4 congruent sectors should all have equal probabilities .
The first may be a boy or a girl. In case of a girl there is two possibilities : GG and GB ...
This is an ill defined sample space though. We in fact, in the general case, have a need to the number of girls and boys in the class and determine our
sample space based on ...
This time the possible outputs are GG, BB and GB which is the same as BG .
Another ill defined sample space concentrating on one-shot outcome, human beings are identifiable not identical . Defining the sample space from this
consideration gives us in real situation wrong probabilities even in case of 4 students in the class, 2G and 2 B.
The purple region is 1/4 ths of the outer ring and has the same area with the yellow region in the middle of ...
Probable sum is the total of dice as explained in "http://gwydir.demon.co.uk/jo/probability/calcdice.htm" . The sample space is shown in ry1 .
Sample space is {{H,H,H},{H,H,T},{H,T,H},{H,T,T},{T,H,H},{T,H,T},{T,T,H},{T,T,T}} which has 8 elements. Every subspace, subset of this is set considered as
an event. We se here the term subspace because the term (sample) space is generally used. What we have is called mathematically a set. To be a space is a norm needed.
If we just consider single element subsets which normally has only a fraction of the total probability 1, we may hardly get the total probability of all events.
Impossible events do have 0 probability .
Sample space is a set and is also a subset of itself. Either {H,H,H}, or {H,H,T}, or {H,T,H}, or {H,T,T}, or {T,H,H}, or {T,H,T}, or {T,T,H} or {T,T,T}
i.e. one of every one-element subsets of the whole space is certainly to happen .
{{H,H,H},{H,H,T},{H,T,H},{H,T,T},{T,H,H},{T,H,T},{T,T,H},{T,T,T}} .
The sample space of the figure above has 8^{3} = 512 elements; whereas the one below has 8 . For further help have please a look at ry1 .
The probability of 1 is 1/12, and of an even is 1/2 olmaktadır. Since the proabilities are disjoint the answer is the sum ... .
First try structures with lesser elements like 2x2 or 3x3.
This is a rather hard question. If not solved, have please a look at the this page page/question-18
("http://www.test-dr.com/education/6th-Grade-Mathematics/probability-5.1.1.html" /question-18) and it's helps.
The probability of less than 5 is 4/12=1/3 .
The probability of an odd is 6/12=1/2 .
These probabilities would be summed if 1 and 3 were not both less than 5 and odd. In this case so doing they will be considered twice. The correct answer
is 4/12 + 6/12 - 2/12 = 8/12 = ... .
The probabilities of both events will be summed and the probability of intersection will be subtracted .
Multiply the probabilities of ... .
Have please a look at ry1 .
The third is little hard to ask, but all are logical questions .
Negation of negative gives positive ...
sample space:{{H,H,H,H},{H,H,H,T},{H,H,T,H},{H,H,T,T},{H,T,H,H},{H,T,H,T},{H,T,T,H},{H,T,T,T},
{T,H,H,H},{T,H,H,T},{T,H,T,H},{T,H,T,T},{T,T,H,H},{T,T,H,T},{T,T,T,H},{T,T,T,T}} .
Total thrown is 400 kg .
The probability of first "d" is 1/10 ths and we have 4 letters .
Probability is always less than or equal to 1 .
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