Question

Two different dice are tossed together. Find the probability that the product of the two numbers on the top of the dice is 6.​

Answer 1

$\huge\mathbb{SOLUTION:-}$

Two dice are tossed

S = [(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 when two dice are tossed = 6 × 6 = 36

Favourable events of getting product as 6 are:

(1 × 6 = 6),(6 × 1 = 6),(2 × 3 = 6),(3 × 2=6)

i.e. (1,6), (6,1), (2,3), (3,2)

• Favorable events of getting product as 6 = 4.

$\therefore$ P(getting product as 6) = $\frac{4}{36}=\frac{1}{9}$

Answer 2

Answer:

Two dices are thrown no probable chances 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)

How many sets possible? Yeah it's 36.

Out of these 36 sets, we know product should be 6.

In what cases 6 can be product?

→ 6 × 1, 1 × 6, 2 × 3, 3 × 2

→ (6, 1); (1, 6); (2, 3); (3, 2)

How many time these sets came? Yeah it's 4.

Probability = 4/36 = 1/9

Answer is 1/9