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

Redeem

Tough Probability

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Thu Oct 15, 2015 2:14 pm

Tough Probability

by late4thing » Wed Oct 05, 2016 5:18 pm
Suppose Mr. Smith goes to Vegas. He knows from previous experiences that the probability he will come out ahead after playing black jack is .3. The probability he will be ahead after playing the roulette wheel is .4. He also knows that his chances of coming out ahead in one game is independent of the other. If he plays both games, find the probability he will end up being ahead in at least one.
Top
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 157
Joined: Mon Aug 16, 2010 7:30 pm
Location: India
Thanked: 65 times
Followed by:3 members

by crackverbal » Thu Oct 06, 2016 1:15 am
Hi late4thing,

Always remember that P(atleast one) = 1 - P(none)

Now the probability of Mr. Smith being ahead in blackjack is P(B) = 0.3. The probability of Mr. Smith not being ahead in blackjack will be P(not B) = 0.7.

The probability of Mr. Smith being ahead in roulette is P(R) = 0.4. The probability of Mr. Smith not being ahead in roulette will be P(not R) = 0.6.

The questions asks us to find the probability of Mr. Smith being ahead in at least one out of blackjack and roulette.

P(being ahead in atleast one) = 1 - P(being ahead in none)
= 1 - (0.7 * 0.6)
= 1 - 0.42
= 0.58

Hope this helps!

CrackVerbal Academics Team
Join Free 4 part MBA Through GMAT Video Training Series here -
https://gmat.crackverbal.com/mba-throug ... video-2018

Enroll for our GMAT Trial Course here -
https://gmatonline.crackverbal.com/

For more info on GMAT and MBA, follow us on @AskCrackVerbal
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Thu Oct 06, 2016 6:45 am
late4thing wrote:Suppose Mr. Smith goes to Vegas. He knows from previous experiences that the probability he will come out ahead after playing blackjack is .3. The probability he will be ahead after playing the roulette wheel is .4. He also knows that his chances of coming out ahead in one game is independent of the other. If he plays both games, find the probability he will end up being ahead in at least one.
crackverbal's approach is the same as the one I'd use. However, we can also solve the question this way:

P(ahead in blackjack) = 0.3
P(NOT ahead in blackjack) = 0.7
P(ahead in roulette) = 0.4
P(NOT ahead in roulette) = 0.6

P(ahead in at least one) = P(ahead in blackjack and NOT ahead in roulette OR NOT ahead in blackjack and ahead in roulette OR ahead in blackjack and ahead in roulette)
= P(ahead in blackjack and NOT ahead in roulette) + P(NOT ahead in blackjack and ahead in roulette) + P(ahead in blackjack and ahead in roulette)
= (0.3 x 0.6) + (0.7 x 0.4) + (0.3 x 0.4)
= 0.18 + 0.28 + 0.12
= 0.58

Cheers,
Brent

RELATED VIDEOS
- The Complement: https://www.gmatprepnow.com/module/gmat ... /video/744
- Probability of Event A OR Event B: https://www.gmatprepnow.com/module/gmat ... /video/747
- Probability of Event A AND Event B: https://www.gmatprepnow.com/module/gmat ... /video/750
- Rewriting Questions: https://www.gmatprepnow.com/module/gmat ... /video/754
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Post new topic Post Reply
• Page 1 of 1