Red Herring?
One green ball, one blue ball, and two red balls are placed in a bowl. I draw two balls simultaneously from the bowl and announce that at least one of them is red.
What is the chance that the other ball I have drawn out is also red?
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4 solutions
Color of two balls is red, so we will take those two red balls as identical balls. Using conditional probability, the answer should be 3 1 . And also given in the question, two balls are drawn simultaneously. So Green-Red, and Red-Green will be considered as one pair, not two different cases. !!!
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I also got 1/3
Dear Sandeep, you can pick green with red ball 1 and green with red ball 2. Hence the two different cases.
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I answered the question using the concept of two red balls being identical due to their color. in that calls green -red is same with both the red balls. And why can't we take two red balls identical ? also the color is red in both the cases. However, I understand what you want to say.
If i've already took one red, then the other ball is green, blue, or red.. so, there's only 1/3 chance that the other ball is red.. then, why the answer should be 1/5??
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i agree.....
conditional probability
Let me word this in the language of Conditional Probability. There are six possible ways to draw 2 balls { R1 R2. R1B1, R2B1, R1G1, R2G1, BG}
Probability of event (A) - drawing one red ball = 5/6
Probability of (A and B) = drawing both red balls = 1/6.
Probability of ( B given A) = Probability (A and B) / Probability (A)
= (1/6) / (5/6) = 1/5...
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What is the difference between drawing simultaneously and drawing without replacement, the two red balls.
I wish all solutions could be written as clearly as this :), rather than with the use of the modern notation. I last did maths 43 yrs ago!!
How could I miss the obvious?
@satyen nabar , is probability your favourite area in maths? (Your problems have been quite good!)
wont we consider the chance of announcing that one ball is red of the two.
confusing one indeed,, if one misses the simultaneously part
Out of total 4 balls, 2 balls can be chosen in 4 C 2 = 6 ways. Since at least one ball must be red, we exclude G B combination leaving total of 5 combinations ( R 1 R 2 , R 1 G , R 1 B , R 2 G , R 2 B ). Only one out of these is with both reds. Hence the final probability is 5 1
Possible combinations of selecting two ball from the available balls are R1R2, R1G2, R1B2, G1R2, G1G2, G1B2, B1R2, B1G2, B1B2. Out of all those possibilities five include at least one red and one possibility is both are red. So required probability is 1/5. Smile!!!
The total of ways to draw 2 balls from 4 is N = 4 × 3 = 1 2 ways. Out of these 12 ways.
- Blue+Green: 2 ways
- Blue+Red: 4 ways
- Green+Red: 4 ways
- Red+Red: 2 ways
After the draw, at least one of the ball is red, the possible outcomes are N 1 = 1 0 . And out of these 10 ways, there are 2 ways both the balls are red. Therefore, the probability is 1 0 2 = 5 1 .
Bayes Thorem:
P(2R¦R) = P(R¦2R) * P(2R) / P(R)
P(R¦2R) = 1 trivially
P(2R) = 1/2 * 1/3 = 1/6
P(R) = 1- P(no R) = 1 - 1/2* 1/3 = 5/6
P(2R¦R) = 1 * (1/6) / (5/6) = 1/5
what if the red balls are identical? i think then the answer will be 1/3
The question states that one red ball was drawn, so there are only three probabilities left, red red, red green, and red blue
Getting two red is 4 × 3 2 × 1 = 1 2 2 = 6 1 .
At least one red is the opposite of no red --"at least" problems almost always use this trick-- i.e. 1 − 1 2 2 = 1 2 1 0 = 6 5 .
Use conditional probability and 6 1 ÷ 6 5 = 5 1 .
(x,y) denotes first ball was color 'x' and second ball was color 'y'
Sample space (conditioned on the given information):
| outcome | # possible ways outcome can occur |
| (r,g) | 2 (... 2 choices for first red ball x 1 choice for second green ball) |
| (r,b) | 2 (... same logic applies) |
| (r,r) | 2 |
| (g,r) | 2 |
| (b,r) | 2 |
Thus, all 5 outcomes are equally likely and only 1 is a "victory", so the probability is 1/5
I disagree. I think the probability should be 2/12 or 1/6 because there are 12 distinct ways of pulling two balls out assuming order matters and only 2 of those ways have both red balls. I don't think announcing that at least one of the balls is red should change the probability of them both being red.
There are 6 possible pairings of the two balls drawn,
Red 1---Red 2
Red 1---Blue
Blue---Red 2
Red 1---Green
Green---Red 2
Green---Blue.
We know the Green ---Blue combination has not been drawn.
This leaves five possible combinations remaining.
Therefore the chance that the Red 1---Red 2 pairing has been drawn are 1 in 5.
Note --- The solution of course would be 1/3, if the balls had been drawn out separately and the color of the first ball announced as Red before the second had been drawn out.
However, as both balls had been drawn together, and then the color of one of the balls announced, the chance is 1/5.
Conditional Probability. There are six possible ways to draw 2 balls { R1 R2. R1B1, R2B1, R1G1, R2G1, BG}
Probability of event (A) - drawing one red ball = 5/6
Probability of (A and B) = drawing both red balls = 1/6.
Probability of ( B given A) = Probability (A and B) / Probability (A)
= (1/6) / (5/6) = 1/5...