Directions:

Two different dice are rolled together.

Based on this information, answer the following questions:

47. The probability that the sum of the two numbers on the top of the dice is at least 10 is:

(a) $\Big(\dfrac{1}{2}\Big)$

(b) $\Big(\dfrac{1}{3}\Big)$

(c) $\Big(\dfrac{1}{4}\Big)$

(d) $\Big(\dfrac{1}{6}\Big)$

48.The probability of getting two different numbers on the two dice is:

(a) $\Big(\dfrac{1}{6}\Big)$

(b) $\Big(\dfrac{5}{6}\Big)$

(c) $\Big(\dfrac{1}{4}\Big)$

(d) $\Big(\dfrac{3}{4}\Big)$

49.The probability that the product of the two numbers on the top of the dice is at most 12 is:

(a) $\Big(\dfrac{1}{6}\Big)$

(b) $\Big(\dfrac{2}{3}\Big)$

(c) $\Big(\dfrac{23}{36}\Big)$

(d) $\Big(\dfrac{25}{36}\Big)$

50.The probability of getting a multiple of 2 on one die and a multiple of 3 on the other is:

(a) $\Big(\dfrac{1}{3}\Big)$

(b) $\Big(\dfrac{5}{18}\Big)$

(c) $\Big(\dfrac{11}{36}\Big)$

(d) $\Big(\dfrac{13}{36}\Big)$

47.When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting the the sum is at least 10,

E = {(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}

The number of favorable outcomes to the event E = 6

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{36} = \dfrac{1}{6}$

Hence, option (d) is the correct option.

48. When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let A be the event of getting the same number on both dice (doubles),

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

Number of outcomes for A = 6

Let E be the event of getting two different numbers on the two dice,

Number of favorable outcomes to the event E = 36 - 6 = 30

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{30}{36} = \dfrac{5}{6}$

Hence, option (b) is the correct option.

49. When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event that the product of the two numbers is at most 12,

E = {(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), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (6, 1), (6, 2)}

The number of favorable outcomes to the event E = 23

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{23}{36}$

Hence, option (c) is the correct option.

50. When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting a multiple of 2 on one die and a multiple of 3 on the other,

E = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (3, 4), (3, 6), (6, 2), (6, 4)}

The number of favorable outcomes to the event E = 11

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{11}{36}$

Hence, option (c) is the correct option.