# Math Help - probability

1. ## probability

In a contest, players must take 3 balls.
There are 3 green balls, 2 blue and one white.

If the third ball is white, the player pass to the second stage.
If the first 2 balls are the same color and the third white, the player pass automatically to the third stage.

A player had automatically passed from first to third stage.

What is the probability of the first 2 balls taken be blue?

I don't have the correct solution

2. ## Working backwards

Since the player went to the 3rd stage, he must have had a white ball on the 3rd draw and 2 balls of the same color on the 1st 2 draws.

What was the probability he had 2 greens on the 1st 2 draws?
Probability 1st ball is green is 3/5.
Probability 2nd ball is green is 2/4.
Total prob = 3/10.

What was the probability he had 2 blues on the 1st 2 draws?
Probability 1st ball is blue is 2/5.
Probability 2nd ball is blue is 1/4.
Total prob = 1/10.

We know we are in one of these 2 situations. The relative probability of blue to green is 1:3, so the probability we had taken 2 blues is 1:4.

3. Originally Posted by qmech
Since the player went to the 3rd stage, he must have had a white ball on the 3rd draw and 2 balls of the same color on the 1st 2 draws.

What was the probability he had 2 greens on the 1st 2 draws?
Probability 1st ball is green is 3/5.
Probability 2nd ball is green is 2/4.
Total prob = 3/10.

What was the probability he had 2 blues on the 1st 2 draws?
Probability 1st ball is blue is 2/5.
Probability 2nd ball is blue is 1/4.
Total prob = 1/10.

We know we are in one of these 2 situations. The relative probability of blue to green is 1:3, so the probability we had taken 2 blues is 1:4.
why do you have to calculate the probability of the 2nd ball?
If you know that the 2nd ball is equal to the 1st, you just have to calc the 1st ball, or no?

4. Hello JoanF
Originally Posted by JoanF
In a contest, players must take 3 balls.
There are 3 green balls, 2 blue and one white.

If the third ball is white, the player pass to the second stage.
If the first 2 balls are the same color and the third white, the player pass automatically to the third stage.

A player had automatically passed from first to third stage.

What is the probability of the first 2 balls taken be blue?

I don't have the correct solution

You haven't told us a vital piece of information: is the first ball returned to the bag (or whatever) after it is drawn? I will assume that it is not.

Suppose that A is the event: The first two balls are blue.

Suppose that B is the event: The player passes to the third stage.

Now the probability that the first ball drawn is blue is $\frac{2}{6}=\frac13$; and the probability that the second is blue (if the first is not returned to the bag) is then $\frac15$. So
$p(A) = \frac13\times \frac15 = \frac{1}{15}$
Similarly the probability that the first two balls drawn are green is $\frac36\times\frac25=\frac15$

If the white ball has not been drawn in the first two draws, the probability that it is drawn on the third draw is $\frac{1}{4}$. This is also $p(B|A)$, the probability that the player passes to the third stage, given that the first two balls were blue.

Now event B can happen in either of two ways: where the third ball is white, with the first two balls either blue or green. So
$p(B) = \frac{1}{15}\times \frac14 + \frac15\times\frac14 = \frac{1}{15}$
Bayes' Theorem says:
$p(A|B) = \frac{p(B|A)p(A)}{p(B)}$
where $p(A|B)$ is the probability that the first two balls were blue, given that the player passed to the third stage (which is what we want). So:
$p(A|B) = \frac{\dfrac14\times\dfrac{1}{15}}{\dfrac{1}{15}}$
$=\frac14$
If the balls are returned to the bag each time, you can use the same method, but make adjustments to the individual probabilities.

Hello JoanFYou haven't told us a vital piece of information: is the first ball returned to the bag (or whatever) after it is drawn? I will assume that it is not.

Suppose that A is the event: The first two balls are blue.

Suppose that B is the event: The player passes to the third stage.

Now the probability that the first ball drawn is blue is $\frac{2}{6}=\frac13$; and the probability that the second is blue (if the first is not returned to the bag) is then $\frac15$. So
$p(A) = \frac13\times \frac15 = \frac{1}{15}$
Similarly the probability that the first two balls drawn are green is $\frac36\times\frac25=\frac15$

If the white ball has not been drawn in the first two draws, the probability that it is drawn on the third draw is $\frac{1}{4}$. This is also $p(B|A)$, the probability that the player passes to the third stage, given that the first two balls were blue.

Now event B can happen in either of two ways: where the third ball is white, with the first two balls either blue or green. So
$p(B) = \frac{1}{15}\times \frac14 + \frac15\times\frac14 = \frac{1}{15}$
Bayes' Theorem says:
$p(A|B) = \frac{p(B|A)p(A)}{p(B)}$
where $p(A|B)$ is the probability that the first two balls were blue, given that the player passed to the third stage (which is what we want). So:
$p(A|B) = \frac{\dfrac14\times\dfrac{1}{15}}{\dfrac{1}{15}}$
$=\frac14$
If the balls are returned to the bag each time, you can use the same method, but make adjustments to the individual probabilities.

No, the first ball is not returned to the bag.

But you already know that he passed to the 3rd stage so there are 2 hypotheses:

(blue=B;green=G;whiteW)

BBW

or

GGB

but when you only know what was the 1st ball taken you automatically know what were the next 2 balls taken, so you only have to calc the probability of the 1st ball. I don't understand this exercice

My calcs:

1st ball is blue: 2/5
now, knowing that the 1st ball is B, the 2nd is B too, so: (2/5) . (1/1)
and you know that the last ball is white so: (2/5) . (1/1) . (1/1)

6. Hello JoanF
Originally Posted by JoanF
No, the first ball is not returned to the bag.

But you already know that he passed to the 3rd stage so there are 2 hypotheses:

(blue=B;green=G;whiteW)

BBW

or

GGB

but when you only know what was the 1st ball taken you automatically know what were the next 2 balls taken, so you only have to calc the probability of the 1st ball. I don't understand this exercice

My calcs:

1st ball is blue: 2/5
now, knowing that the 1st ball is B, the 2nd is B too, so: (2/5) . (1/1)
and you know that the last ball is white so: (2/5) . (1/1) . (1/1)

But it doesn't work like that, does it? You can't on the one hand say: the probability that the first ball is blue is $\tfrac25$ (as if you have no other information), and then suddenly jump forward with probabilites of 1 for the second and third balls, as if you secretly knew all the time that the player has passed to the third stage. Either you know what has happened, or you don't.
Why $\tfrac25$ anyway? Are you saying that when we draw the first ball there isn't really a white ball there, and we just have 2 blue balls and 3 green ones? Clearly, that's nonsense! It makes every bit as much sense to say that since the first ball was either blue or green, the probability of each colour is one-half!
The probability that the first two balls are blue $= \frac26\times\frac15=\frac{1}{15}$
The probability that the first two balls are green $= \frac36\times\frac25=\frac{3}{15}$ (which I have deliberately left 'uncancelled')
Now all these probabilities do is to predict the likely outcomes 'in the long run'. So 'in the long run' we should expect to get two green balls 3 times as often as two blue balls. Therefore, if someone passes immediately from stage 1 to stage 3, they will do so having drawn two green balls three times as often as having drawn two blue ones. So, in the long run, on every four occasions that this happens, once it will be because the player drew two blue balls, and three times because they drew two green ones. Thus the probability that it happened because of the two blue balls is $\tfrac14$.