Question

Two dice are rolled simultaneously. Find the probability of the following events.
(1) The sum of the digits on the upper surface is a prime number.
(2) The product of the digits on the upper surface is 12.
(3) The digit on the first die is greater than the digit on the second die.​

Answer 1

Step-by-step explanation:

P(A)= 5\12

P(B)= 1\9

P(C)= 5\12

Answer 2

Given: Two dice are rolled simultaneously.

To find: The probability of the following events.

(1) The sum of the digits on the upper surface is a prime number.

(2) The product of the digits on the upper surface is 12.

(3) The digit on the first die is greater than the digit on the second die.​

Solution: The probability of the following events.

(1) The sum of the digits on the upper surface is a prime number is 7/36.

(2) The product of the digits on the upper surface is 12 is 1/9.

(3) The digit on the first die is greater than the digit on the second die 5/12.​

Whenever we calculate the probability of a particular event, we divide the number of favourable outcomes by the total number of outcomes. Since two dice are rolled simultaneously, the total number of outcomes for the three probabilities must be 36 (6*6).

(1)

For the prime number 2, there is only one combination: (1,1). For the prime number 3, there can be two combinations: (1,2) and (2,1). For the prime number 5, there can be four combinations: (1,4), (4,1), (2,3) and (3,2). Thus, in total, the number of possibilities are

$1+2+4=7$

Thus, the probability is

$P(1) = \frac{7}{36}$

(2)

The combinations that can give a product of 12 are (2,6), (6,2), (3,4) and (4,3). There are 4 combinations. Thus, the probability is

$P(2) = \frac{4}{36}$

        $=\frac{1}{9}$

(3)

The combinations such that the digit on the first die is greater than the digit on the second die are: (1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,5), (4,6) and (5,6). There are 15 combinations. Thus, the probability is

$P(3) = \frac{15}{36}$

        $= \frac{5}{12}$

Therefore, the probability of the following events.

(1) The sum of the digits on the upper surface is a prime number is 7/36.

(2) The product of the digits on the upper surface is 12 is 1/9.

(3) The digit on the first die is greater than the digit on the second die 5/12.​