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

Redeem

Probability

Expert replies
by kartikshah » Fri Jul 27, 2012 6:07 pm
What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits?

A. 271/900

B. 27/100

C. 7/25

D. 1/9

E. 1/10

Apart from providing solution to the above question, could someone guide me to a good resource on probability, permutations and combinations? I'm going to appear for my exam early next week but I still don't feel confident about this topic!!
Join the discussion
Source: — Problem Solving |

by eagleeye » Fri Jul 27, 2012 6:16 pm
kartikshah wrote:What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits?

A. 271/900

B. 27/100

C. 7/25

D. 1/9

E. 1/10

Apart from providing solution to the above question, could someone guide me to a good resource on probability, permutations and combinations? I'm going to appear for my exam early next week but I still don't feel confident about this topic!!
Total numbers from 100 to 999 = 999-100+1 = 900
first digit cant be 0.
Digits with 7 missing = (8 options for 1st digit)*(9 options for 2nd)*(9 options for 3rd)
= 8*9*9

Required probability = 1 - 8*9*9/(900) = 1- 18/25 = 7/25.

Unfortunately I don't have a specified resource, but you should check out Brent's GMATPrepNow. He posts really good probability solutions on the forum.
Join the discussion

by kartikshah » Fri Jul 27, 2012 6:22 pm
I didn't get why you subtracted 648/900 from 1
When we say
d1 = 8 ways to choose; doesn't that include 7
d2 = 9 ways to choose; doesn't that include 7
d3 = 9 ways to choose; doesn't that include 7

?
Please explain.
Join the discussion

by eagleeye » Fri Jul 27, 2012 6:31 pm
kartikshah wrote:I didn't get why you subtracted 648/900 from 1
When we say
d1 = 8 ways to choose; doesn't that include 7
d2 = 9 ways to choose; doesn't that include 7
d3 = 9 ways to choose; doesn't that include 7

?
Please explain.
No. None of them include 7. Total number of possible digits for d2 and d3 = 10 (0,1,2,3,4,5,6,7,8,9). If I de-select 7, I get 9 digits as you saw.
Does this work?
Join the discussion

by kartikshah » Fri Jul 27, 2012 6:32 pm
Yea, you're right!
I had not counted zero as a possible digit for d2 and d3!!
Join the discussion