Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 27 May 2017, 00:57

### 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

# A fair coin is tossed 6 times . what is the probability that

Author Message
Director
Joined: 07 Jun 2004
Posts: 612
Location: PA
Followers: 7

Kudos [?]: 794 [0], given: 22

A fair coin is tossed 6 times . what is the probability that [#permalink]

### Show Tags

16 Jan 2005, 18:39
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A fair coin is tossed 6 times . what is the probability that exactly 2 heads will show.
VP
Joined: 18 Nov 2004
Posts: 1436
Followers: 2

Kudos [?]: 39 [0], given: 0

### Show Tags

16 Jan 2005, 21:06
6C2(1/2)^6 = 15/64
Director
Joined: 21 Sep 2004
Posts: 608
Followers: 1

Kudos [?]: 35 [0], given: 0

### Show Tags

16 Jan 2005, 21:31
used binomial theorem for this one..
nCr* (p)^r (q)^n-r
probability of getting a head is 1/2=p
probability of NOT getting a head is 1/2=q
n=6
r=2

6C2 * (1/2)^2 * (1/2)^6-2
= 6C2*(1/2)^6
=15/64
Director
Joined: 07 Jun 2004
Posts: 612
Location: PA
Followers: 7

Kudos [?]: 794 [0], given: 22

### Show Tags

17 Jan 2005, 06:53
you guys are correct
Intern
Joined: 12 Aug 2004
Posts: 33
Followers: 0

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

### Show Tags

18 Jan 2005, 14:05
I'm completely lost on this one, i've been studying for about 4 months and took some time off, now i'm paying for this "vacation", can someone show me the math used to get this answer?

Thanks
Intern
Joined: 20 Dec 2004
Posts: 25
Followers: 0

Kudos [?]: 3 [1] , given: 0

### Show Tags

18 Jan 2005, 15:35
1
KUDOS
jeremy02 wrote:
I'm completely lost on this one, i've been studying for about 4 months and took some time off, now i'm paying for this "vacation", can someone show me the math used to get this answer?

Thanks

The probability of a particular event occurring is the number of outcomes that result in that particular event divided by the total number of possible outcomes.

-In this problem the total number of possible outcomes:
Since there are 2 possible outcomes for each coin toss, (2*2*2*2*2*2) = 64 which is the total number of possible outcomes.

-Now you can use the combination formula to determine the number of outcomes of exactly 2 of the coins landing on heads out of 6 flips.

C(n,r) =n!/(r!(n-r)!) where n = the number of n objects (n=6 flips) taken r (r=2; 2 out of 6 flips landing on heads) at a time.

6!/(2!(6-2)!) = 6!(2!(4!)= 15.

Once again, the probability of a particular event occurring is the number of outcomes that result in that particular event divided by the total number of possible outcomes.

# of outcomes that result in 2 flips out of 6 landing on heads = 15
total # of possible outcomes = 64

Manager
Joined: 28 Jul 2004
Posts: 59
Location: Na
Followers: 1

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

### Show Tags

30 Jan 2005, 13:54
Basic Question:
Can we consider:
P[Getting exactly 2 heads]=1 - P[Getting exactly 4 tails]
=1 - 6C4(1/2)^4
=1 - (15)/(16)
= 1/16
+++
I know I am missing something fundamental.Can someone point out what I am missing?Thank you.Rgds,

Anna

banerjeea_98 wrote:
6C2(1/2)^6 = 15/64

_________________

We can crack the exam together

SVP
Joined: 03 Jan 2005
Posts: 2236
Followers: 16

Kudos [?]: 342 [0], given: 0

### Show Tags

30 Jan 2005, 16:10
Anna, you have two mistakes. First
P(exactly two heads) = P(exactly four tails)
Not 1-P(exactly four tails)

Second, when you calculate P(exactly four tails), you still need to multiple C(6,4) by (1/2)^6 instead of (1/2)^4, since there are six throws.
Director
Joined: 19 Nov 2004
Posts: 558
Location: SF Bay Area, USA
Followers: 4

Kudos [?]: 209 [0], given: 0

### Show Tags

30 Jan 2005, 17:30
It might be easier to use Bernouli's formula like vprabhala did to get this one quickly in the exam. Time kills!
Manager
Joined: 28 Jul 2004
Posts: 59
Location: Na
Followers: 1

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

### Show Tags

30 Jan 2005, 19:27
Hongu Hu,
Makes sense.Thank you.Rgds,

Anna

HongHu wrote:
Anna, you have two mistakes. First
P(exactly two heads) = P(exactly four tails)
Not 1-P(exactly four tails)

Second, when you calculate P(exactly four tails), you still need to multiple C(6,4) by (1/2)^6 instead of (1/2)^4, since there are six throws.

_________________

We can crack the exam together

30 Jan 2005, 19:27
Display posts from previous: Sort by