Sasi Math (Revised for 1024)

Scroll:Probability >> Addition theorem on probability >> ps (4500)

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 basket contains 90 apples and 40 oranges out of which 7 apples and 6 oranges are rotten. If a person takes out one fruit at random, find the probability that the fruit is either an apple or a good fruit.

Answer:_______________

A basket contains 80 apples and 30 oranges out of which 6 apples and 4 oranges are rotten. If a person takes out one fruit at random, find the probability that the fruit is either an apple or a good fruit.

Answer:_______________

A basket contains 80 apples and 10 oranges out of which 6 apples and 4 oranges are rotten. If a person takes out one fruit at random, find the probability that the fruit is either an apple or a good fruit.

Answer:_______________

A basket contains 80 apples and 40 oranges out of which 9 apples and 6 oranges are rotten. If a person takes out one fruit at random, find the probability that the fruit is either an apple or a good fruit.

Answer:_______________

A basket contains 90 apples and 10 oranges out of which 5 apples and 3 oranges are rotten. If a person takes out one fruit at random, find the probability that the fruit is either an apple or a good fruit.

Answer:_______________

A basket contains 70 apples and 10 oranges out of which 9 apples and 3 oranges are rotten. If a person takes out one fruit at random, find the probability that the fruit is either an apple or a good fruit.

Answer:_______________

A basket contains 70 apples and 30 oranges out of which 7 apples and 6 oranges are rotten. If a person takes out one fruit at random, find the probability that the fruit is either an apple or a good fruit.

Answer:_______________

A basket contains 90 apples and 20 oranges out of which 7 apples and 4 oranges are rotten. If a person takes out one fruit at random, find the probability that the fruit is either an apple or a good fruit.

Answer:_______________

A basket contains 80 apples and 40 oranges out of which 7 apples and 5 oranges are rotten. If a person takes out one fruit at random, find the probability that the fruit is either an apple or a good fruit.

Answer:_______________

10)

A basket contains 60 apples and 10 oranges out of which 7 apples and 6 oranges are rotten. If a person takes out one fruit at random, find the probability that the fruit is either an apple or a good fruit.

Answer:_______________

Answer: $\frac{62}{65}$

SOLUTION 1 :

Total number of fruit = Apple 90 + orange 40

n(S) = 130.

Let A be the event of getting a fruit which ia apple.

n(A) = 90 .

P(A) = n(A) / n(S) = $\frac{90}{130.}$

Let B be the event of getting a good fruit.

Total rotten furits = 7 apples + 6 oranges = 13

Total good furit = Total fruit - Rotten fruits.

= 130 - 13

n(B) = 117

P(B) = n(B) / n(S) = $\frac{117}{130.}$

A and B are not mutually excusive events.

Let A∩B = 83 ( Numbers of good apples is 83 )

P(A∩B) = n(A∩B) / n(S) = $\frac{83}{130}$ .

P(A or B) = n(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{90}{130}$ + $\frac{117}{130}$ - $\frac{83}{130}$ = 124 / 130

= $\frac{124}{130}$

= $\frac{62}{65}$

Answer: $\frac{53}{55}$

SOLUTION 1 :

Total number of fruit = Apple 80 + orange 30

n(S) = 110.

Let A be the event of getting a fruit which ia apple.

n(A) = 80 .

P(A) = n(A) / n(S) = $\frac{80}{110.}$

Let B be the event of getting a good fruit.

Total rotten furits = 6 apples + 4 oranges = 10

Total good furit = Total fruit - Rotten fruits.

= 110 - 10

n(B) = 100

P(B) = n(B) / n(S) = $\frac{100}{110.}$

A and B are not mutually excusive events.

Let A∩B = 74 ( Numbers of good apples is 74 )

P(A∩B) = n(A∩B) / n(S) = $\frac{74}{110}$ .

P(A or B) = n(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{80}{110}$ + $\frac{100}{110}$ - $\frac{74}{110}$ = 106 / 110

= $\frac{106}{110}$

= $\frac{53}{55}$

Answer: $\frac{43}{45}$

SOLUTION 1 :

Total number of fruit = Apple 80 + orange 10

n(S) = 90.

Let A be the event of getting a fruit which ia apple.

n(A) = 80 .

P(A) = n(A) / n(S) = $\frac{80}{90.}$

Let B be the event of getting a good fruit.

Total rotten furits = 6 apples + 4 oranges = 10

Total good furit = Total fruit - Rotten fruits.

= 90 - 10

n(B) = 80

P(B) = n(B) / n(S) = $\frac{80}{90.}$

A and B are not mutually excusive events.

Let A∩B = 74 ( Numbers of good apples is 74 )

P(A∩B) = n(A∩B) / n(S) = $\frac{74}{90}$ .

P(A or B) = n(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{80}{90}$ + $\frac{80}{90}$ - $\frac{74}{90}$ = 86 / 90

= $\frac{86}{90}$

= $\frac{43}{45}$

Answer: $\frac{19}{20}$

SOLUTION 1 :

Total number of fruit = Apple 80 + orange 40

n(S) = 120.

Let A be the event of getting a fruit which ia apple.

n(A) = 80 .

P(A) = n(A) / n(S) = $\frac{80}{120.}$

Let B be the event of getting a good fruit.

Total rotten furits = 9 apples + 6 oranges = 15

Total good furit = Total fruit - Rotten fruits.

= 120 - 15

n(B) = 105

P(B) = n(B) / n(S) = $\frac{105}{120.}$

A and B are not mutually excusive events.

Let A∩B = 71 ( Numbers of good apples is 71 )

P(A∩B) = n(A∩B) / n(S) = $\frac{71}{120}$ .

