Question

two different dice are tossed together. find the probability that the product of the two number on the top of the dice is 6.​

Answer 1

Solution: 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

Answer 2

Answer:

Step-by-step explanation:

Given :-

Two different dice are tossed together.

To Find :-

The probability

Solution :-

Two dice are thrown, the outcome is listed as

    1     ;  2    ;  3    ;  4   ;  5    ;   1

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

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

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

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

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

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

Total Number of possible outcomes = 36

The favourable outcome is denoted by E and are (1, 6), (2, 3), (3, 2) and (6, 1)

Number of favourable outcomes = 4

P E = $\sf \frac{Favourable \: Number \: of \: outcomes}{Total \: Number \: of \: otcomes}$

P E = $\sf \frac{4}{36}$

P E = $\bf \frac{1}{9}$

Hence, the probability is 1/9.