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(i) There are $4$ red and $6$ black balls. A ball is drawn at random, its colour is observed and this ball with another two balls of same colour are returned. Now, if a ball is drawn at random, what is the probability that the ball is red?

MY WORK :

If the ball drawn at first is red, then the probability that the last one is red: $$\frac{6}{12}$$

But, if the ball drawn at first us black, then the probability of the last one to be red: $$\frac{4}{12}$$

So, the probability is: $$\frac{6}{12}+\frac{4}{12}$$ $$=\frac{5}{6}$$ ...

But, my answer doesn't match. Why?

(ii) $6$ points are taken inside a circle . What is the probability that the points lie in the semi circle?

MY WORK :

For a particular point, the probability is: $$\frac{\text{Area of semi circle}}{\text{Area of circle}}$$ $$=\frac{1}{2}$$

So, for $6$ points, the probability becomes: $$\frac{1}{2^6}$$

Am I correct ?

N. F. Taussig
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Entrepreneur
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1 Answers1

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The probability of the second ball being red is the sum of probabilities of choosing a red ball both times and choosing a black ball first then a red ball.

$\frac{4}{10}\cdot \frac{6}{12} + \frac{6}{10}\cdot \frac{4}{12} = \frac{48}{120} = \frac{2}{5}$

Your answer would be correct for the second question if you defined the half of the circle before selecting the points. This isn't the case.

It matters not where the first point is placed and assuming this point is within a specific semi circle, say upper or lower, or left or right, after that its a sequence of probabilities that the remaining five points are within that semicircle. Because the first point can be any one of $6$ the probability is:

$$6\cdot (\frac{1}{2})^5$$

Phil H
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  • Can you please help with the second one? – Entrepreneur Jul 10 '18 at 17:49
  • See my revised answer including the second question. – Phil H Jul 10 '18 at 19:48
  • It's not clear why the first point should be at the centre of the semicircle -- why not at some other point? Also, the six events thus defined aren't mutually exclusive, so you can't just add their probabilities. If you consider the semicircle that's to one side (say, the clockwise side) of the point, the six events become mutually exclusive, and your result follows. – joriki Jul 11 '18 at 09:45
  • See also https://math.stackexchange.com/questions/325141. – joriki Jul 11 '18 at 09:54
  • @joriki The 1st pt at the center defines a semi circle into which the other 5 pts each have a prob of 1/2 to be in. If this isn't defined surely the prob is >1/2 for the 2nd point. Thinking about this a little more, in that case it wouldn't matter where the 2nd is placed. And the third pt also has a prob >1/2 of being within a semi circle containing all 3 with a prob of 1 if the first 2 are exactly opposite....tricky. From the previous discussion I see that the semi circle chosen is either the upper or lower semi circle depending on where the first pt is placed. That makes more sense. – Phil H Jul 11 '18 at 14:02
  • I think the key point is to ensure that the events are mutually exclusive and exhaust the possibilities for all the points to be in a semicircle. That's achieved by this way of defining the semicircles. – joriki Jul 11 '18 at 14:17
  • Interesting. To achieve that key point, the orientation of the semi circle split must be adjusted after every additional point to contain all previous points (if possible).. Otherwise a situation can occur with 3 points in an upper semi circle and 3 points in a lower, yet all 6 points are in a left semi circle split by a vertical line through the center. Is that correct or am I missing something? I see this topic went through a lot of previous discussion so sorry if I am just rehashing old stuff. – Phil H Jul 11 '18 at 17:16