problem for 10 points on fundamental counting &sample space and probability?

Question

concider the experiment of tossing 2 coins. list the sample space of this experiment, and the probability associated with each element of the sample space. i also need the rules and procedures you followed to develope your answer please help !!!...

Answer 1

when two coins are tossed you may get HH,HT,TH,TT
so the sample space is 4 if you give importance to whether the first coin shows head or the second
probability associated with
minimum 1 head=3/4
minimum 2 heads=1/4
minimum 1 tail=3/4
minimum 2 tails=1/4
only one tail=2/4=>1/2
only on head=2/4=>1/2
probability is defined as the no of favourable events divided by the sample space
favourable event is what you want or specify and the samplw space is all possible outcomes of the experiment

Answer 2

when 2 coins are tossed there will be 2 squared ie 4 chances for them

HH - 1/4
HT-1/4
TT=I/4.
TH- 1/4
THE RULES R QUITE SIMPLE. since there are 4 different ways the coins can fall each chance would have a 14th probability.
hope this is useful.

Answer 3

that is particularly a complicated one yet there are 4 diverse consequences that there is a white chip picked very last, so that you want to artwork out the probability for all the outcomes (rrrww, rrwrw, rrwrw, wrrrw and rwrrw) and then upload the fraction jointly, the answer might want to come to 33/100 and twenty

Answer 4

Hi Richard,

First, you need to completely define the possibilities for your experiments. For instance, are "heads" (h) and "tails" (t) the only possibilities, or could you have something else occur such as "edge" (e)? If so, the set of possibilities for coin 1 would be {h1,t1,e1} and for coin 2 would be {h2,t2,e2}. (If there were other possibilities, you would need to include them, too.)

Next, you need to make the sample space for the experiment. The two main rules for this are that the elements of the sample space are mutually exclusive and collectively exhaustive. "Mutually exclusive" means that the occurrence of any one element means all the other elements cannot occur. "Collectively exhaustive" means that at least one element must occur. For your two-coin-toss experiment, the sample space using the possibilities defined above is given by the following table.

s1: (h1, h2)
s2: (h1, t2)
s3: (h1, e2)
s4: (t1, h2)
s5: (t1, t2)
s6: (t1, e2)
s7: (e1, h2)
s8: (e1, t2)
s9: (e1, e2)

Next, you need to specify probabilities for your experiments. Although it may be reasonable to assert that the results of the two coin flips are independent, you need to be explicit about this; otherwise the general case is that the results are dependent. The general case for coin 1 (c1) and coin 2 (c2):

Let
P(c1=h1)=ph, P(c1=t1)=pt, P(c1=e1)=pe;
P(c2=h2|c1=h1)=qhh, P(c2=t2|c1=h1)=qth, P(c2=e2|c1=h1)=qeh;
P(c2=h2|c1=t1)=qht, P(c2=t2|c1=t1)=qtt, P(c2=e2|c1=t1)=qet;
P(c2=h2|c1=e1)=qhe, P(c2=t2|c1=e1)=qte, P(c2=e2|c1=e1)=qee;
where ph + pt + pe = 1, qhh + qth + qeh = 1, qht + qtt + qet = 1, qhe + qte + qee = 1, and all p's and q's are between 0 and 1, inclusive.

Then
P(s1) = ph x qhh,
P(s2) = ph x qth,
P(s3) = ph x qeh,
P(s4) = pt x qht,
P(s5) = pt x qtt,
P(s6) = pt x qet,
P(s7) = pe x qhe,
P(s8) = pe x qte,
P(s9) = pe x qee.

============================
All of the above answers your question. Below describes special cases.

If you assert that the results of the coin flips are independent, then
P(c2=h2|c1=h1) = P(c2=h2|c1=t1) = P(c2=h2|c1=e1);
P(c2=t2|c1=h1) = P(c2=t2|c1=t1) = P(c2=t2|c1=e1);
P(c2=e2|c1=h1) = P(c2=e2|c1=t1) = P(c2=e2|c1=e1),

so you can simplify the probabilities for coin 2 by letting
P(c2=h2)=qh, P(c2=t2)=qt, P(c2=e2)=qe, giving

P(s1) = ph x qh,
P(s2) = ph x qt,
P(s3) = ph x qe,
P(s4) = pt x qh,
P(s5) = pt x qt,
P(s6) = pt x qe,
P(s7) = pe x qh,
P(s8) = pe x qt,
P(s9) = pe x qe.

The next special case is to simplify the sample space by defining the result of the coins flips as {h1, t1'} and {h2, t2'}, respectively, where t1' and t2' include the possibilities of both tails and edge (i.e., t1' and t2' can be thought of as "not heads"). The sample space then becomes:

s1': (h1, h2)
s2': (h1, t2')
s3': (t1', h2)
s4': (t1', t2')

Probabilities for this simplified sample space are specified in similar manner as explained above. Note well, it is not correct to automatically assign equal probabilities to each element of the sample space. This is obvious when the sample space includes the possibility of "edge." However, even in the simplified sample space where "edge" is subsumed, you could have a biased coin whose weight or shape could make one outcome more likely than another.

Answer 5

HH - 0.5 * 0.5 = 0.25
HT - 0.5 * 0.5 = 0.25
TH - 0.5 * 0.5 = 0.25
TT - 0.5 * 0.5 = 0.25