Question

What are the odds in favour of obtaining a sum of 7 when a pair of dice is tossed once?

Answer 1

$\frac{6}{36 \: } \: or \: \frac{1}{6}$

Answer 2

P(E)=$\dfrac{1}{6}$

Step-by-step explanation:

When a pair of dice is tossed once.

All 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), (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) and (6, 6).

Total number of possible outcomes = 36

Let E be the event of getting sum is 7.

Favourable outcomes are:

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

Number of Favourable outcomes = 6

∴ P(E)=$\dfrac{Number of Favourable outcome}{Total number of possible outcomes}$

$=\dfrac{6}{36} =\dfrac{1}{6}$

Hence, P(E)=$\dfrac{1}{6}$