Sasi Math (Revised for 1024)

Scroll:Probability >> Addition theorem on probability >> ps (4512)

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 two digit number is formed with the digits 2,3,4 ( repetition is allowed). Find the probablity that the number is divisible by 3 or 4.

Answer:_______________

Each individual letter of the word " ACCOMMODATION " is written in a piece of paper, and all 13 pieces of papers are placed ina jar. If one piece of paper is selected at random from the jar, find the probability that

(i) ... the letter 'A' or 'O' is selected.

(ii) ... the letter 'M' or 'C' is selected.

The letter A or O is selected is

The letter M or C is selected is

Answer:_______________

A two digit number is formed with the digits 2,5,9 ( repetition is allowed). Find the probablity that the number is divisible by 2 or 5.

Answer:_______________

Each individual letter of the word " COMMISSIONER" is written in a piece of paper, and all 12 pieces of papers are placed ina jar. If one piece of paper is selected at random from the jar, find the probability that

(i) ... the letter 'S' or 'O' is selected.

(ii) ... the letter 'M' or 'I' is selected.

The letter S or O is selected is

The letter M or I is selected is

Answer:_______________

A two digit number is formed with the digits 2,5,9 ( repetition is allowed). Find the probablity that the number is divisible by 2 or 9.

Answer:_______________

A two digit number is formed with the digits 2,3,4 ( repetition is allowed). Find the probablity that the number is divisible by 2 or 4.

Answer:_______________

A two digit number is formed with the digits 3,5,7 ( repetition is allowed). Find the probablity that the number is divisible by 5 or 7.

Answer:_______________

A two digit number is formed with the digits 3,5,7 ( repetition is allowed). Find the probablity that the number is divisible by 3 or 5.

Answer:_______________

A two digit number is formed with the digits 3,5,7 ( repetition is allowed). Find the probablity that the number is divisible by 3 or 7.

Answer:_______________

10)

A two digit number is formed with the digits 2,3,4 ( repetition is allowed). Find the probablity that the number is divisible by 3 or 2.

Answer:_______________

A two digit number is formed with the digits 2,3,4 ( repetition is allowed). Find the probablity that the number is divisible by 3 or 4.

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

SOLUTION 1 :

Sample space

S = { 22, 23, 24, 32, 33, 34, 42, 43, 44 }

n(S) = 9 .

Let A be the event of getting a number divisible by 3.

A = { 24, 33, 42 }

n(A) = 3.

P(A) = $\frac{n(A)}{n(S)}$

P(A) = $\frac{3}{9}$ .

Let B be the event of getting a number divisible by 4.

B = { 24, 32, 44 }

n(B) = 3. A and B are mutually exclusive.

P(B) = $\frac{n(B)}{n(S)}$

P(B) = $\frac{3}{9}$ .

P(A∪B) = P(A) + P(B)

= $\frac{3}{9}$ + $\frac{3}{9}$ = $\frac{6}{9}$ = $\frac{2}{3}$

The probability that the number is divisible by 2 or 4 = $\frac{2}{3}$ .

(i) ... the letter 'A' or 'O' is selected.

(ii) ... the letter 'M' or 'C' is selected.

The letter A or O is selected is Answer: $\frac{5}{13}$

The letter M or C is selected is Answer: $\frac{4}{13}$

SOLUTION 1 :

There are 13 letters in the word ACCOMMODATION .

n(S) = 13 .

Let A be the event of getting the letter A

n(A) = 2 .

P(A) = $\frac{n(A)}{n(S)}$

P(A) = $\frac{2}{13}$ .

Let B be the event of getting the letter O

n(B) = 3. A and B are mutually exclusive.

P(B) = $\frac{n(B)}{n(S)}$

P(B) = $\frac{3}{13.}$

P(A∪B) = P(A) + P(B) ( A∩B =âˆ®)

= $\frac{2}{13}$ + $\frac{3}{13}$ = $\frac{5}{13}$ .

The probability that the letter A (or) O is selected is $\frac{5}{13}$ .

Let C be the event of getting the letter M.

n(C) = 2

P(C) = $\frac{n(C)}{n(S)}$

P(C) = $\frac{2}{13}$

Let D be the event of getting the letter C

n(D) = 2

P(D) = $\frac{n(D)}{n(S)}$

P(D) = $\frac{2}{13}$ .

