Question

In a class of 35, students are numbered from 1 to 35. The ratio of boys to girls is 4:3. The roll numbers of students begin with boys and end with girls. Find the probability that a student selected is either a boy with a prime roll number or a girl with a composite roll number or an even roll number.

Answer 1

Sample space (S) = {1, 2, 3,… ,35}

n(S) = 35

Total number of students = 35

Number of boys = \(\frac 47\) × 35 = 20

[Boys Numbers = {1, 2, 3,…, 20}]

Number of girls = \(\frac 37\) × 35 = 15

[Girls Numbers = { 21, 22, …, 35}]

Let A be the event of getting a boy roll number with prime number

A = {2, 3, 5, 7, 11, 13, 17, 19}

n(A) = 8

P(A) = \(\frac{n(A)}{n(S)} = \frac 8{35}\)

Let B be the event of getting girls roll numbers with composite numbers.

B = {21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35}

n(B) = 12 

P(B) = \(\frac{n(B)}{n(S)} = \frac {12}{35}\)

Let C be the event of getting an even roll number.

C = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34}

n(C) = 17

P(C) = \(\frac{n(C)}{n(S)} = \frac {17}{35}\)

n(A⋂B) = 0 

⇒ P(A⋂B) = 0

B⋂C = {22, 24, 26, 28, 30, 32, 34}

n(B⋂C) = 7

P(B⋂C) = \(\frac{n(B\cap C)}{n(S)} = \frac 7{35}\)

A⋂C = {2}

n(A⋂C) = 1

P(A⋂C) = \(\frac{n(A\cap C)}{n(S)} = \frac 1{35}\)

A⋂B⋂C = {}

n(A⋂B⋂C) = 0

P(A⋂B⋂C) = \(\frac{n(A\cap B\cap C)}{n(S)} \) = 0

P(A∪B∪C) = P(A) + P(B) + P(C) - P(A⋂B) - P(B⋂C) - P(A⋂C) + P(A⋂B⋂C)

\(= \frac{8}{35} + \frac{12}{35} + \frac{17}{35} -0 - \frac 7{35} - \frac 1{35}+ 0\)

\(= \frac{8 + 12 + 17 - 7 - 1}{35}\)

\(= \frac{29}{35}\)

Hence, the required probability is \(\frac{29}{35}\).

Answer 2

n(S) = 35

Student with even roll number - E