probability question

I am completely confused. I cant think on how to solve this but it might just be (5C2 +5C2)/10C2. But I am not sure.
 
It turns out my answer is right. But I cant comprehend it too well.
 
Sonal7 said:
It turns out my answer is right. But I cant comprehend it too well.
Click to expand...
I am not sure what might convince you but this may help.
\(B_1;~B_2;~B_3\\B_1;~W_2;~B_3\\W_1;~W_2;~W_3\\W_1;~B_2;~W_3\)
That is a list of the possible events for the first three events in which the first and third ball are the same colour.
The probability of the first is \(\dfrac{5}{10}\cdot\dfrac{4}{9}\cdot\dfrac{3}{8}\)
The probability of the second is \(\dfrac{5}{10}\cdot\dfrac{5}{9}\cdot\dfrac{4}{8}\)
Add those two together and double it to get \(\dfrac{4}{9}\).
 
pka said:
I am not sure what might convince you but this may help.
\(B_1;~B_2;~B_3\\B_1;~W_2;~B_3\\W_1;~W_2;~W_3\\W_1;~B_2;~W_3\)
That is a list of the possible events for the first three events in which the first and third ball are the same colour.
The probability of the first is \(\dfrac{5}{10}\cdot\dfrac{4}{9}\cdot\dfrac{3}{8}\)
The probability of the second is \(\dfrac{5}{10}\cdot\dfrac{5}{9}\cdot\dfrac{4}{8}\)
Add those two together and double it to get \(\dfrac{4}{9}\).
Click to expand...
Very nice :) - also the use of symmetry.
 
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Sonal7 said:
I am completely confused. I cant think on how to solve this but it might just be (5C2 +5C2)/10C2. But I am not sure.
Click to expand...
Let's consider whether this is correct work (which it might not be though it gives the right answer, as you know).

Your expression gives the number of ways to choose either two black balls, or two white balls, over the number of ways to choose 2 balls without regard to color.

All you need to do is to justify ignoring the middle ball entirely. Can you do so? Or, can you write a different expression that doesn't ignore the middle ball?
 
This is such a well know example that is used to dislodge some misconceptions.
What is the probability that the first and last balls are the same colour?
Five black balls & five white balls. If the first & last are both black then the other five white and three black can be arranged in
\(\dfrac{8!}{5!\cdot3!}=\mathcal{C}^8_5\) ways. Where as there are a total \(\mathcal{C}_5^{10}\) ways to arrange the ten balls.
SEE HERE Double that it equals \(\dfrac{4}{9}\) Do you want to guess the probability that fourth and eighth balls are the same colour?
 
pka said:
This is such a well know example that is used to dislodge some misconceptions.
What is the probability that the first and last balls are the same colour?
Five black balls & five white balls. If the first & last are both black then the other five white and three black can be arranged in
\(\dfrac{8!}{5!\cdot3!}=\mathcal{C}^8_5\) ways. Where as there are a total \(\mathcal{C}_5^{10}\) ways to arrange the ten balls.
SEE HERE Double that it equals \(\dfrac{4}{9}\) Do you want to guess the probability that fourth and eighth balls are the same colour?
Click to expand...
It should be the same right? 4/9
 
Dr.Peterson said:
Let's consider whether this is correct work (which it might not be though it gives the right answer, as you know).

Your expression gives the number of ways to choose either two black balls, or two white balls, over the number of ways to choose 2 balls without regard to color.

All you need to do is to justify ignoring the middle ball entirely. Can you do so? Or, can you write a different expression that doesn't ignore the middle ball?
Click to expand...
I had not realised what you meant by middle balls, I do now. Thank you.
 
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