Sasi Math (Revised for 1024)

Scroll:Probability >> Addition theorem on probability >> saq (4496)

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.

If A and B are two events such that P(A) = $\frac{1}{5}$ , P(B) = $\frac{5}{2}$ and P(A∪B) = $\frac{2}{8}$ , then find P(A∩B).

P(A∩B) =

Answer:_______________

If A and B are two events such that P(A) = $\frac{2}{3}$ , P(B) = $\frac{5}{2}$ and P(A∪B) = $\frac{2}{8}$ , then find P(A∩B).

P(A∩B) =

Answer:_______________

If A and B are two events such that P(A) = $\frac{2}{3}$ , P(B) = $\frac{3}{2}$ and P(A∪B) = $\frac{3}{8}$ , then find P(A∩B).

P(A∩B) =

Answer:_______________

If A and B are two events such that P(A) = $\frac{1}{5}$ , P(B) = $\frac{3}{4}$ and P(A∪B) = $\frac{2}{8}$ , then find P(A∩B).

P(A∩B) =

Answer:_______________

If A and B are two events such that P(A) = $\frac{2}{3}$ , P(B) = $\frac{3}{2}$ and P(A∪B) = $\frac{2}{8}$ , then find P(A∩B).

P(A∩B) =

Answer:_______________

If A and B are two events such that P(A) = $\frac{1}{3}$ , P(B) = $\frac{5}{6}$ and P(A∪B) = $\frac{2}{8}$ , then find P(A∩B).

P(A∩B) =

Answer:_______________

If A and B are two events such that P(A) = $\frac{1}{3}$ , P(B) = $\frac{5}{4}$ and P(A∪B) = $\frac{3}{8}$ , then find P(A∩B).

P(A∩B) =

Answer:_______________

If A and B are two events such that P(A) = $\frac{1}{3}$ , P(B) = $\frac{3}{4}$ and P(A∪B) = $\frac{2}{8}$ , then find P(A∩B).

P(A∩B) =

Answer:_______________

If A and B are two events such that P(A) = $\frac{2}{5}$ , P(B) = $\frac{3}{2}$ and P(A∪B) = $\frac{2}{8}$ , then find P(A∩B).

P(A∩B) =

Answer:_______________

10)

If A and B are two events such that P(A) = $\frac{1}{5}$ , P(B) = $\frac{3}{4}$ and P(A∪B) = $\frac{3}{8}$ , then find P(A∩B).

P(A∩B) =

Answer:_______________

If A and B are two events such that P(A) = $\frac{1}{5}$ , P(B) = $\frac{5}{2}$ and P(A∪B) = $\frac{2}{8}$ , then find P(A∩B).

P(A∩B) = Answer: $\frac{49}{20}$

SOLUTION 1 :

Given:

P(A) = $\frac{1}{5}$ , P(B) = $\frac{5}{2}$ and P(A∪B) = $\frac{2}{8}$ .

By Addition theorem on probability.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∩B) = P(A) + P(B) - P(A∪B)

= $\frac{1}{5}$ + $\frac{5}{2}$ - $\frac{2}{8}$

= 16 + 200 - 20 ÷ 80

= 216 - 20 ÷ 80

= $\frac{196}{80}$ = $\frac{49}{20}$

If A and B are two events such that P(A) = $\frac{2}{3}$ , P(B) = $\frac{5}{2}$ and P(A∪B) = $\frac{2}{8}$ , then find P(A∩B).

P(A∩B) = Answer: $\frac{35}{12}$

SOLUTION 1 :

Given:

P(A) = $\frac{2}{3}$ , P(B) = $\frac{5}{2}$ and P(A∪B) = $\frac{2}{8}$ .

By Addition theorem on probability.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∩B) = P(A) + P(B) - P(A∪B)

= $\frac{2}{3}$ + $\frac{5}{2}$ - $\frac{2}{8}$

= 32 + 120 - 12 ÷ 48

= 152 - 12 ÷ 48

= $\frac{140}{48}$ = $\frac{35}{12}$

If A and B are two events such that P(A) = $\frac{2}{3}$ , P(B) = $\frac{3}{2}$ and P(A∪B) = $\frac{3}{8}$ , then find P(A∩B).

P(A∩B) = Answer: 4 $\frac{3}{2}$ 4

SOLUTION 1 :

Given:

P(A) = $\frac{2}{3}$ , P(B) = $\frac{3}{2}$ and P(A∪B) = $\frac{3}{8}$ .

By Addition theorem on probability.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∩B) = P(A) + P(B) - P(A∪B)

= $\frac{2}{3}$ + $\frac{3}{2}$ - $\frac{3}{8}$

= 32 + 72 - 18 ÷ 48

= 104 - 18 ÷ 48

= $\frac{86}{48}$ = 4 $\frac{3}{2}$ 4

If A and B are two events such that P(A) = $\frac{1}{5}$ , P(B) = $\frac{3}{4}$ and P(A∪B) = $\frac{2}{8}$ , then find P(A∩B).

P(A∩B) = Answer: $\frac{7}{10}$

SOLUTION 1 :

Given:

P(A) = $\frac{1}{5}$ , P(B) = $\frac{3}{4}$ and P(A∪B) = $\frac{2}{8}$ .

