Sasi Math (Revised for 1024)

Scroll:Probability >> probability >> ps (4492)

Written Instructions:

For each Multiple Choice Question (MCQ), four options are given. One of them is the correct answer. Make your choice (1,2,3 or 4). Write your answers in the brackets provided..

For each Short Answer Question(SAQ) and Long Answer Question(LAQ), write your answers in the blanks provided.

Leave your answers in the simplest form or correct to two decimal places.

A jar contains 160 marbles each of the colours blue, green and white. The probility of drawing a blue marbles is $\frac{2}{8}$ and the probability of drawing a green marble is $\frac{5}{16.}$ How many white marbles does the jar contain

Answer:_______________

A jar contains 144 marbles each of the colours blue, green and white. The probility of drawing a blue marbles is $\frac{1}{4}$ and the probability of drawing a green marble is $\frac{7}{8.}$ How many white marbles does the jar contain

Answer:_______________

A jar contains 144 marbles each of the colours blue, green and white. The probility of drawing a blue marbles is $\frac{2}{12}$ and the probability of drawing a green marble is $\frac{5}{24.}$ How many white marbles does the jar contain

Answer:_______________

A jar contains 140 marbles each of the colours blue, green and white. The probility of drawing a blue marbles is $\frac{2}{10}$ and the probability of drawing a green marble is $\frac{7}{20.}$ How many white marbles does the jar contain

Answer:_______________

A jar contains 108 marbles each of the colours blue, green and white. The probility of drawing a blue marbles is $\frac{2}{6}$ and the probability of drawing a green marble is $\frac{5}{12.}$ How many white marbles does the jar contain

Answer:_______________

A jar contains 100 marbles each of the colours blue, green and white. The probility of drawing a blue marbles is $\frac{1}{10}$ and the probability of drawing a green marble is $\frac{7}{20.}$ How many white marbles does the jar contain

Answer:_______________

A jar contains 120 marbles each of the colours blue, green and white. The probility of drawing a blue marbles is $\frac{1}{6}$ and the probability of drawing a green marble is $\frac{5}{12.}$ How many white marbles does the jar contain

Answer:_______________

A jar contains 176 marbles each of the colours blue, green and white. The probility of drawing a blue marbles is $\frac{1}{8}$ and the probability of drawing a green marble is $\frac{7}{16.}$ How many white marbles does the jar contain

Answer:_______________

A jar contains 126 marbles each of the colours blue, green and white. The probility of drawing a blue marbles is $\frac{1}{9}$ and the probability of drawing a green marble is $\frac{5}{18.}$ How many white marbles does the jar contain

Answer:_______________

10)

A jar contains 112 marbles each of the colours blue, green and white. The probility of drawing a blue marbles is $\frac{1}{4}$ and the probability of drawing a green marble is $\frac{7}{8.}$ How many white marbles does the jar contain

Answer:_______________

Answer: 70

SOLUTION 1 :

A jar contains 160 marbles each of which is blue, green and white.

n(S) = 160.

w.k.t P(S) = 1.

Given:

Probability of drawing blue marble, P(B) = $\frac{2}{8}$ .

Probability of drawing green marbles, P(G) = $\frac{5}{16.}$

Let X be the number of white marbles, n(W) = x.

⇒ P(G) = n(G) / n(S) = $\frac{x}{160}$

⇒ P(S) = 1.

p(B) + P(G) + P(W) = 1

$\frac{2}{8}$ + $\frac{5}{16}$ + $\frac{x}{160}$ = 1

40 + 50 + x ÷ 160 = 1 LCM = 160

90 + x = 160

x = 160 - 90

x = 70.

Number of white balls = 70

Answer: -18

SOLUTION 1 :

A jar contains 144 marbles each of which is blue, green and white.

n(S) = 144.

w.k.t P(S) = 1.

Given:

Probability of drawing blue marble, P(B) = $\frac{1}{4}$ .

Probability of drawing green marbles, P(G) = $\frac{7}{8.}$

Let X be the number of white marbles, n(W) = x.

⇒ P(G) = n(G) / n(S) = $\frac{x}{144}$

⇒ P(S) = 1.

p(B) + P(G) + P(W) = 1

$\frac{1}{4}$ + $\frac{7}{8}$ + $\frac{x}{144}$ = 1

36 + 126 + x ÷ 144 = 1 LCM = 144

162 + x = 144

x = 144 - 162

x = -18.

Number of white balls = -18

Answer: 90

SOLUTION 1 :

A jar contains 144 marbles each of which is blue, green and white.

n(S) = 144.

w.k.t P(S) = 1.

Given:

Probability of drawing blue marble, P(B) = $\frac{2}{12}$ .

Probability of drawing green marbles, P(G) = $\frac{5}{24.}$

Let X be the number of white marbles, n(W) = x.

⇒ P(G) = n(G) / n(S) = $\frac{x}{144}$

⇒ P(S) = 1.

p(B) + P(G) + P(W) = 1

$\frac{2}{12}$ + $\frac{5}{24}$ + $\frac{x}{144}$ = 1

24 + 30 + x ÷ 144 = 1 LCM = 144

54 + x = 144

x = 144 - 54

x = 90.

Number of white balls = 90

Answer: 63

SOLUTION 1 :

A jar contains 140 marbles each of which is blue, green and white.

n(S) = 140.

w.k.t P(S) = 1.

