question_answer

Two different dice are thrown together. Find the probability that the numbers obtained

(i) have a sum less than 7

(ii) have a product less than 16

(iii) is a doublet of odd numbers.

Answer:

Total possible outcomes in each case \[=6\times 6=36\]
(i) Have a sum less than 7,
Possible outcomes 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)
(5, 1)
\[n(E)=15\]
So, probability \[=\frac{15}{36}=\frac{5}{12}\]
(ii) Have a product less than 16,
Possible outcomes 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) (6, 2)
\[n(E)=25\]
So, probability \[=\frac{25}{36}\]
(iii) Is a doublet of odd no.,
Possible outcomes are
(1, 1), (3, 3), (5, 5)
\[n(E)=3\]
P (doublet of odd no.) \[=\frac{3}{36}=\frac{1}{12}\]