By Addition theorem on probability.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∩B) = P(A) + P(B) - P(A∪B)

= $\frac{1}{5}$ + $\frac{3}{4}$ - $\frac{2}{8}$

= 32 + 120 - 40 ÷ 160

= 152 - 40 ÷ 160

= $\frac{112}{160}$ = $\frac{7}{10}$

If A and B are two events such that P(A) = $\frac{2}{3}$ , P(B) = $\frac{3}{2}$ and P(A∪B) = $\frac{2}{8}$ , then find P(A∩B).

P(A∩B) = Answer: $\frac{23}{12}$

SOLUTION 1 :

Given:

P(A) = $\frac{2}{3}$ , P(B) = $\frac{3}{2}$ and P(A∪B) = $\frac{2}{8}$ .

By Addition theorem on probability.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∩B) = P(A) + P(B) - P(A∪B)

= $\frac{2}{3}$ + $\frac{3}{2}$ - $\frac{2}{8}$

= 32 + 72 - 12 ÷ 48

= 104 - 12 ÷ 48

= $\frac{92}{48}$ = $\frac{23}{12}$

If A and B are two events such that P(A) = $\frac{1}{3}$ , P(B) = $\frac{5}{6}$ and P(A∪B) = $\frac{2}{8}$ , then find P(A∩B).

P(A∩B) = Answer: $\frac{11}{12}$

SOLUTION 1 :

Given:

P(A) = $\frac{1}{3}$ , P(B) = $\frac{5}{6}$ and P(A∪B) = $\frac{2}{8}$ .

By Addition theorem on probability.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∩B) = P(A) + P(B) - P(A∪B)

= $\frac{1}{3}$ + $\frac{5}{6}$ - $\frac{2}{8}$

= 48 + 120 - 36 ÷ 144

= 168 - 36 ÷ 144

= $\frac{132}{144}$ = $\frac{11}{12}$

If A and B are two events such that P(A) = $\frac{1}{3}$ , P(B) = $\frac{5}{4}$ and P(A∪B) = $\frac{3}{8}$ , then find P(A∩B).

P(A∩B) = Answer: $\frac{29}{24}$

SOLUTION 1 :

Given:

P(A) = $\frac{1}{3}$ , P(B) = $\frac{5}{4}$ and P(A∪B) = $\frac{3}{8}$ .

By Addition theorem on probability.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∩B) = P(A) + P(B) - P(A∪B)

= $\frac{1}{3}$ + $\frac{5}{4}$ - $\frac{3}{8}$

= 32 + 120 - 36 ÷ 96

= 152 - 36 ÷ 96

= $\frac{116}{96}$ = $\frac{29}{24}$

If A and B are two events such that P(A) = $\frac{1}{3}$ , P(B) = $\frac{3}{4}$ and P(A∪B) = $\frac{2}{8}$ , then find P(A∩B).

P(A∩B) = Answer: $\frac{5}{6}$

SOLUTION 1 :

Given:

P(A) = $\frac{1}{3}$ , P(B) = $\frac{3}{4}$ and P(A∪B) = $\frac{2}{8}$ .

By Addition theorem on probability.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∩B) = P(A) + P(B) - P(A∪B)

= $\frac{1}{3}$ + $\frac{3}{4}$ - $\frac{2}{8}$

= 32 + 72 - 24 ÷ 96

= 104 - 24 ÷ 96

= $\frac{80}{96}$ = $\frac{5}{6}$

If A and B are two events such that P(A) = $\frac{2}{5}$ , P(B) = $\frac{3}{2}$ and P(A∪B) = $\frac{2}{8}$ , then find P(A∩B).

P(A∩B) = Answer: 3 $\frac{3}{2}$ 0

SOLUTION 1 :

Given:

P(A) = $\frac{2}{5}$ , P(B) = $\frac{3}{2}$ and P(A∪B) = $\frac{2}{8}$ .

By Addition theorem on probability.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∩B) = P(A) + P(B) - P(A∪B)

= $\frac{2}{5}$ + $\frac{3}{2}$ - $\frac{2}{8}$

= 32 + 120 - 20 ÷ 80

= 152 - 20 ÷ 80

= 13 $\frac{2}{8}$ 0 = 3 $\frac{3}{2}$ 0

10)

If A and B are two events such that P(A) = $\frac{1}{5}$ , P(B) = $\frac{3}{4}$ and P(A∪B) = $\frac{3}{8}$ , then find P(A∩B).

P(A∩B) = Answer: 2 $\frac{3}{4}$ 0

SOLUTION 1 :

Given:

P(A) = $\frac{1}{5}$ , P(B) = $\frac{3}{4}$ and P(A∪B) = $\frac{3}{8}$ .

By Addition theorem on probability.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∩B) = P(A) + P(B) - P(A∪B)

= $\frac{1}{5}$ + $\frac{3}{4}$ - $\frac{3}{8}$

= 32 + 120 - 60 ÷ 160

= 152 - 60 ÷ 160

= $\frac{92}{160}$ = 2 $\frac{3}{4}$ 0