Easy Probability Question 3?

There are six possible ways to draw two pens from the population of 4.

(Bb, BR, BG, bR, bG, RG)

4 nCr 2 = 6

Among these six possibilities, only one has two blue pens (Bb), and all but one (RG) have at least one blue pen. So one out of five picks having one blue pen will also have a second blue pen.

You can also use the slightly more complicated method of Bayes' theorem, which states that the probability of one event, given a second, can be computed by the probability of the individual events. You'd want to find the probability that two blue pens have been drawn (A) given that one blue pen has been drawn (B).

P(A|B) = P(B|A) P(A) / P(B)

or "the probability of A given B is equal to the probability of B given A, times the probability of A, divided by the probability of B."

P(A) = 1:6

P(B) = 5:6

P(B|A) = 1 (if you drew both blue pens, you also drew one blue pen)

1/6 * 1

-------

5/6

gives you, again 0.2 or 1:5.

Well,

the pencils are named B1, B2, R and G (for green)

we have a total of 4C2 = 4*3/2 = 6 possibilities of drawing two pencils

whereby :

the event (both are blue) = (B1 and B2)

P( both are blue ) = 1/6

the event ( 1 is blue ) = (B1 and R) U (B1 and G) U (B2 and R) U (B2 and G)

therefore :

knowing that one of the pencil is blue, we have 5 possible events

and 1 among them is favorable : (B1 and B2)

so :

P( both are blue | one among them is blue) = 1/5

hope it' ll help !!

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P(B!A) = 1/3 [1B, 1R, 1G]