Question 488990: If I drop three coins on a table, what is the probability of finding
1. exactly 2 heads
2. at most 1 head
3. at least 2 heads
4. at least 1 head
sample space = 8
I came up with HHT, HTH, THH,HTT,THT,TTH,TTT,HHH.
state the probability of each in the form of fraction and percent.
For 1,2,3 I came up with 1/4 or .25, and for 4 = 1/6.
Have I done this correctly, or can you show me a better way to show this??
Is this a form of "classical probability"??
Thank you.
Found 2 solutions by chessace, Theo:
Answer by chessace(471)   (Show Source):
You can put this solution on YOUR website! Good that you tried.
For something this small, it's fine just to list all possibilities, but this should be done systematically to avoid missing or double counting.
I.e., HHH HHT HTH HTT THH THT TTH TTT (count H=0 and T=1 and this is counting in binary from 0 to 7).
Possibly because your list was random, you "didn't count perfectly".
1. Has 3 hits, P = 3/8
2. Has 4 hits, P = 1/2
3. Is exactly like 2 because H and T are symetric. P = 1/2
4. It is impossible to get any probability out of this example = 1/6; P = 7/8.
There is indeed a better way to do this.
Note that none of the 4 questions care about the order, which is reasonable since the coins were dropped all at once.
Thus you can expand (H+T)^3 [works for any "3", e.g., 5 coins --> (H+T)^5]
HHH + 3HHT + 3HTT + TTT (not in any coin order, just H written before T).
Meaning: There is only one way to get all H; 3 ways to get HHT = 3 ways to pick which coin is the T.
From this you just look [and add]:
1. 2nd term: 3 --> 3/8
2. 3rd and 4th terms: 3+1 = 4 --> 1/2
3. 1st and 2nd terms: 4
4. 1st 3 terms: 1+3+3 = 7
Yes, this is "classic probability" although I haven't heard that term.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! p(heads) = .5
p(tails) = .5
since the probability of success and failure are the same, then the probability for each occurrence is the same.
that probability is .5*.5*.5 = .125
what differs is the ways in which each occurrence can happen.
3 heads can only happen in 1 way.
3 tails can only happen in 1 way.
1 head and 2 tails can happen in 3 ways.
they are:
htt
tht
tth
2 heads and 1 tail can happen in 3 ways.
they are:
hht
hth
thh
the formula for determining the ways they can happen is the combination formula.
that formula is
nCx = n! / (x! * (n-x)!))
you are drawing 3 coins and you want exactly 1 head.
the formula for that becomes:
n = 3
x = 1
nCx = 3! / (1! * 2!) = 3! / 2! = (3 * 2!) / 2! = 3
3! means 3 * 2 * 1
2! means 2 * 1
1! means 1
the total probability of all possible occurrences has to equal to 1.
for your 3 coins, here's how the probabilities would be calculated.
probability of 0 heads would be:
3C0 * (.5)^3 = 3! / (0! * 3!) * .125 = 3! / 3! * .125 = 1 * .125 = .125
0! is always equal to 1
1! is always equal to 1
probability of 1 head would be:
3C1 * (.5)^3 = 3! / (1! * 2!) * .125 = 3! / 2! * .125 = 3 * .125 = .375
probability of 2 heads would be:
3C2 * (.5)^2 = 3! / (2! * 1!) * .125 = 3! / 2! * .125 = 3 * .125 = .375
probability of 3 heads would be:
3C3 * (.5)^2 = 3! / (3! * 0!) * .125 = 3! / 3! * .125 = 1 * .125 = .125
your total probability is:
.125 + .125 + .375 + .375 = .250 + .750 = 1.0
this is what it should be so we calculated correctly.
the general formula is:
nCx * p^x * q^(n-x)
p is the probability of success
q is the probability of failure.
in your experiment, the probability of success and the probability of failure were the same so it didn't matter.
in other case, they will be different.
suppose the coin was weighted so the probability of heads was .7 and the probability of tails was .3
the same problem would give you the following probabilities.
probability of exactly 0 heads = 1 * .7^0 * .3^3 = .027
probability of exactly 1 head = 3 * .7^1 * .3^2 = .189
probability of exactly 2 heads = 3 * .7^2 * .3^1 = .441
probability of exactly 3 heads = 1 * .7^3 = .343
the total probability is still equal to 1.
the number of ways each occurrence can happen is the same.
the probability of each occurrence, however, is different.
the answer to your questions from the data for equal probabilities is:
If I drop three coins on a table, what is the probability of finding
1. exactly 2 heads
exactly 2 heads would be .375
2. at most 1 head
at most 1 head would be exactly 0 and exactly 1 which would be .125 + .375 = .5
3. at least 2 heads
at least 2 heads would be exactly 2 and exactly 3 which would be .375 and .125 = .5
4. at least 1 head
at least 1 head would be exactly 1 and exactly 2 and exactly 3 which would be .375 + .375 + .125 = .875
not that this is also 1 minus exactly 0 which would be equal to 1 - .125 = .875.
.125 = 1/8
.375 = 3/8
i think you tallied the possibilities correctly but may not have tallied the probabilities correctly.
doing it the way you did it, i got:
HHH
HHT
HTH
THH
HTT
THT
TTH
TTT
this part seems to be ok.
looking at this set of data, however, i get the following.
probability of exactly 0 heads is 1/8 (TTT)
probability of exactly 1 head is 3/8 (HTT, THT, TTH)
probability of exactly 2 heads is 3/8 (HHT, HTH, THH)
probability of exactly 3 heads is 1/8 (HHH)
the answers to the questions are:
1. exactly 2 heads = 3/8
2. at most 1 head = 4/8
3. at least 2 heads = 4/8
4. at least 1 head = 7/8
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