Sasi Math (Revised for 1024)

Scroll:Probability >> Addition theorem on probability >> ps (4499)

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 bag contains 70 bolts and 100 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer:_______________

A bag contains 60 bolts and 150 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer:_______________

A bag contains 70 bolts and 140 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer:_______________

A bag contains 40 bolts and 100 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer:_______________

A bag contains 40 bolts and 150 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer:_______________

A bag contains 60 bolts and 170 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer:_______________

A bag contains 40 bolts and 180 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer:_______________

A bag contains 80 bolts and 140 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer:_______________

A bag contains 60 bolts and 140 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer:_______________

10)

A bag contains 30 bolts and 180 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer:_______________

A bag contains 70 bolts and 100 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer: $\frac{12}{17}$

SOLUTION 1 :

A bag contains 70 bolts and 100 nuts.

n(S) = 170.

If half of the bolts ( 70 bolts ) and half of the nuts ( 50 nuts ) are rusted.

Let A be the event of an item is chosen which is rusted item.

No.of ruted bolts = $\frac{1}{2}$ (70) = 35.

No.of rusted nuts = $\frac{1}{2}$ (100) = 50.

n(A) = 35 + 50 = 85 .

P(A) = n(A) / n(S) = $\frac{85}{170.}$

Let B be the event of an item is chosen which is bolt.

n(B) = 70

P(B) = n(B) / n(S) = $\frac{70}{170.}$

Let A∩B be the event of an item which is rusted bolt.

n(A∩B) = 35

P(A∩B) = n(AB) / n(S) = $\frac{35}{170}$ .

Sinces A and B are not mutully excusive,

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{85}{170}$ + $\frac{70}{170}$ - $\frac{35}{170}$ = 120 / 170

= $\frac{120}{170}$

= $\frac{12}{17}$

A bag contains 60 bolts and 150 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer: $\frac{9}{14}$

SOLUTION 1 :

A bag contains 60 bolts and 150 nuts.

n(S) = 210.

If half of the bolts ( 60 bolts ) and half of the nuts ( 75 nuts ) are rusted.

Let A be the event of an item is chosen which is rusted item.

No.of ruted bolts = $\frac{1}{2}$ (60) = 30.

No.of rusted nuts = $\frac{1}{2}$ (150) = 75.

n(A) = 30 + 75 = 105 .

P(A) = n(A) / n(S) = $\frac{105}{210.}$

Let B be the event of an item is chosen which is bolt.

n(B) = 60

P(B) = n(B) / n(S) = $\frac{60}{210.}$

Let A∩B be the event of an item which is rusted bolt.

n(A∩B) = 30

P(A∩B) = n(AB) / n(S) = $\frac{30}{210}$ .

Sinces A and B are not mutully excusive,

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{105}{210}$ + $\frac{60}{210}$ - $\frac{30}{210}$ = 135 / 210

= $\frac{135}{210}$

= $\frac{9}{14}$

A bag contains 70 bolts and 140 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer: $\frac{2}{3}$

SOLUTION 1 :

A bag contains 70 bolts and 140 nuts.

n(S) = 210.

If half of the bolts ( 70 bolts ) and half of the nuts ( 70 nuts ) are rusted.

Let A be the event of an item is chosen which is rusted item.

No.of ruted bolts = $\frac{1}{2}$ (70) = 35.

No.of rusted nuts = $\frac{1}{2}$ (140) = 70.

n(A) = 35 + 70 = 105 .

P(A) = n(A) / n(S) = $\frac{105}{210.}$

Let B be the event of an item is chosen which is bolt.

n(B) = 70

P(B) = n(B) / n(S) = $\frac{70}{210.}$

Let A∩B be the event of an item which is rusted bolt.

n(A∩B) = 35

P(A∩B) = n(AB) / n(S) = $\frac{35}{210}$ .

