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

Redeem

probability

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Newbie | Next Rank: 10 Posts
Posts: 8
Joined: Sat Aug 25, 2012 9:43 pm

probability

by srivathsan » Sat Sep 08, 2012 7:26 am
there are 100 cards numbered from 1 to 100 . If three cards are drawn at random and with replacement, what is the probability that the sum of three numbers on the cards so selected will be odd?

a.1/4
b.3/8
c.1/2
d.5/8
e.3/4
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 502
Joined: Tue Jun 03, 2008 11:36 pm
Thanked: 99 times
Followed by:21 members

by vk_vinayak » Sat Sep 08, 2012 8:23 am
srivathsan wrote:there are 100 cards numbered from 1 to 100 . If three cards are drawn at random and with replacement, what is the probability that the sum of three numbers on the cards so selected will be odd?

a.1/4
b.3/8
c.1/2
d.5/8
e.3/4
Not sure if this method is correct.

Two possibilities will make the sum odd: All three are ODD -or- two ODD and one even.

Possibility of all three odd: (50/100) * (50/100) *(50/100) = 1/8
Possibility of two odd and one even: (50/100) * (50/100) *(50/100) = 1/8

Total: (1/8)+(1/8) = 1/4. Choise A.
- VK

I will (Learn. Recognize. Apply)
Top
Post new topic Post Reply
• Page 1 of 1