Directions:

A bag contains 5 yellow, 6 red, 3 white and n black balls. The probability of drawing a white ball from the bag is $\Big(\dfrac{1}{7}\Big)$.

Based on this information, answer the following questions:

51.How many black balls are there in the bag?
(a) 6
(b) 7
(c) 8
(d) 9

52.If a ball is picked at random from the bag, what is the probability that the chosen ball is not black?

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

(b) $\Big(\dfrac{2}{7}\Big)$

(c) $\Big(\dfrac{4}{7}\Big)$

(d) $\Big(\dfrac{6}{7}\Big)$

53.If a ball is picked at random from the bag, what is the probability that it is either yellow or black?

(a) $\Big(\dfrac{3}{7}\Big)$

(b) $\Big(\dfrac{4}{7}\Big)$

(c) $\Big(\dfrac{6}{7}\Big)$

(d) $\Big(\dfrac{5}{14}\Big)$

54.If 5 more blue balls are added to the bag and one red ball is removed from it, what is the probability of picking up a red ball if a ball is picked up at random from the bag?

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

(b) $\Big(\dfrac{1}{5}\Big)$

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

(d) $\Big(\dfrac{1}{7}\Big)$

51. Total number of balls in bag = 5 yellow + 6 red + 3 white + n black = 14 + n

Let E be the event of drawing a white ball.

The number of favorable outcomes (white balls) = 3

$\therefore P(E) = \dfrac{\text{Number of white balls}}{\text{Total number of balls}} \\[1em] \dfrac{1}{7} = \dfrac{3}{14 + n} \\[1em] 14 + n = 21 \\[1em] n = 7.$

Hence, option (b) is the correct option.

52. Total number of balls in bag = 21

Let E be the event that the ball is not black.

Number of favorable outcomes (yellow + red + white) = 5 + 6 + 3 = 14

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{14}{21} = \dfrac{2}{3}$

Hence, option (a) is the correct option.

53. Total number of outcomes = 21

Let E be the event that the ball is either yellow or black.

Number of favorable outcomes (5 yellow + 7 black) = 12

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{21} = \dfrac{4}{7}$

Hence, option (b) is the correct option.

54.Total number of balls in bag = 21

After adding 5 blue balls and removing 1 red ball,

Total number of balls in bag = 25

Number of red balls remaining = 5

Let E be the event of picking a red ball.

The number of favorable outcomes = 5

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{25} = \dfrac{1}{5}$

Hence, option (b) is the correct option.