P(C or D) = P(C∪D) ( C∩D = âˆ®)

= P(C) + P(D)

= $\frac{2}{13}$ + $\frac{2}{13}$ = $\frac{4}{13}$ .

The probability that the letter C or D is selected is $\frac{4}{13}$ .

A two digit number is formed with the digits 2,5,9 ( repetition is allowed). Find the probablity that the number is divisible by 2 or 5.

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

SOLUTION 1 :

Sample space

S = { 22, 25, 29, 52, 55, 59, 92, 95, 99 }

n(S) = 9 .

Let A be the event of getting a number divisible by 2.

A = { 22,52,92 }

n(A) = 3 .

P(A) = $\frac{n(A)}{n(S)}$

P(A) = $\frac{3}{9}$ .

Let B be the event of getting a number divisible by 5

B = { 25, 55, 95 }

n(B) = 3. A and B are mutually exclusive.

P(B) = $\frac{n(B)}{n(S)}$

P(B) = $\frac{3}{9}$ .

P(A∪B) = P(A) + P(B)

= $\frac{3}{9}$ + $\frac{3}{9}$ = $\frac{6}{9}$ = $\frac{2}{3}$ .

The probability that the number is divisible by 2 or 5 = $\frac{2}{3}$ .

(i) ... the letter 'S' or 'O' is selected.

(ii) ... the letter 'M' or 'I' is selected.

The letter S or O is selected is Answer: $\frac{1}{3}$

The letter M or I is selected is Answer: $\frac{1}{3}$

SOLUTION 1 :

There are 12 letters in the word COMMISSIONER .

n(S) = 12 .

Let A be the event of getting the letter S.

n(A) = 2 .

P(A) = $\frac{n(A)}{n(S)}$

P(A) = $\frac{2}{12}$ .

Let B be the event of getting the letter O

n(B) = 2. A and B are mutually exclusive.

P(B) = $\frac{n(B)}{n(S)}$

P(B) = $\frac{2}{12}$ .

P(A∪B) = P(A) + P(B) ( A∩B =âˆ®)

= $\frac{2}{12}$ + $\frac{2}{12}$ = $\frac{4}{12}$ = $\frac{1}{3}$

The probability that the letter S (or) O is selected is $\frac{1}{3}$ .

Let C be the event of getting the letter M.

n(C) = 2

P(C) = $\frac{n(C)}{n(S)}$

P(C) = $\frac{2}{12}$

Let D be the event of getting the letter I.

n(D) = 2

P(D) = $\frac{n(D)}{n(S)}$

P(D) = $\frac{2}{12}$ .

P(C or D) = P(C∪D) ( C∩D = âˆ®)

= P(C) + P(D)

= $\frac{2}{12}$ + $\frac{2}{12}$ = $\frac{4}{12}$ = $\frac{1}{3}$ ..

The probability that the letter M or I is selected is $\frac{1}{3}$ .

A two digit number is formed with the digits 2,5,9 ( repetition is allowed). Find the probablity that the number is divisible by 2 or 9.

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

SOLUTION 1 :

Sample space

S = { 22, 25, 29, 52, 55, 59, 92, 95, 99 }

n(S) = 9 .

Let A be the event of getting a number divisible by 2.

A = { 22,52,92 }

n(A) = 3 .

P(A) = $\frac{n(A)}{n(S)}$

P(A) = $\frac{3}{9}$ .

Let B be the event of getting a number divisible by 9

B = { 99 }

n(B) = 1. A and B are mutually exclusive.

P(B) = $\frac{n(B)}{n(S)}$

P(B) = $\frac{1}{9.}$

P(A∪B) = P(A) + P(B)

= $\frac{3}{9}$ + $\frac{1}{9}$ = $\frac{4}{9}$ .

The probability that the number is divisible by 2 or 9 = $\frac{4}{9}$ .

A two digit number is formed with the digits 2,3,4 ( repetition is allowed). Find the probablity that the number is divisible by 2 or 4.

Answer: 1

SOLUTION 1 :

Sample space

S = { 22, 23, 24, 32, 33, 34, 42, 43, 44 }

n(S) = 9 .

Let A be the event of getting a number divisible by 2.

A = { 22, 24, 32, 34, 42, 44 }

n(A) = 6 .

P(A) = $\frac{n(A)}{n(S)}$

P(A) = $\frac{6}{9}$ .

