Probability of repeated trials : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 07:58

### 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 June
Open Detailed Calendar

# Probability of repeated trials

Author Message
Senior Manager
Joined: 31 Jul 2006
Posts: 440
Followers: 3

Kudos [?]: 44 [0], given: 0

### Show Tags

08 Aug 2006, 16:51
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

guys I am looking for a formula to solve this problem:

Coin tossed 4 times. What is probability that atleast 3 out of 4 tosses will come up heads? (answer is 5/16)

kaplan did it by writing out all combinations of H&Ts. I know that's going to be a PITA on exam day for me, so I'd rather have a neat formula handy.

Why doesn't this formula work for me?

.....................N!
BDF = ------------------ * p^M * (1 â€” p)^N â€” M
.............(N â€” M)! * M!

BDF = the binomial distribution probability;
p = the individual probability of the phenomenon (p = 0.5 to get heads in coin tossing);
M = the exact number of successes
N = the number of trials

Thx!

Last edited by Nsentra on 08 Aug 2006, 16:55, edited 1 time in total.
 Kaplan GMAT Prep Discount Codes Veritas Prep GMAT Discount Codes Manhattan GMAT Discount Codes
Manager
Joined: 08 May 2004
Posts: 158
Location: Pittsburgh
Followers: 1

Kudos [?]: 0 [0], given: 0

### Show Tags

08 Aug 2006, 16:54
I have the same problem! How in the heck are you going to right it all out on exam day without wasting 2-3 minutes on a problem?
I have never seen it asked like that on GMAT prep..
so I wouldln't worry too too much..
Senior Manager
Joined: 31 Jul 2006
Posts: 440
Followers: 3

Kudos [?]: 44 [0], given: 0

### Show Tags

08 Aug 2006, 16:58
jamesrwright3 wrote:
I have the same problem! How in the heck are you going to right it all out on exam day without wasting 2-3 minutes on a problem?
I have never seen it asked like that on GMAT prep..
so I wouldln't worry too too much..

I appreciate Kaplan for putting some stuff for me in baby terms, but sometimes by not explaining everything behind the answer they are preventing me from seeing the real math behind the solution
Manager
Joined: 26 Jun 2006
Posts: 152
Followers: 1

Kudos [?]: 6 [0], given: 0

### Show Tags

08 Aug 2006, 17:06
Why am not getting 5/16? In fact, I am getting 1-5/16=11/16... Confused

Pron(0 or 1 heads) = ?

Total number of possible combinationsL 2^4=16

0 heads: 1 combination (when all tails)
1 head, 3 tails: 4 combinations

Prob(0 or 1) = (1+4)/16 = 5/16

Prob(3 or more)=1-5/16=11/16

???
Senior Manager
Joined: 31 Jul 2006
Posts: 440
Followers: 3

Kudos [?]: 44 [0], given: 0

### Show Tags

08 Aug 2006, 17:19
v1rok wrote:
Why am not getting 5/16? In fact, I am getting 1-5/16=11/16... Confused

Pron(0 or 1 heads) = ?

Total number of possible combinationsL 2^4=16

0 heads: 1 combination (when all tails)
1 head, 3 tails: 4 combinations

Prob(0 or 1) = (1+4)/16 = 5/16

Prob(3 or more)=1-5/16=11/16

???

They are saying there are 5 possible outcomes with alteast 3 heads in 4 tosses

hhht
hhth
hthh
thhh
hhhh

the word 'atleast' is prob. why my probability formula is not working out

so 5/16

Last edited by Nsentra on 08 Aug 2006, 17:20, edited 1 time in total.
Manager
Joined: 26 Jun 2006
Posts: 152
Followers: 1

Kudos [?]: 6 [0], given: 0

### Show Tags

08 Aug 2006, 17:19
Nevermind, I found where I got derailed:

P(3 or more)= 1 - P(0,1,or2)

He-he, looks like I need some rest...
Senior Manager
Joined: 31 Jul 2006
Posts: 440
Followers: 3

Kudos [?]: 44 [0], given: 0

### Show Tags

08 Aug 2006, 17:22
v1rok wrote:
Nevermind, I found where I got derailed:

P(3 or more)= 1 - P(0,1,or2)

He-he, looks like I need some rest...

Is this the only way to solve this problem? Write out all possible outcomes? there is no algebraic formula?
Manager
Joined: 26 Jun 2006
Posts: 152
Followers: 1

Kudos [?]: 6 [0], given: 0

### Show Tags

08 Aug 2006, 17:24
Exactly, nissan! I was working too quickly (trying to fit in those 2 minutes!) and accidently skipped one case...

