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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 prime roll number or a girl with composite roll number or an even roll number.
Solution
Sample space (S) = {1, 2, 3, …, 35}
n(S) = 35
Total number of students = 35
Number of boys = `4/7 xx 35`
= 20 ...[Boys Numbers = {1, 2, 3, …, 20}]
Number of girls = `3/7 xx 35`
= 15 ...[Girls Numbers = {21, 22, …, 35}]
Let A be the event of getting a boy role number with prime number
A = {2, 3, 5, 7, 11, 13, 17, 19}
n(A) = 8
P(A) = `("n"("A"))/("n"("S")) = 8/35`
Let B be the event of getting girls roll number with composite number.
B = {21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35}
n(B) = 12
P(B) = `("n"("B"))/("n"("S")) = 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) = `("n"("C"))/("n"("S")) = 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) = `("n"("B" ∩ "C"))/("n"("S")) = 7/35`
(A ∩ C) = {2}
n(A ∩ C) = 1
P(A ∩ C) = `("n"("A" ∩ "C"))/("n"("S")) = 1/35`
A ∩ B ∩ C = { }
n(A ∩ B ∩ C) = 0
P(A ∩ B ∩ C) = 0
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(A ∩ C) + P(A ∩ B ∩ C)
= `8/35 + 12/35 + 17/35 - 0 - 7/35 - 1/35 + 0`
= `8/35 + 12/35 + 17/35 - 8/35`
= `(8 + 12 + 17 - 8)/35`
= `29/35`
Probability of getting roll number is `29/35`
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