- #1

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I am reading an online stats book, and there is the following question, which I solved incorrectly, and I think I understand what is my mistake, but I will be grateful for your explanation, if I have incorrectly detected the logic behind my mistake. I am weak at math (trying to improve it on my own - mainly the process of thinking and analyzing math problems), so I am coming here for help.)

Also, as I was analyzing my mistake, I came up with some other ones by thinking about other possible questions about choosing a red object for this particular problem. I am sorry for lengthy and a bit tangled explanation below.

**Problem:**

One of the following 30 items is chosen at random. What is the probability it is an X given that it is red? Put your answer in decimal form.

They give a picture, on which there are 7 red items (zeros and X's) and there are 3 red X-s. Also, there are two more colors - blue and pink, each containing Xs and 0s. The total number of X (red, blue, pink) = 13.

**My solution:**

I have solved the problem by first finding the probability of choosing a red object out of 30 objects:

P(red) = 7/30

And then multiplying this by the probability that X comes up out of these red ones:

P(X | red) = 3/7

Hence, the answer I gave was: P = P(red) x P(X|red) = 0.1

The book give the answer P(X | red) = 3/7 = 0.43

**My explanation of my mistake**- please, let me know if it is correct, or not:

**(1)**P = P(red) x P(X|red) = 0.1 this approach and hence the answer are wrong because

P(red) means that we first compute the probability of "taking out" one red object out of 30 objects, i.e. that is out first event.

And if then we multiply it by P(X | red), it means that we are trying to compute the second event of choosing any (not only red) X out of the now 29 objects, and, if that was a question, the correct approach would be:

P(red) x P(X | red) = (7 / 30) x (13 / 29)

But, when I wrote that, I realized that 13/29 might also be incorrect, because 7/30 might include the red X, and hence when computing the second event there might be less than 13 Xs left. Oops, I am stumbled here.

**(2)**On the other hand, if first we need to choose any red object, and then choose only a red X, the correct approach would be:

P(red) x P(red X | red) = (7 / 30) x (3 / 6) but here again a question - if we choose a red object in the first event and the probability is P(red) = (7 / 30), then P(red X | red) = (3 / 6) might be incorrect in the numerator as there might already be less than 3 red Xs, correct?

Thank you very much.