Two different dice are thrown together what is the probability of getting a total of 6 or 7

SOLUTION:

The outcomes when two dice are thrown together are

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

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

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

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

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

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

Total number of outcomes = 36 (i) Let A be the event of getting the numbers whose sum is less than 7.

The outcomes in favour of event A are (1, 1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2) and (5,1).

Number of favourable outcomes = 15

∴ P(A ) = Number of favourable outcomesTotal number of outcomes=$\frac{15}{36}$=$\frac{5}{12}$

(ii) Let B be the event of getting the numbers whose product is less than 16. The outcomes in favour of event B are (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (6,1) and (6,2).

Number of favourable outcomes = 25

∴ P(B ) = Number of favourable outcomes/Total number of outcomes=$\frac{25}{36}$ (iii) Let C be the event of getting the numbers which are doublets of odd numbers. The outcomes in favour of event C are (1,1), (3,3) and (5,5). Number of favourable outcomes = 3

∴ P(C ) = Number of favourable outcomes/Total number of outcomes=$\frac{3}{36}$=$\frac{1}{12}$

If two different dice are thrown together, they have numbers 1, 2, 3, 4, 5 and 6 and 1, 2, 3, 4, 5 and 6 on them.

Total number of outcomes -

S = $\left[\right(1,1);(1,2);(1,3);(1,4);(1,5);(1,6);(2,1);(2.2);(2,3);(2,4);(2,5);(2,6);(3,1);(3,2);(3,3);(3,4);(3,5);(3,6);(4,1);(4,2);(4,3);(4,4);(4,5);(4,6);(5,1);(5,2);(5,3);(5,4);(5,5);(5,6);(6,1);(6,2);(6,3);(6,4);(6,5);(6,6\left)\right]$

n(s) = 36

(i) A : the sum of the numbers appeared is less than 7.

Favourable outcomes: $(1,1);(1,2);(1,3);(1,4);(1,5);(2,1);\left(2.2\right);(2,3);(2,4);(3,1);(3,2);(3,3);(4,1);(4,2);(5,1)$

n(A) = 15

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{15}{36}=\frac{5}{12}$

(ii) B: the product of the numbers appeared is less than 18.

Favourable outcomes: $(1,1);(1,2);(1,3);(1,4);(1,5);(1,6);(2,1);(2,2);(2,3);(2,4);(2,5);(2,6);(3,1);(3,2);(3,3);(3,4);(3,5);(4,1);(4,2);(4,3);(4,4);(5,1);(5,2);(5,3);(6,1);(6,2)$

n(B) = 26

$p\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{26}{36}=\frac{13}{18}$

Mathematics

Secondary School Mathematics X

Standard X