Hard probability concept

[Primarygun]

Sometimes, we may draw a tree diagram with domains (the columns) as (1st, 2nd,3rd) with the rows having "Yes" or "No".

For instance, find the P that both of the two children in your home have done their homework.

Then, we can apply the above case.

 

Now, consider a case that seven balls are in a bag.

One ball will be taken out.

There are four red balls, 2 blue balls, and 1 green balls.

Find P(G).

Obviously, it is 1/7 if we set domains as Type of ball.

What about we set the domains as

Red ball, Green Ball, Blue ball ,with the rows of "Yes" or "No"?

We can find that P(Red yes)=4/7. while P(Red NO)=3/7, Similarly we can find out the other P.

Now when doing the question,

Use P(Red No, Blue No, Green Yes).

The P is different from the obvious method, can anyone explain it?

[aommaster]

The reason for this is that you are looking for the probability:

The ball is not a red and not "the ball is green"

Then, you draw another ball agaiin after replacement and say "The ball is not blue". Again, you are not alooking for a green ball.

 

I hope this helps!

[Bignose]

In order to calculate the "not" quantities, you have to set up the conditional probabilities correctly.

 

You got started on the right foot. The Probability of Not Red = P(NR) is 3/7.

 

Now, you have to look at the remainder of the distribution:

 

Probability of Blue given that the drawn ball is Not Red = P(B | NR) = 2/3

which is also equal to Probability of Not Green given Not Red = P(NG | NR)

Probability of Green given Not Red = P(G | NR) = P(NB | NR) = 1/3

 

The vertical bar is read "conditional that", that is P(G | NR) is probability of green given that the ball is not red.

 

Now, if are given that the ball is not Red and Not Blue, you have to look at the remaining subpopulation. P(G | NR & NB) = 1 since there is only a green ball left.

 

Now, the total Probability of green is

 

P(G | NR & NB)*P(NB | NR)*P(NR) = 1*(1/3)*(3/7) = 1/7 just as calculated in the first place.

[Primarygun]

In order to calculate the "not" quantities, you have to set up the conditional probabilities correctly.

 

You got started on the right foot. The Probability of Not Red = P(NR) is 3/7.

 

Now, you have to look at the remainder of the distribution:

 

Probability of Blue given that the drawn ball is Not Red = P(B | NR) = 2/3

which is also equal to Probability of Not Green given Not Red = P(NG | NR)

Probability of Green given Not Red = P(G | NR) = P(NB | NR) = 1/3

 

The vertical bar is read "conditional that", that is P(G | NR) is probability of green given that the ball is not red.

 

Now, if are given that the ball is not Red and Not Blue, you have to look at the remaining subpopulation. P(G | NR & NB) = 1 since there is only a green ball left.

 

Now, the total Probability of green is

 

P(G | NR & NB)*P(NB | NR)*P(NR) = 1*(1/3)*(3/7) = 1/7 just as calculated in the first place.

I hope someone can help me again.

I was not close to the proper approach until your appearance.

Now, I notice something important.

For tossing a head and tail, the "domains", a term I refer to, affect each other under no circumstance.

But, here the case, taking out one red ball surely affect the P of a Green ball, so they are dependent.

1.We can always use your approach for both dependent and independent events, right? But for indepedent events, we can chose not to use this method while the conditional P must be applied for the latter, right?

If my thoughts are totally correct, I think I've got it at that moment.

[Primarygun]

Perhaps the second method is not the common one, but I may sometimes use it instead.

May anyone help me?

[Bignose]

The definition of independent probabilities is that irregardless of the conditions in that probability, the probability remains the same.

 

In symbols. Events A & B are independent if P(A|B) = P(A)

and P(B|A) = P(B).

 

Successive flips of a (fair) coin are independent. That is, it does not matter how many heads there has been in a row, the next flip P(H) = 0.5 and P(T)=0.5. Next time you go to the casino you can no confidently make fun of the people who bet on the roulette table "Oh, there have been 4 blacks in a row, so red MUST be coming up."

 

Sampling of the balls in your example can be independent if you replace the drawn out ball after each pick. With replacement, the probabilities remain 4/7, 2/7, and 1/7 for R,B,G respectively. The conditional probability in my 1st response has to be used since you gave additional information that the ball drawn out is not red. So, you have to look at the subpopulations with the given information that the drawn ball is not red.

 

If you sample the balls without replacement, then once again you have to start using conditional probablilties. For example, we can calculate all the probabilites if the 1st ball is red. If the 1st is R, then the remaining population is 3R, 2B, 1G. So, the probabilites in order are 3/6=1/2, 2/6=1/3 and 1/6.

 

so Probability of drawing a red ball given that the 1st drawn ball is red=

P(R|R) = 1/2

P(B|R) = 1/3

P(G|R) = 1/6

 

and so on for all the cases.

 

Finally, a lot of these things are explained in introductory probability books. There are a lot of good ones out there, with lots more examples than just this. I'd encourage you to go check them out if you want to understand this al lot more fully.