Question

In a class of 35, students are numbered from 1 to 35. The ratio of boys and 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 prime roll number or a girl with composite roll number or an even roll number.

• A. $\frac{19}{44}$
• B. $\frac{24}{38}$
• C. $\frac{47}{65}$
• D. $\frac{29}{35}$

Answer 1

The correct option is D $\frac{29}{35}$

Sample space (S) = {1, 2, 3,… ,35}
n(S) = 35
Total number of students = 35
Number of boys $=\frac{4}{7}\times 35=20$
[Boys Numbers = {1, 2, 3,…, 20}]
Number of girls $=\frac{3}{7}\times 35=1$5
[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) $=\frac{n\left(A\right)}{n\left(S\right)}=\frac{8}{35}$
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) $=\frac{n\left(B\right)}{n\left(S\right)}=\frac{12}{35}$
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) $=\frac{n\left(C\right)}{n\left(S\right)}=\frac{17}{35}$
n(A⋂B) = 0 $\Rightarrow$P(A⋂B)=0
B⋂C={22, 24, 26, 28, 30, 32, 34}
n(B⋂C) = 7
P(B⋂C)$=\frac{n(B\bigcap C)}{n\left(S\right)}=\frac{7}{35}$
A⋂C={2}
n(A⋂C) = 1
P(A⋂C)$=\frac{n(A\bigcap C)}{n\left(S\right)}=\frac{1}{35}$
A⋂B⋂C={}
n(A⋂B⋂C) = 0
P(A⋂B⋂C)$=\frac{n(A\bigcap B\bigcap C)}{n\left(S\right)}=0$
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 $\frac{29}{35}.$