Question

A bucket contains exactly 3 marble, one red, one blue and one green.

A person arbitrarily pulls out each marble one at a time.

What is the probability that the last marble removed is non-red?

Answer in the form of a fully reduced fraction.

Answer 1

Here ,following actions are taking place.

1. Marbles are getting pulled without replacement.

2. Different colored marbles are selected each time.

Now , for the third marble to be non red (i.e, blue or green) , either first or second ball should be red.

Let us assume that first ball is red

Then its probability of being red is 1/3

Let us assume that second ball is red

Then first ball should be either blue or green and second ball should be red.

First ball blue/green probability =2/3 (two favourable cases required out of 3)

Second ball red =1/2 (one favourable case required out of two cases)

Thus net probabilty of red ball in second draw =2/3*1/2 =1/3

Thus overall probabilty = Probability of first ball red +probability of second ball red

=1/3+1/3 =2/3

=0.66667

=66.67 %

Thus there is a two third chance that the third ball will be on red.

This conclusion can also be drawn from a simple assertion that for the third ball to be blue or green,

the probability will be 2/3 ( two favourable cases needed out of three possible cases)