ are \ same)=\dfrac{ 6}{36}=\dfrac{1}{6}$$

v.

$$(1, 3), (2, 4), (3, 5), (4, 6), (3, 1), (4, 2), (5, 3), (6, 4)$$

$$P(difference\ is\ two)=\dfrac{8}{36}=\dfrac{2}{9}$$