Sinces A and B are not mutully excusive,

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{105}{210}$ + $\frac{70}{210}$ - $\frac{35}{210}$ = 140 / 210

= $\frac{140}{210}$

= $\frac{2}{3}$

A bag contains 40 bolts and 100 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer: $\frac{9}{14}$

SOLUTION 1 :

A bag contains 40 bolts and 100 nuts.

n(S) = 140.

If half of the bolts ( 40 bolts ) and half of the nuts ( 50 nuts ) are rusted.

Let A be the event of an item is chosen which is rusted item.

No.of ruted bolts = $\frac{1}{2}$ (40) = 20.

No.of rusted nuts = $\frac{1}{2}$ (100) = 50.

n(A) = 20 + 50 = 70 .

P(A) = n(A) / n(S) = $\frac{70}{140.}$

Let B be the event of an item is chosen which is bolt.

n(B) = 40

P(B) = n(B) / n(S) = $\frac{40}{140.}$

Let A∩B be the event of an item which is rusted bolt.

n(A∩B) = 20

P(A∩B) = n(AB) / n(S) = $\frac{20}{140}$ .

Sinces A and B are not mutully excusive,

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{70}{140}$ + $\frac{40}{140}$ - $\frac{20}{140}$ = 90 / 140

= $\frac{90}{140}$

= $\frac{9}{14}$

A bag contains 40 bolts and 150 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer: $\frac{23}{38}$

SOLUTION 1 :

A bag contains 40 bolts and 150 nuts.

n(S) = 190.

If half of the bolts ( 40 bolts ) and half of the nuts ( 75 nuts ) are rusted.

Let A be the event of an item is chosen which is rusted item.

No.of ruted bolts = $\frac{1}{2}$ (40) = 20.

No.of rusted nuts = $\frac{1}{2}$ (150) = 75.

n(A) = 20 + 75 = 95 .

P(A) = n(A) / n(S) = $\frac{95}{190.}$

Let B be the event of an item is chosen which is bolt.

n(B) = 40

P(B) = n(B) / n(S) = $\frac{40}{190.}$

Let A∩B be the event of an item which is rusted bolt.

n(A∩B) = 20

P(A∩B) = n(AB) / n(S) = $\frac{20}{190}$ .

Sinces A and B are not mutully excusive,

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{95}{190}$ + $\frac{40}{190}$ - $\frac{20}{190}$ = 115 / 190

= $\frac{115}{190}$

= $\frac{23}{38}$

A bag contains 60 bolts and 170 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer: $\frac{29}{46}$

SOLUTION 1 :

A bag contains 60 bolts and 170 nuts.

n(S) = 230.

If half of the bolts ( 60 bolts ) and half of the nuts ( 85 nuts ) are rusted.

Let A be the event of an item is chosen which is rusted item.

No.of ruted bolts = $\frac{1}{2}$ (60) = 30.

No.of rusted nuts = $\frac{1}{2}$ (170) = 85.

n(A) = 30 + 85 = 115 .

P(A) = n(A) / n(S) = $\frac{115}{230.}$

Let B be the event of an item is chosen which is bolt.

n(B) = 60

P(B) = n(B) / n(S) = $\frac{60}{230.}$

Let A∩B be the event of an item which is rusted bolt.

n(A∩B) = 30

P(A∩B) = n(AB) / n(S) = $\frac{30}{230}$ .

Sinces A and B are not mutully excusive,

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{115}{230}$ + $\frac{60}{230}$ - $\frac{30}{230}$ = 145 / 230

= $\frac{145}{230}$

= $\frac{29}{46}$

A bag contains 40 bolts and 180 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer: $\frac{13}{22}$

SOLUTION 1 :

A bag contains 40 bolts and 180 nuts.

n(S) = 220.

If half of the bolts ( 40 bolts ) and half of the nuts ( 90 nuts ) are rusted.

Let A be the event of an item is chosen which is rusted item.

No.of ruted bolts = $\frac{1}{2}$ (40) = 20.

No.of rusted nuts = $\frac{1}{2}$ (180) = 90.

n(A) = 20 + 90 = 110 .

