BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

DS on Probability

Expert replies
by [email protected] » Tue Jan 17, 2012 7:50 pm
A bowl is filled with consecutively numbered tiles from 1 to x. Joe pulls out a tile and uses it to contruct sequence Q, which consists of 10 consecutive integers starting with the number drawn. If Joe then selects one number from Sequence Q, what is the probability that the selected number is a multiple of 3?

1) The last number in sequence Q is a prime number that is less than 20.

2) x (is less than or equal to) 10
Join the discussion
Source: — Data Sufficiency |

by shankar.ashwin » Tue Jan 17, 2012 8:46 pm
okay you have a set of 10 numbers and want to know the probability that one number picked is divisible by 3.

Consider these cases,

Set (1,2,3,4,5,6,7,8,9,10) - starting with non-multiple of 3

P(divisible by 3) = 3/10

Set (2,3,4,5,6,7,8,9,10,11) - starting with non-multiple of 3

P(divisible by 3) = 3/10

Set (3,4,5,6,7,8,9,10,11,12) - starting with multiple of 3

P(divisible by 3) = 4/10

So, the question can be rephrased as "Is x divisible by 3?"

Statement 1:

last number is prime..

primes >10 - 11,13,17 and 19

and sequence would begin with 2,4,8 and 10 respectively. So in each case probability remains 3/10 - Sufficient

Statement 2

x can either be a multiple of 3 when probability is 4/10 or x can begin with non-multiple of 3 when probability is 3/10 - Insufficient

A IMO
Join the discussion

by [email protected] » Tue Jan 17, 2012 8:55 pm
shankar.ashwin wrote:okay you have a set of 10 numbers and want to know the probability that one number picked is divisible by 3.

Consider these cases,

Set (1,2,3,4,5,6,7,8,9,10) - starting with non-multiple of 3

P(divisible by 3) = 3/10

Set (2,3,4,5,6,7,8,9,10,11) - starting with non-multiple of 3

P(divisible by 3) = 3/10

Set (3,4,5,6,7,8,9,10,11,12) - starting with multiple of 3

P(divisible by 3) = 4/10

So, the question can be rephrased as "Is x divisible by 3?"

Statement 1:

last number is prime..

primes >10 - 11,13,17 and 19

and sequence would begin with 2,4,8 and 10 respectively. So in each case probability remains 3/10 - Sufficient

Statement 2

x can either be a multiple of 3 when probability is 4/10 or x can begin with non-multiple of 3 when probability is 3/10 - Insufficient

A IMO

yes, answer is A. Thanks!
Join the discussion

by [email protected] » Mon Mar 19, 2012 1:57 am
Yes a good question I got it as A but after a lot of time...
IT IS TIME TO BEAT THE GMAT

LEARNING, APPLICATION AND TIMING IS THE FACT OF GMAT AND LIFE AS WELL... KEEP PLAYING!!!

Whenever you feel that my post really helped you to learn something new, please press on the 'THANK' button.
Join the discussion

by killer1387 » Mon Mar 19, 2012 2:09 am
statement 1:
possible endings are 11,13,17,19

for 11:
2-11--> probability (P)=3/10
4-13--> 3/10
8-17--> 3/10
10-19--> 3/10

sufficient

Statement 2:

insufficient

A
Join the discussion