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

Question

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

Please explain

SigEpUCI · Answer

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)

ywilfred · Answer

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

ywilfred · Answer

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
2) calculating probability and adding
3) bernoulli's eqn

SigEpUCI · Answer

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?

damit · Answer

sigep your approach is correct

Anonymous · Answer

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?

Anonymous · Answer

I mean 32 for denominator

Anonymous · Answer

Is the answer 5C3+5C2/2^5.....5/8?????

SigEpUCI · Answer

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.

Anonymous · Answer

So the answer is 10/32....5/16?

Anonymous · Answer

So the answer is 10/32....5/16?

hardworker_indian · Answer

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

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

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