SOLUTION: An urn contains 3 red and 4 green balls. 4 balls are drawn, one after another subject to the following rule. If the ball drawn is red we keep is out of the urn; if its green we put

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Question 1026005: An urn contains 3 red and 4 green balls. 4 balls are drawn, one after another subject to the following rule. If the ball drawn is red we keep is out of the urn; if its green we put it back. Find the probability that the 4th ball drawn is red.
Found 2 solutions by Fombitz, Edwin McCravy:
Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!
Look at the possible 16 outcomes and count the ones that have a red ball for 4th draw,
RRRR <--- Couldn't happen since only there is not a fourth red ball
RRRG
RRGR <--
RRGG
RGRR <--
RGRG
RGGR <--
RGGG
GRRR <--
GRRG
GRGR <--
GRGG
GGRR <--
GGRG
GGGR <--
GGGG
There are 7 possible outcomes with a fourth draw of red, so then the probability would be,
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
The other tutor paid no attention to the rule:

If the ball drawn is red we keep is out of the urn;
if it's green we put it back.
He wrongly assumed that all had the same probability.

There are 7 ways to draw out 4 with the 4th one R:
RRGR,RGRR,RGGR,GRRR,GRGR,GGRR, and GGGR


1. RRGR
We begin with 3 R's and 4 G's.
The probability of drawing the 1st one R is 3/7.
If we do, there remain 2 R's and 4 G's.
The probability of drawing the 2nd one R is 2/6.
If we do, there remain 1 R and 4 G's.
The probability of drawing the 3rd one G is 4/5.
If we do we put it back, and there remain 1 R & 4 G's.
The probability of drawing the 4th one R is 1/5.
P(RRGR) = (3/7)(2/6)(4/5)(1/5) = 4/175

2. RGRR
We begin with 3 R's and 4 G's.
The probability of drawing the 1st one R is 3/7.
If we do, there remain 2 R's and 4 G's.
The probability of drawing the 2nd one G is 4/6.
If we do we put it back, and there remain 2 R's & 4 G's.
The probability of drawing the 3rd one R is 2/6.
If we do, there remain 1 R and 4 G's.
The probability of drawing the 4th one R is 1/5.
P(RRGR) = (3/7)(4/6)(2/6)(1/5) = 2/105

3. RGGR
We begin with 3 R's and 4 G's.
The probability of drawing the 1st one R is 3/7.
If we do, there remain 2 R's and 4 G's.
The probability of drawing the 2nd one G is 4/6.
If we do we put it back, and there remain 2 R's & 4 G's.
The probability of drawing the 3rd one G is 4/6.
If we do we put it back, and there remain 2 R's & 4 G's.
The probability of drawing the 4th one R is 2/6.
P(RGGR) = (3/7)(4/6)(4/6)(2/6) = 4/63

4. GRRR 
We begin with 3 R's and 4 G's.
The probability of drawing the 1st one G is 4/7.
If we do we put it back, and there remain 3 R's & 4 G's.
The probability of drawing the 2nd one R is 3/7.
If we do, there remain 2 R's and 4 G's.
The probability of drawing the 3rd one R is 2/6.
If we do, there remain 1 R and 4 G's.
The probability of drawing the 4th one R is 1/5.
P(RGGR) = (4/7)(3/7)(2/6)(1/5) = 4/245

5. GRGR
We begin with 3 R's and 4 G's.
The probability of drawing the 1st one G is 4/7.
If we do we put it back, and there remain 3 R's & 4 G's.
The probability of drawing the 2nd one R is 3/7.
If we do, there remain 2 R's and 4 G's.
The probability of drawing the 3rd one G is 4/6.
If we do we put it back, and there remain 2 R's & 4 G's.
The probability of drawing the 4th one R is 2/6.
P(GRGR) = (4/7)(3/7)(4/6)(2/6) = 8/147.

6. GGRR
We begin with 3 R's and 4 G's.
The probability of drawing the 1st one G is 4/7.
If we do we put it back, and there remain 3 R's & 4 G's.
The probability of drawing the 2nd one G is 4/7.
If we do we put it back, and there remain 3 R's & 4 G's.
The probability of drawing the 3rd one R is 3/7.
If we do, there remain 2 R's and 4 G's.
The probability of drawing the 4th one R is 2/6.
P(GRGR) = (4/7)(4/7)(3/7)(2/6) = 16/343.

7. GGGR
We begin with 3 R's and 4 G's.
The probability of drawing the 1st one G is 4/7.
If we do we put it back, and there remain 3 R's & 4 G's.
The probability of drawing the 2nd one G is 4/7.
If we do we put it back, and there remain 3 R's & 4 G's.
The probability of drawing the 3rd one G is 4/7.
If we do we put it back, and there remain 3 R's & 4 G's.
The probability of drawing the 4th one R is 3/7.
P(GRGR) = (4/7)(4/7)(4/7)(3/7) = 192/2401.

4/175+2/105+4/63+4/245+8/147+16/343+192/2401 = 163558/540225

A little over 30% of the time. 

Edwin
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