P(A) = n(A) / n(S) = $\frac{110}{220.}$

Let B be the event of an item is chosen which is bolt.

n(B) = 40

P(B) = n(B) / n(S) = $\frac{40}{220.}$

Let A∩B be the event of an item which is rusted bolt.

n(A∩B) = 20

P(A∩B) = n(AB) / n(S) = $\frac{20}{220}$ .

Sinces A and B are not mutully excusive,

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{110}{220}$ + $\frac{40}{220}$ - $\frac{20}{220}$ = 130 / 220

= $\frac{130}{220}$

= $\frac{13}{22}$

A bag contains 80 bolts and 140 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer: $\frac{15}{22}$

SOLUTION 1 :

A bag contains 80 bolts and 140 nuts.

n(S) = 220.

If half of the bolts ( 80 bolts ) and half of the nuts ( 70 nuts ) are rusted.

Let A be the event of an item is chosen which is rusted item.

No.of ruted bolts = $\frac{1}{2}$ (80) = 40.

No.of rusted nuts = $\frac{1}{2}$ (140) = 70.

n(A) = 40 + 70 = 110 .

P(A) = n(A) / n(S) = $\frac{110}{220.}$

Let B be the event of an item is chosen which is bolt.

n(B) = 80

P(B) = n(B) / n(S) = $\frac{80}{220.}$

Let A∩B be the event of an item which is rusted bolt.

n(A∩B) = 40

P(A∩B) = n(AB) / n(S) = $\frac{40}{220}$ .

Sinces A and B are not mutully excusive,

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{110}{220}$ + $\frac{80}{220}$ - $\frac{40}{220}$ = 150 / 220

= $\frac{150}{220}$

= $\frac{15}{22}$

A bag contains 60 bolts and 140 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer: $\frac{13}{20}$

SOLUTION 1 :

A bag contains 60 bolts and 140 nuts.

n(S) = 200.

If half of the bolts ( 60 bolts ) and half of the nuts ( 70 nuts ) are rusted.

Let A be the event of an item is chosen which is rusted item.

No.of ruted bolts = $\frac{1}{2}$ (60) = 30.

No.of rusted nuts = $\frac{1}{2}$ (140) = 70.

n(A) = 30 + 70 = 100 .

P(A) = n(A) / n(S) = $\frac{100}{200.}$

Let B be the event of an item is chosen which is bolt.

n(B) = 60

P(B) = n(B) / n(S) = $\frac{60}{200.}$

Let A∩B be the event of an item which is rusted bolt.

n(A∩B) = 30

P(A∩B) = n(AB) / n(S) = $\frac{30}{200}$ .

Sinces A and B are not mutully excusive,

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{100}{200}$ + $\frac{60}{200}$ - $\frac{30}{200}$ = 130 / 200

= 1 $\frac{30}{200}$

= $\frac{13}{20}$

10)

A bag contains 30 bolts and 180 nuts . Half of the bolts and half of the nuts are ruted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.

Answer: $\frac{4}{7}$

SOLUTION 1 :

A bag contains 30 bolts and 180 nuts.

n(S) = 210.

If half of the bolts ( 30 bolts ) and half of the nuts ( 90 nuts ) are rusted.

Let A be the event of an item is chosen which is rusted item.

No.of ruted bolts = $\frac{1}{2}$ (30) = 15.

No.of rusted nuts = $\frac{1}{2}$ (180) = 90.

n(A) = 15 + 90 = 105 .

P(A) = n(A) / n(S) = $\frac{105}{210.}$

Let B be the event of an item is chosen which is bolt.

n(B) = 30

P(B) = n(B) / n(S) = $\frac{30}{210.}$

Let A∩B be the event of an item which is rusted bolt.

n(A∩B) = 15

P(A∩B) = n(AB) / n(S) = $\frac{15}{210}$ .

Sinces A and B are not mutully excusive,

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{105}{210}$ + $\frac{30}{210}$ - $\frac{15}{210}$ = 120 / 210

= $\frac{120}{210}$

= $\frac{4}{7}$