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

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Director
Joined: 07 Jun 2004
Posts: 612
Location: PA
A fair coin is tossed 6 times . what is the probability that [#permalink]

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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: 1433

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16 Jan 2005, 21:06
6C2(1/2)^6 = 15/64
Director
Joined: 21 Sep 2004
Posts: 607

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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

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

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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

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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: 58
Location: Na

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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: 2233

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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: 556
Location: SF Bay Area, USA

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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: 58
Location: Na

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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
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