Answer to Question #91018 in Statistics and Probability for gatare james

Question

A bag contain ten marbles which can be described by colour and whether it is numbered or lettered as follows
4 are white (W) and lettered (L)
2 are white (w) and numbered (N)
3 are yellow (y) and lettered (L)
1 is yellow (Y) and numbered (N)
One marble is randomly drawn from the bag and it is found to be white, what is the probability that is numbered.

Answer 1

For the white marbles we have:

4 lettered

2 numbered

6 total

Probability to get a white A marble is given by

$$P(A) = \frac{N_A}{N_{\mathrm{total}}}$$

where N_A is the number of white A marbles and N_total is the total number of white marbles.

4/6 to find that the white marble is lettered

2/6 to find that the white marble is numbered

The answer: 2/6=1/3 (regardless the distribution of the yellow marbles, for we did not need it at all)

We can also obtain this result from the definition of conditional probability:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

Here A stands for the marble is numbered, B - the marble is white.

We have:

6 white marbles (B)

2 white and numbered ones (A and B)

10 in total

using the first expression, we obtain

$$P(B) = \frac{N_B}{N_{\mathrm{total}}} = \frac{6}{10} = \frac{3}{5}$$ $$P(A \cap B) = \frac{N_{\mathrm{A\, and\, B}}}{N_{\mathrm{total}}} =\frac{2}{10} = \frac{1}{5}$$

and

$$P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{1}{5} : \frac{3}{5} = \frac{1}{3}$$