Let B be the event of getting a number divisible by 4

B = { 24, 32, 44 }

n(B) = 3. A and B are mutually exclusive.

P(B) = $\frac{n(B)}{n(S)}$

P(B) = $\frac{3}{9.}$

P(A∪B) = P(A) + P(B)

= $\frac{6}{9}$ + $\frac{3}{9}$ = $\frac{9}{9}$ = 1

The probability that the number is divisible by 2 or 4 = 1.

A two digit number is formed with the digits 3,5,7 ( repetition is allowed). Find the probablity that the number is divisible by 5 or 7.

Answer: $\frac{5}{9}$

SOLUTION 1 :

Sample space

S = { 33, 35, 37, 53, 55, 57, 73, 75, 77 }

n(S) = 9 .

Let A be the event of getting a number divisible by 5.

A = { 35, 55, 75, }

n(A) = 3 .

P(A) = $\frac{n(A)}{n(S)}$

P(A) = $\frac{3}{9}$ .

Let B be the event of getting a number divisible by 7

B = { 35, 77 }

n(B) = 2. A and B are mutually exclusive.

P(B) = $\frac{n(B)}{n(S)}$

P(B) = $\frac{2}{9.}$

P(A∪B) = P(A) + P(B)

= $\frac{3}{9}$ + $\frac{2}{9}$ = $\frac{5}{9}$

The probability that the number is divisible by 5 or 7 = $\frac{5}{9}$ .

A two digit number is formed with the digits 3,5,7 ( repetition is allowed). Find the probablity that the number is divisible by 3 or 5.

Answer: $\frac{5}{9}$

SOLUTION 1 :

Sample space

S = { 33, 35, 37, 53, 55, 57, 73, 75, 77 }

n(S) = 9 .

Let A be the event of getting a number divisible by 3.

A = { 33, 75, }

n(A) = 2 .

P(A) = $\frac{n(A)}{n(S)}$

P(A) = $\frac{2}{9}$ .

Let B be the event of getting a number divisible by 5

B = { 35, 55, 75 }

n(B) = 3. A and B are mutually exclusive.

P(B) = $\frac{n(B)}{n(S)}$

P(B) = $\frac{3}{9.}$

P(A∪B) = P(A) + P(B)

= $\frac{2}{9}$ + $\frac{3}{9}$ = $\frac{5}{9}$

The probability that the number is divisible by 3 or 5 = $\frac{5}{9}$

A two digit number is formed with the digits 3,5,7 ( repetition is allowed). Find the probablity that the number is divisible by 3 or 7.

Answer: $\frac{2}{9}$

SOLUTION 1 :

Sample space

S = { 33, 35, 37, 53, 55, 57, 73, 75, 77 }

n(S) = 9 .

Let A be the event of getting a number divisible by 3.

A = { 33, 75, }

n(A) = 2 .

P(A) = $\frac{n(A)}{n(S)}$

P(A) = $\frac{2}{9}$ .

Let B be the event of getting a number divisible by 7

B = { 35, 77 }

n(B) = 2. A and B are mutually exclusive.

P(B) = $\frac{n(B)}{n(S)}$

P(B) = $\frac{2}{9}$ .

P(A∪B) = P(A) + P(B)

= $\frac{2}{9}$ + $\frac{2}{9}$ = $\frac{4}{9}$

The probability that the number is divisible by 3 or 7 = $\frac{4}{9}$ .

10)

A two digit number is formed with the digits 2,3,4 ( repetition is allowed). Find the probablity that the number is divisible by 3 or 2.

Answer: 1

SOLUTION 1 :

Sample space

S = { 22, 23, 24, 32, 33, 34, 42, 43, 44 }

n(S) = 9 .

Let A be the event of getting a number divisible by 3.

A = { 24, 33, 42 }

n(A) = 3.

P(A) = $\frac{n(A)}{n(S)}$

P(A) = $\frac{3}{9}$ .

Let B be the event of getting a number divisible by 2.

B = { 22, 24, 32, 34, 42, 44 }

n(B) = 6. A and B are mutually exclusive.

P(B) = $\frac{n(B)}{n(S)}$

P(B) = $\frac{6}{9.}$

P(A∪B) = P(A) + P(B)

= $\frac{3}{9}$ + $\frac{6}{9}$ = $\frac{9}{9}$ = 1.

The probability that the number is divisible by 3 or 2 = 1.