Home
Class 12
MATHS
Statement-1: Out of 21 tickets with numb...

Statement-1: Out of 21 tickets with number 1 to 21, 3 tickets are drawn at random, the chance that the numbers on them are in AP is `(10)/(133)`.
Statement-2: Out of (2n+1) tickets consecutively numbered three are drawn at ranodm, the chance that the number on them are in AP is (4n-10)/`(4n^(2)-1)`.

A

Statement-1 is true, Statement-2 is true: Statement-2 is a correct explanation for Statement-1

B

Statement-1 is true, Statement-2 is true: Statement-2 is not a correct explanation for Statement-1

C

Statement-1 is true, Statement-2 is false

D

Statement-1 is false, Statement-2 is true

Text Solution

Verified by Experts

The correct Answer is:
c
Promotional Banner

Topper's Solved these Questions

  • PROBABILITY

    ARIHANT MATHS|Exercise Exercise (Subjective Type Questions)|15 Videos

  • PROBABILITY

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|55 Videos

  • PROBABILITY

    ARIHANT MATHS|Exercise Probability Exercise 5: Match Type Questions|4 Videos

  • PERMUTATIONS AND COMBINATIONS

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|28 Videos

  • PRODUCT OF VECTORS

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|52 Videos

Similar Questions

Explore conceptually related problems

Out of (2n+1) tickets consecutively numbered, three are drawn at random. Find the chance that the numbers on them are in AP.

A bag contains 20 tickets numbered 1 to 20. Two tickets are drawn at random . Find the probability that both the numbers on the ticket are prime.

12 tickets are numbered from 1 to 12 one ticket is drawn at random, then the probability of the number to be divisible by 2 or 3 is:

20 cards are numbered from 1 to 20. One card is then drawn at random. What is the probability that the number on the card will be : a multiple of 4 ?

A box contains tickets numbered 1 to 20.3 tickets are drawn from the box with replacement. The probability that the largest number on the tickets is 7, is

20 cards are numbered from 1 to 20. One card is then drawn at random. What is the probability that the number on the card will be : not a multiple of 4 ?

Tickets are numbered from 1 to 10. two tickets are drawn one after the other at random. Find the probability that the number on one of the tickets is a multiple of 5 and on the other a multiple of 4.

From a set of 17 cards, numbered 1, 2, 3, 4, .... ,16, 17, one is drawn at random. Show that the chance that its number is divisible by 3 or 7 is 7/17 .

A bag contains 4 tickets numbered 00, 01, 10 and 11. Four tickets are chosen at random with replacement, the probability that the sum of numbers on the tickets is 23 is

if 6n tickets numbered 0,1,2,....... 6n-1 are placed in a bag and three are drawn out , show that the chance that the sum of the numbers on then is equal to 6n is (3n)/((6n-1)(6n-2))