GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Jan 2019, 22:21

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### The winning strategy for a high GRE score

January 17, 2019

January 17, 2019

08:00 AM PST

09:00 AM PST

Learn the winning strategy for a high GRE score — what do people who reach a high score do differently? We're going to share insights, tips and strategies from data we've collected from over 50,000 students who used examPAL.
• ### Free GMAT Strategy Webinar

January 19, 2019

January 19, 2019

07:00 AM PST

09:00 AM PST

Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

# The probability is 0.6 that an “unfair” coin will turn up

Author Message
TAGS:

### Hide Tags

Director
Joined: 25 Oct 2008
Posts: 506
Location: Kolkata,India
The probability is 0.6 that an “unfair” coin will turn up  [#permalink]

### Show Tags

Updated on: 11 Mar 2012, 13:07
6
00:00

Difficulty:

35% (medium)

Question Stats:

71% (01:34) correct 29% (01:44) wrong based on 412 sessions

### HideShow timer Statistics

The probability is 0.6 that an “unfair” coin will turn up tails on any given toss. If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails?

A. 0.064
B. 0.36
C. 0.64
D. 0.784
E. 0.936

Where's my error:
P(1)+P(2)+P(3)
6/10.4/10.4/10 + 6/10.6/10.4/10 +6/10.6/10.6/10
=0.516

_________________

http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

Originally posted by tejal777 on 02 Nov 2009, 04:51.
Last edited by Bunuel on 11 Mar 2012, 13:07, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 52232

### Show Tags

11 Mar 2012, 13:05
3
3
tejal777 wrote:
The probability is 0.6 that an “unfair” coin will turn up tails on any given toss. If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails?

Where's my error:
P(1)+P(2)+P(3)
6/10.4/10.4/10 + 6/10.6/10.4/10 +6/10.6/10.6/10
=0.516

The problem with your solution is that 1 or 2 tails can occur in several ways:

THH can occur in 3 ways: THH, HTH and HHT (basically it's # of arrangements of 3 letters THH out of whihc 2 H's are identical, which is 3!/2!=3). So, you should have P(THH)=0.6*0.4^2*3!/2!;
Similarly TTH can also occur in 3 way: TTH, THT and HTT. So, you should have P(TTH)=0.6^2*0.4*3!/2!.
Notice that we don't have the same issue with TTT, since it can occur only in one way: TTT.

$$P(T\geq{1})=0.6*0.4^2*\frac{3!}{2!}+0.6^2*0.4*\frac{3!}{2!}+0.6^3=0.936$$.

Though this question can be solved with an easier approach: $$P(T\geq{1})=1-P(H=3)=1-0.4^3=0.936$$ (the probability of at least one tails is 1 minus the probability of zero tails).

Hope it's clear.
_________________
Manager
Joined: 29 Oct 2009
Posts: 196
GMAT 1: 750 Q50 V42

### Show Tags

02 Nov 2009, 05:09
8
1
solve it this way:

probability that at least one of the tosses will turn up tails = 1 - probability that all will be heads

= 1 - (0.4*0.4*0.4)

= 1 - 0.064

= 0.936

where you went wrong:

probability that 1 is tails = (.6*.4*.4)+(.4*.6*.4)+(.4*.4*.6)

similarly for probability that 2 is tails.. consider all the cases..

in your solution, you have considered only one case each for both of the above..

even if you had solved it correctly, this method is longer and more complicated.. solve it as shown in my post above.. easier and quicker..

hope this helps!

cheers!
_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders!
http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html

##### General Discussion
VP
Joined: 05 Mar 2008
Posts: 1399

### Show Tags

Updated on: 02 Nov 2009, 05:17
I'm getting. Someone correct me if i'm wrong

First time:
probability of getting tails is .6 = .6

2nd time:
probability of getting heads is (.4)(.6) = .24

3rd time:
probability of getting tails is (.4)(.4)(.6) = .144

Probability of flipping unfair coin and getting tails is .984

Originally posted by lagomez on 02 Nov 2009, 05:05.
Last edited by lagomez on 02 Nov 2009, 05:17, edited 2 times in total.
Manager
Status: mba here i come!
Joined: 07 Aug 2011
Posts: 206

### Show Tags

11 Mar 2012, 12:46
required probability = 1 - p(3 heads)

$$= 1 - (\frac{2}{5})^3 = \frac{117}{125}$$
_________________

press +1 Kudos to appreciate posts

Intern
Joined: 27 Oct 2014
Posts: 23
GMAT 1: 680 Q49 V34
GPA: 3.8
Re: The probability is 0.6 that an “unfair” coin will turn up  [#permalink]

### Show Tags

08 Dec 2014, 07:07
Wy we do not multiply 0.4^3by 3!/2! ?
So as a result we will have = 1- 0.4^3*3!/2!=0.808
I do understand that this is incorrect answer, but could not explain to myself what is wrong with it.
Math Expert
Joined: 02 Sep 2009
Posts: 52232
Re: The probability is 0.6 that an “unfair” coin will turn up  [#permalink]

### Show Tags

08 Dec 2014, 07:16
1
Van4ez wrote:
Wy we do not multiply 0.4^3by 3!/2! ?
So as a result we will have = 1- 0.4^3*3!/2!=0.808
I do understand that this is incorrect answer, but could not explain to myself what is wrong with it.

Factorial correction (multiplying by x!/(y!...)) account for different ways an event can occur. For example, THH can occur in 3 ways: THH, HTH and HHT, so we are multiplying by 3!/2! = 3 (3!/2! is the number of permutations of 3 letters THH where 2 T's are the same).

TTT case can occur only in one way Tails, Tails, Tails, so there is no need for factorial correction for this case.

Hope it's clear.

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 9420
Re: The probability is 0.6 that an “unfair” coin will turn up  [#permalink]

### Show Tags

12 Jan 2018, 21:00
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: The probability is 0.6 that an “unfair” coin will turn up &nbs [#permalink] 12 Jan 2018, 21:00
Display posts from previous: Sort by