Given:

Probability of drawing blue marble, P(B) = $\frac{2}{10}$ .

Probability of drawing green marbles, P(G) = $\frac{7}{20.}$

Let X be the number of white marbles, n(W) = x.

⇒ P(G) = n(G) / n(S) = $\frac{x}{140}$

⇒ P(S) = 1.

p(B) + P(G) + P(W) = 1

$\frac{2}{10}$ + $\frac{7}{20}$ + $\frac{x}{140}$ = 1

28 + 49 + x ÷ 140 = 1 LCM = 140

77 + x = 140

x = 140 - 77

x = 63.

Number of white balls = 63

Answer: 27

SOLUTION 1 :

A jar contains 108 marbles each of which is blue, green and white.

n(S) = 108.

w.k.t P(S) = 1.

Given:

Probability of drawing blue marble, P(B) = $\frac{2}{6}$ .

Probability of drawing green marbles, P(G) = $\frac{5}{12.}$

Let X be the number of white marbles, n(W) = x.

⇒ P(G) = n(G) / n(S) = $\frac{x}{108}$

⇒ P(S) = 1.

p(B) + P(G) + P(W) = 1

$\frac{2}{6}$ + $\frac{5}{12}$ + $\frac{x}{108}$ = 1

36 + 45 + x ÷ 108 = 1 LCM = 108

81 + x = 108

x = 108 - 81

x = 27.

Number of white balls = 27

Answer: 55

SOLUTION 1 :

A jar contains 100 marbles each of which is blue, green and white.

n(S) = 100.

w.k.t P(S) = 1.

Given:

Probability of drawing blue marble, P(B) = $\frac{1}{10}$ .

Probability of drawing green marbles, P(G) = $\frac{7}{20.}$

Let X be the number of white marbles, n(W) = x.

⇒ P(G) = n(G) / n(S) = $\frac{x}{100}$

⇒ P(S) = 1.

p(B) + P(G) + P(W) = 1

$\frac{1}{10}$ + $\frac{7}{20}$ + $\frac{x}{100}$ = 1

10 + 35 + x ÷ 100 = 1 LCM = 100

45 + x = 100

x = 100 - 45

x = 55.

Number of white balls = 55

Answer: 50

SOLUTION 1 :

A jar contains 120 marbles each of which is blue, green and white.

n(S) = 120.

w.k.t P(S) = 1.

Given:

Probability of drawing blue marble, P(B) = $\frac{1}{6}$ .

Probability of drawing green marbles, P(G) = $\frac{5}{12.}$

Let X be the number of white marbles, n(W) = x.

⇒ P(G) = n(G) / n(S) = $\frac{x}{120}$

⇒ P(S) = 1.

p(B) + P(G) + P(W) = 1

$\frac{1}{6}$ + $\frac{5}{12}$ + $\frac{x}{120}$ = 1

20 + 50 + x ÷ 120 = 1 LCM = 120

70 + x = 120

x = 120 - 70

x = 50.

Number of white balls = 50

Answer: 77

SOLUTION 1 :

A jar contains 176 marbles each of which is blue, green and white.

n(S) = 176.

w.k.t P(S) = 1.

Given:

Probability of drawing blue marble, P(B) = $\frac{1}{8}$ .

Probability of drawing green marbles, P(G) = $\frac{7}{16.}$

Let X be the number of white marbles, n(W) = x.

⇒ P(G) = n(G) / n(S) = $\frac{x}{176}$

⇒ P(S) = 1.

p(B) + P(G) + P(W) = 1

$\frac{1}{8}$ + $\frac{7}{16}$ + $\frac{x}{176}$ = 1

22 + 77 + x ÷ 176 = 1 LCM = 176

99 + x = 176

x = 176 - 99

x = 77.

Number of white balls = 77

Answer: 77

SOLUTION 1 :

A jar contains 126 marbles each of which is blue, green and white.

n(S) = 126.

w.k.t P(S) = 1.

Given:

Probability of drawing blue marble, P(B) = $\frac{1}{9}$ .

Probability of drawing green marbles, P(G) = $\frac{5}{18.}$

Let X be the number of white marbles, n(W) = x.

⇒ P(G) = n(G) / n(S) = $\frac{x}{126}$

⇒ P(S) = 1.

p(B) + P(G) + P(W) = 1

$\frac{1}{9}$ + $\frac{5}{18}$ + $\frac{x}{126}$ = 1

14 + 35 + x ÷ 126 = 1 LCM = 126

49 + x = 126

x = 126 - 49

x = 77.

Number of white balls = 77

10)

Answer: -14

SOLUTION 1 :

A jar contains 112 marbles each of which is blue, green and white.

n(S) = 112.

w.k.t P(S) = 1.

Given:

Probability of drawing blue marble, P(B) = $\frac{1}{4}$ .

Probability of drawing green marbles, P(G) = $\frac{7}{8.}$

Let X be the number of white marbles, n(W) = x.

⇒ P(G) = n(G) / n(S) = $\frac{x}{112}$

⇒ P(S) = 1.

p(B) + P(G) + P(W) = 1

$\frac{1}{4}$ + $\frac{7}{8}$ + $\frac{x}{112}$ = 1

28 + 98 + x ÷ 112 = 1 LCM = 112

126 + x = 112

x = 112 - 126

x = -14.

Number of white balls = -14