Anyway, you don't have to write down all possible combinations of heads and tails.

You just need to remember that 2^4 is total number of combinations (=16)
and then count only those cases where we have eaither 4 heads (1 such case) or 3 heads (4 cases). So,

Prob=5/16
Senior Manager
Joined: 31 Jul 2006
Posts: 440
Followers: 3

Kudos [?]: 44 [0], given: 0

### Show Tags

08 Aug 2006, 17:27
v1rok wrote:
Exactly, nissan! I was working too quickly (trying to fit in those 2 minutes!) and accidently skipped one case...

I know for a fact that is what is going to happen to me on a test. Thanks for giving another thought on how to do this though.
Manager
Joined: 26 Jun 2006
Posts: 152
Followers: 1

Kudos [?]: 6 [0], given: 0

### Show Tags

08 Aug 2006, 17:31
Nsentra wrote:
v1rok wrote:
Exactly, nissan! I was working too quickly (trying to fit in those 2 minutes!) and accidently skipped one case...

I know for a fact that is what is going to happen to me on a test. Thanks for giving another thought on how to do this though.

Cheer up! That's exactly why you are here, isn't it. A little bit of practicing and you will get confident enough so that none of the tricks GMAT trows at you would make you blink...
CEO
Joined: 20 Nov 2005
Posts: 2911
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 24

Kudos [?]: 273 [0], given: 0

### Show Tags

08 Aug 2006, 22:57
Prob of getting exactly three heads = 4C3 * (1/2)^3 * (1/2)^1 = 4/16

Prob of getting exactly 4 heads = 4C4 * (1/2)^4 = 1/16

Total prob = 1/16 + 4/16 = 5/16
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Last edited by ps_dahiya on 11 Aug 2006, 10:18, edited 1 time in total.
Senior Manager
Joined: 31 Jul 2006
Posts: 440
Followers: 3

Kudos [?]: 44 [0], given: 0

### Show Tags

09 Aug 2006, 09:58
ps_dahiya wrote:
Prob of getting exactly three heads = 4C3 * (1/2)^3 * (1/2)^1 = 4/16

Prob of getting exactly 4 heads = 4C4 * (1/2)^4 = 1/16

Total prob = 1/16 + 4/16 = 5/16

Perfect! Thanks! (great quote too)
Manager
Joined: 03 Jul 2006
Posts: 176
Followers: 1

Kudos [?]: 25 [0], given: 0

### Show Tags

10 Aug 2006, 13:07
ps_dahiya,

could you explain the formula that you used. seems simple
CEO
Joined: 20 Nov 2005
Posts: 2911
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 24

Kudos [?]: 273 [0], given: 0

### Show Tags

10 Aug 2006, 13:15
rkatl wrote:
ps_dahiya,

could you explain the formula that you used. seems simple

Its is explained here by the math goddess HongHu:

http://www.gmatclub.com/phpbb/viewtopic.php?t=14706
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Director
Joined: 17 Jul 2006
Posts: 714
Followers: 1

Kudos [?]: 12 [0], given: 0

### Show Tags

11 Aug 2006, 09:21
PS_dahiya,
There is a typo in your reply. Please don't mistake me. I know you are so good. Just to help others to understand ....

Probability of winning head or tail in tossing coin is 1/2

Prob of getting exactly three heads = 4C3 * (1/2)^3 * (1/3)^1 = 4/16

Prob of getting exactly 4 heads = 4C4 * (1/2)^4 = 1/16

Total prob = 1/16 + 4/16 = 5/16.

Manager
Joined: 12 May 2006
Posts: 116
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

11 Aug 2006, 10:19
How I approached the problem:

16 possible outcomes altogether (4*4)

Getting at least 3:
Getting all 4-
4C4 = 1

Getting 3 out of 4:
4C3 = 4

4+1 = 5

5/16 [/u]
CEO
Joined: 20 Nov 2005
Posts: 2911
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 24

Kudos [?]: 273 [0], given: 0

### Show Tags

11 Aug 2006, 10:19
PS_dahiya,
There is a typo in your reply. Please don't mistake me. I know you are so good. Just to help others to understand ....

Probability of winning head or tail in tossing coin is 1/2

Prob of getting exactly three heads = 4C3 * (1/2)^3 * (1/3)^1 = 4/16

Prob of getting exactly 4 heads = 4C4 * (1/2)^4 = 1/16

Total prob = 1/16 + 4/16 = 5/16.

Thanks buddy. I have corrected the error.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

11 Aug 2006, 10:19
Display posts from previous: Sort by