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# I am having trouble with symetric probability. So if a coin

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I am having trouble with symetric probability. So if a coin [#permalink]  15 Aug 2004, 11:48
I am having trouble with symetric probability.

So if a coin is tossed 5 times what is the probability of getting 3 H and 2T?

is that the same as getting 4H and 1T (since it is .5 for each?)

i.e. 3h, 2t = .5*.5*.5*.5*.5

and 4h , 1t = .5*.5*.5*.5*.5

Senior Manager
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For symmetric probability, it helps to think about probabilities of outcomes (when unordered) as combination problems.

For instance, for the probability of getting 3H and 2T,

think of it as

number of outcomes with 3H & 2T / Total number of outcomes

in this case it would be 5 choose 3 / 2^5

the five choose 3 (5!/ 3!2!) is the number of ways you can pick 3 of the coins (as presumably assign heads to them)
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3 heads and 2 tails would be

P(3 heads) = 1/2*1/2*1/2 = 1/8 (independent events)
P(2 tails) = 1/2*1/2 = 1/4 (again, independent events)

P(3 heads + 2 tails) = 1/8 + 1/4 = 3/8
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i forgot to add, 3 heads and 2 tails means just that. if the questions states at least 3 heads and 2 tails, then that would be different.

I think you might work this problem out a number of ways:

1) # of outcomes/# of favorable outcome
3) bernoulli's eqn
Senior Manager
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ywilfred wrote:
3 heads and 2 tails would be

P(3 heads) = 1/2*1/2*1/2 = 1/8 (independent events)
P(2 tails) = 1/2*1/2 = 1/4 (again, independent events)

P(3 heads + 2 tails) = 1/8 + 1/4 = 3/8

I disagree with this.

If the question was phrased "What is the probability flipping a coin 3 times and getting 3 heads, or flipping it twice and getting 2 tails", then your answer would be correct.

Anyone else want to comment?
Manager
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Joined: 31 Dec 1969
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If someone wouldn't mind, could you run through the process of exactly how they would calculate the numerator. I know the denominator is 2^5 or 36, but how does one calculate exactly 3 heads and 2 two tails?
Joined: 31 Dec 1969
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GMAT 1: 710 Q49 V0
GMAT 2: 740 Q40 V50
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Kudos [?]: 88 [0], given: 86982

I mean 32 for denominator
Joined: 31 Dec 1969
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GMAT 1: 710 Q49 V0
GMAT 2: 740 Q40 V50
GMAT 3: 700 Q48 V38
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Kudos [?]: 88 [0], given: 86982

Senior Manager
Joined: 25 Jul 2004
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Kudos [?]: 7 [0], given: 0

think of all the possible sequences you can get that would have 3 heads,

HHHTT
HHTHT
HHTTH
HTHHT
HTHTH
HTTHH
THTHH
TTHHH
THHHT
THHTH

in each of the above examples, we have five coins, and we are choosing 3 of the five coins to be heads. This is how we arrive at the number of ways equalling 5 Choose 3.

5C3 = (5! / (3!2!)) = 10.
Joined: 31 Dec 1969
Location: India
Concentration: General Management, Strategy
GMAT 1: 710 Q49 V0
GMAT 2: 740 Q40 V50
GMAT 3: 700 Q48 V38
GPA: 3.6
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Kudos [?]: 88 [0], given: 86982

Joined: 31 Dec 1969
Location: India
Concentration: General Management, Strategy
GMAT 1: 710 Q49 V0
GMAT 2: 740 Q40 V50
GMAT 3: 700 Q48 V38
GPA: 3.6
Followers: 0

Kudos [?]: 88 [0], given: 86982

Director
Joined: 20 Jul 2004
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As SigEpUCI said,

Pb = no of fav events/no of possible events

no of possible events = 2 x 2 x 2 x 2 x 2 = 2^5.
(each of the flips may result in a H ot T)

no of fav events = 5C3.
We have to Choose (combination) 3 places for three Hs among the 5 empty spots. Once done, the two other empty spots will be occupoed by the two Ts. To select three spots from 5 = 5C3

Pm = 5C3/2^5 = 10/32 = 5/16
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