Q. 45.0( 1 Vote )

A box contain 100

Answer :

Given, total number of cards in the box is 100 with each card marked with numbers from 1 to 100.

Therefore, we know that from 1 to 100 , number of single digit numbers = 1 to 9 = 9


Also, number of two-digit numbers from 1 to 100 = 10 to 99 = 90


Only three digit number from 1 to 100 is 100.


So, number of three-digit numbers from 1 to 100 = 1


Numbers divisible by 8 from 1 to 100 are 8,16,24,32,40,48,56,64,72,80,88,96 = 12


Numbers which are multiple of 9 are 9,18,27,36,45,54,63,72,81,90,99 = 11


Numbers which are multiple of 5 are 5,10,15,20,……….100 = 20


Now, let the event that a single digit number is drawn from the box be A.


Event that card bearing a two-digit number is drawn from the box be B.


Event that a card bearing a three-digit number is drawn from the box be C.


Event that a card bearing a number divisible by 8 is drawn from the box be D.


Event that a card bearing a number which is multiple of 9 is drawn from the box be E.


Event that a card bearing a number which is multiple of 5 is drawn from the box be F


Number of favourable outcomes of event A = 9


Number of favourable outcomes of event B = 90


Number of favourable outcomes of event C = 1


Number of favourable outcomes of event D = 12


Number of favourable outcomes of event E = 11


Number of favourable outcomes of event F = 20


Total number of favourable outcomes = 100


(i) Probability of the card bearing a single digit number = Probability of occurrence of event A = P(A)


P(A) = 0.09


(ii) Probability of the card bearing a two-digit number = Probability of occurrence of event B = P(B)


P(B) = 0.9


(iii) Probability of the card bearing a three-digit number = Probability of occurrence of event C = P(C)


P(C) = 0.01


(iv) Probability of the card bearing a number divisible by 8 = Probability of occurrence of event D = P(D)


P(D) = 0.12


(v) Probability of the card bearing a number which is multiple 9 = Probability of occurrence of event E = P(E)


P(E) = 0.11


(vi) Probability of the card bearing a number which is multiple 5 = Probability of occurrence of event F = P(F)


P(F) = 0.2


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

The probability oGujarat Board Mathematics

The probability tGujarat Board Mathematics

A game of chance Gujarat Board Mathematics

A box contain 100Gujarat Board Mathematics

Marks obtained byGujarat Board Mathematics