Sasi Math (Revised for 1024)

Scroll:Probability >> Addition theorem on probability >> saq (4497)

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 P(A) = $\frac{1}{4}$ , P(B) = $\frac{7}{4}$ , P(A∪B) = 1.

Find (i) P(A∩B)

Answer:_______________

If P(A) = $\frac{3}{4}$ , P(B) = $\frac{7}{8}$ , P(A∪B) = 1.

Find (i) P(A∩B)

Answer:_______________

If P(A) = $\frac{3}{2}$ , P(B) = $\frac{7}{8}$ , P(A∪B) = 1.

Find (i) P(A∩B)

Answer:_______________

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

Find (i) P(A∩B)

Answer:_______________

If P(A) = $\frac{3}{2}$ , P(B) = $\frac{7}{6}$ , P(A∪B) = 1.

Find (i) P(A∩B)

Answer:_______________

If P(A) = $\frac{1}{2}$ , P(B) = $\frac{7}{8}$ , P(A∪B) = 1.

Find (i) P(A∩B)

Answer:_______________

If P(A) = $\frac{3}{2}$ , P(B) = $\frac{7}{10}$ , P(A∪B) = 1.

Find (i) P(A∩B)

Answer:_______________

If P(A) = $\frac{1}{4}$ , P(B) = $\frac{7}{8}$ , P(A∪B) = 1.

Find (i) P(A∩B)

Answer:_______________

If P(A) = $\frac{3}{4}$ , P(B) = $\frac{7}{4}$ , P(A∪B) = 1.

Find (i) P(A∩B)

Answer:_______________

10)

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

Find (i) P(A∩B)

Answer:_______________

If P(A) = $\frac{1}{4}$ , P(B) = $\frac{7}{4}$ , P(A∪B) = 1.

Find (i) P(A∩B)

Answer: $\frac{1}{1}$

SOLUTION 1 :

Given:

P(A) = $\frac{1}{4}$ , P(B) = $\frac{7}{4}$ and P(A∪B) = 1.

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)

(i) ...... P(A∩B) = $\frac{1}{4}$ + $\frac{7}{4}$ - 1

= 1 + 7 - 4 ÷ 4

= 8 - 4 ÷ 4

P(A∩B) = $\frac{4}{4}$ = $\frac{1}{1}$

If P(A) = $\frac{3}{4}$ , P(B) = $\frac{7}{8}$ , P(A∪B) = 1.

Find (i) P(A∩B)

Answer: $\frac{5}{8}$

SOLUTION 1 :

Given:

P(A) = $\frac{3}{4}$ , P(B) = $\frac{7}{8}$ and P(A∪B) = 1.

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)

(i) ...... P(A∩B) = $\frac{3}{4}$ + $\frac{7}{8}$ - 1

= 6 + 7 - 8 ÷ 8

= 13 - 8 ÷ 8

P(A∩B) = $\frac{5}{8}$ = $\frac{5}{8}$

If P(A) = $\frac{3}{2}$ , P(B) = $\frac{7}{8}$ , P(A∪B) = 1.

Find (i) P(A∩B)

Answer: $\frac{11}{8}$

SOLUTION 1 :

Given:

P(A) = $\frac{3}{2}$ , P(B) = $\frac{7}{8}$ and P(A∪B) = 1.

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)

(i) ...... P(A∩B) = $\frac{3}{2}$ + $\frac{7}{8}$ - 1

= 12 + 7 - 8 ÷ 8

= 19 - 8 ÷ 8

P(A∩B) = $\frac{11}{8}$ = $\frac{11}{8}$

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

Find (i) P(A∩B)

Answer: $\frac{3}{8}$

SOLUTION 1 :

Given:

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

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)

(i) ...... P(A∩B) = $\frac{3}{4}$ + $\frac{5}{8}$ - 1

= 6 + 5 - 8 ÷ 8

= 11 - 8 ÷ 8

P(A∩B) = $\frac{3}{8}$ = $\frac{3}{8}$

If P(A) = $\frac{3}{2}$ , P(B) = $\frac{7}{6}$ , P(A∪B) = 1.

Find (i) P(A∩B)

Answer: $\frac{5}{3}$

SOLUTION 1 :

Given:

P(A) = $\frac{3}{2}$ , P(B) = $\frac{7}{6}$ and P(A∪B) = 1.

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)

(i) ...... P(A∩B) = $\frac{3}{2}$ + $\frac{7}{6}$ - 1

= 9 + 7 - 6 ÷ 6

= 16 - 6 ÷ 6

P(A∩B) = $\frac{10}{6}$ = $\frac{5}{3}$

If P(A) = $\frac{1}{2}$ , P(B) = $\frac{7}{8}$ , P(A∪B) = 1.

Find (i) P(A∩B)

Answer: $\frac{3}{8}$

SOLUTION 1 :

Given:

P(A) = $\frac{1}{2}$ , P(B) = $\frac{7}{8}$ and P(A∪B) = 1.

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)

(i) ...... P(A∩B) = $\frac{1}{2}$ + $\frac{7}{8}$ - 1

= 4 + 7 - 8 ÷ 8

= 11 - 8 ÷ 8

P(A∩B) = $\frac{3}{8}$ = $\frac{3}{8}$

If P(A) = $\frac{3}{2}$ , P(B) = $\frac{7}{10}$ , P(A∪B) = 1.

Find (i) P(A∩B)

Answer: $\frac{6}{5}$

SOLUTION 1 :

Given:

P(A) = $\frac{3}{2}$ , P(B) = $\frac{7}{10}$ and P(A∪B) = 1.

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)

(i) ...... P(A∩B) = $\frac{3}{2}$ + $\frac{7}{10}$ - 1

= 15 + 7 - 10 ÷ 10

= 22 - 10 ÷ 10

P(A∩B) = $\frac{12}{10}$ = $\frac{6}{5}$

If P(A) = $\frac{1}{4}$ , P(B) = $\frac{7}{8}$ , P(A∪B) = 1.

Find (i) P(A∩B)

Answer: $\frac{1}{8}$

SOLUTION 1 :

Given:

P(A) = $\frac{1}{4}$ , P(B) = $\frac{7}{8}$ and P(A∪B) = 1.

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)

(i) ...... P(A∩B) = $\frac{1}{4}$ + $\frac{7}{8}$ - 1

= 2 + 7 - 8 ÷ 8

= 9 - 8 ÷ 8

P(A∩B) = $\frac{1}{8}$ = $\frac{1}{8}$

If P(A) = $\frac{3}{4}$ , P(B) = $\frac{7}{4}$ , P(A∪B) = 1.

Find (i) P(A∩B)

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

SOLUTION 1 :

Given:

P(A) = $\frac{3}{4}$ , P(B) = $\frac{7}{4}$ and P(A∪B) = 1.

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)

(i) ...... P(A∩B) = $\frac{3}{4}$ + $\frac{7}{4}$ - 1

= 3 + 7 - 4 ÷ 4

= 10 - 4 ÷ 4

P(A∩B) = $\frac{6}{4}$ = $\frac{3}{2}$

10)

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

Find (i) P(A∩B)

Answer: $\frac{1}{1}$

SOLUTION 1 :

Given:

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

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)

(i) ...... P(A∩B) = $\frac{3}{4}$ + $\frac{5}{4}$ - 1

= 3 + 5 - 4 ÷ 4

= 8 - 4 ÷ 4

P(A∩B) = $\frac{4}{4}$ = $\frac{1}{1}$