P(A or B) = n(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{80}{120}$ + $\frac{105}{120}$ - $\frac{71}{120}$ = 114 / 120

= $\frac{114}{120}$

= $\frac{19}{20}$

Answer: $\frac{97}{100}$

SOLUTION 1 :

Total number of fruit = Apple 90 + orange 10

n(S) = 100.

Let A be the event of getting a fruit which ia apple.

n(A) = 90 .

P(A) = n(A) / n(S) = $\frac{90}{100.}$

Let B be the event of getting a good fruit.

Total rotten furits = 5 apples + 3 oranges = 8

Total good furit = Total fruit - Rotten fruits.

= 100 - 8

n(B) = 92

P(B) = n(B) / n(S) = $\frac{92}{100.}$

A and B are not mutually excusive events.

Let A∩B = 85 ( Numbers of good apples is 85 )

P(A∩B) = n(A∩B) / n(S) = $\frac{85}{100}$ .

P(A or B) = n(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{90}{100}$ + $\frac{92}{100}$ - $\frac{85}{100}$ = 97 / 100

= $\frac{97}{100}$

Answer: $\frac{77}{80}$

SOLUTION 1 :

Total number of fruit = Apple 70 + orange 10

n(S) = 80.

Let A be the event of getting a fruit which ia apple.

n(A) = 70 .

P(A) = n(A) / n(S) = $\frac{70}{80.}$

Let B be the event of getting a good fruit.

Total rotten furits = 9 apples + 3 oranges = 12

Total good furit = Total fruit - Rotten fruits.

= 80 - 12

n(B) = 68

P(B) = n(B) / n(S) = $\frac{68}{80.}$

A and B are not mutually excusive events.

Let A∩B = 61 ( Numbers of good apples is 61 )

P(A∩B) = n(A∩B) / n(S) = $\frac{61}{80}$ .

P(A or B) = n(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{70}{80}$ + $\frac{68}{80}$ - $\frac{61}{80}$ = 77 / 80

= $\frac{77}{80}$

Answer: $\frac{47}{50}$

SOLUTION 1 :

Total number of fruit = Apple 70 + orange 30

n(S) = 100.

Let A be the event of getting a fruit which ia apple.

n(A) = 70 .

P(A) = n(A) / n(S) = $\frac{70}{100.}$

Let B be the event of getting a good fruit.

Total rotten furits = 7 apples + 6 oranges = 13

Total good furit = Total fruit - Rotten fruits.

= 100 - 13

n(B) = 87

P(B) = n(B) / n(S) = $\frac{87}{100.}$

A and B are not mutually excusive events.

Let A∩B = 63 ( Numbers of good apples is 63 )

P(A∩B) = n(A∩B) / n(S) = $\frac{63}{100}$ .

P(A or B) = n(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{70}{100}$ + $\frac{87}{100}$ - $\frac{63}{100}$ = 94 / 100

= $\frac{94}{100}$

= $\frac{47}{50}$

Answer: $\frac{53}{55}$

SOLUTION 1 :

Total number of fruit = Apple 90 + orange 20

n(S) = 110.

Let A be the event of getting a fruit which ia apple.

n(A) = 90 .

P(A) = n(A) / n(S) = $\frac{90}{110.}$

Let B be the event of getting a good fruit.

Total rotten furits = 7 apples + 4 oranges = 11

Total good furit = Total fruit - Rotten fruits.

= 110 - 11

n(B) = 99

P(B) = n(B) / n(S) = $\frac{99}{110.}$

A and B are not mutually excusive events.

Let A∩B = 83 ( Numbers of good apples is 83 )

P(A∩B) = n(A∩B) / n(S) = $\frac{83}{110}$ .

P(A or B) = n(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{90}{110}$ + $\frac{99}{110}$ - $\frac{83}{110}$ = 106 / 110

= $\frac{106}{110}$

= $\frac{53}{55}$

Answer: $\frac{23}{24}$

SOLUTION 1 :

Total number of fruit = Apple 80 + orange 40

n(S) = 120.

Let A be the event of getting a fruit which ia apple.

n(A) = 80 .

P(A) = n(A) / n(S) = $\frac{80}{120.}$

Let B be the event of getting a good fruit.

Total rotten furits = 7 apples + 5 oranges = 12

Total good furit = Total fruit - Rotten fruits.

= 120 - 12

n(B) = 108

P(B) = n(B) / n(S) = $\frac{108}{120.}$

A and B are not mutually excusive events.

Let A∩B = 73 ( Numbers of good apples is 73 )

P(A∩B) = n(A∩B) / n(S) = $\frac{73}{120}$ .

P(A or B) = n(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{80}{120}$ + $\frac{108}{120}$ - $\frac{73}{120}$ = 115 / 120

= $\frac{115}{120}$

= $\frac{23}{24}$

10)

Answer: $\frac{32}{35}$

SOLUTION 1 :

Total number of fruit = Apple 60 + orange 10

n(S) = 70.

Let A be the event of getting a fruit which ia apple.

n(A) = 60 .

P(A) = n(A) / n(S) = $\frac{60}{70.}$

Let B be the event of getting a good fruit.

Total rotten furits = 7 apples + 6 oranges = 13

Total good furit = Total fruit - Rotten fruits.

= 70 - 13

n(B) = 57

P(B) = n(B) / n(S) = $\frac{57}{70.}$

A and B are not mutually excusive events.

Let A∩B = 53 ( Numbers of good apples is 53 )

P(A∩B) = n(A∩B) / n(S) = $\frac{53}{70}$ .

P(A or B) = n(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B) .

= $\frac{60}{70}$ + $\frac{57}{70}$ - $\frac{53}{70}$ = 64 / 70

= $\frac{64}{70}$

= $\frac{32}{35}$