# 3.10 Probability concepts -- probability  (Page 2/2)

 Page 2 / 2

Note that this process will always give you a number between 0 (no results match) and 1 (all results match). Probabilities are always between 0 (for something that never happens) and 1 (for something that is guaranteed to happen).

But what does it really mean to say that “the probability is 1/6?” You aren’t going to get 1/6 of a song. One way to make this result more concrete is to imagine that you run the machine on its “Randomize” setting 100 times. You should expect to get non-slow loud rock songs 1 out of every 6 times; roughly 17 songs will match that description. This gives us another way to express the answer: there is a 17% probability of any given song matching this description.

## The multiplication rule

We can look at the above problem another way.

What is the chance that any given, randomly selected song will be non-slow? $\frac{2}{3}$ . That is to say, 2 out of every three randomly chosen songs will be non-slow.

Now... out of those $\frac{2}{3}$ , how many will be loud? Half of them. The probability that a randomly selected song is both non-slow and loud is half of $\frac{2}{3}$ , or $\frac{1}{2}×\frac{2}{3}$ , or $\frac{1}{3}$ .

And now, out of that $\frac{1}{3}$ , how many will be rock? Again, half of them: $\frac{1}{2}×\frac{1}{3}$ . This leads us back to the conclusion we came to earlier: 1/6 of randomly chosen songs will be non-slow, loud, rock songs. But it also gives us an example of a very general principle that is at the heart of all probability calculations:

When two events are independent, the probability that they will bothoccur is the probability of one, multiplied by the probability of the other.

What does it mean to describe two events as “independent?” It means that they have no effect on each other. In real life, we know that rock songs are more likely to be fast and loud than slow and quiet. Our machine, however, keeps all three categories independent: choosing “Rock” does not make a song more likely to be fast or slow, loud or quiet.

In some cases, applying the multiplication rule is very straightforward. Suppose you generate two different songs: what is the chance that they will both be fast songs? The two songs are independent of each other, so the chance is $\frac{1}{3}×\frac{1}{3}=\frac{1}{9}$ .

Now, suppose you generate five different songs. What is the chance that they will all be fast? $\frac{1}{3}×\frac{1}{3}×\frac{1}{3}×\frac{1}{3}×\frac{1}{3}$ , or $\left(\frac{1}{3}{\right)}^{5}$ , or 1 chance in 243. Not very likely, as you might suspect!

Other cases are less obvious. Suppose you generate five different songs. What is the probability that none of them will be a fast song? The multiplication rule tells us only how to find the probability of “this and that”; how can we apply it to this question?

The key is to reword the question, as follows. What is the chance that the first song will not be fast, and the second song will not be fast, and the third song will not be fast, and so on? Expressed in this way, the question is a perfect candidate for the multiplication rule. The probability of the first song being non-fast is $\frac{2}{3}$ . Same for the second, and so on. So the probability is $\left(\frac{2}{3}{\right)}^{5}$ , or 32/243, or roughly 13%.

Based on this, we can easily answer another question: if you generate five different songs, what is the probability that at least one of them will be fast? Once again, the multiplication rule does not apply directly here: it tells us “this and that,” not “this or that.” But we can recognize that this is the opposite of the previous question. We said that 13% of the time, none of the songs will be fast. That means that the other 87% of the time, at least one of them will!

a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
how did I we'll learn this
f(x)= 2|x+5| find f(-6)
f(n)= 2n + 1
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Got questions? Join the online conversation and get instant answers!