# Probability: Random Selection

Two balls are drawn in succession from a box containing 4 red and 2 white balls.

a) What is the probability of drawing a red ball on the second draw if the first one is not put back in the box after it is drawn?

b)What is the probability of drawing a red ball on the second draw if the first one is put back in the box after it is drawn?

## Solution This solution is **FREE** courtesy of BrainMass!

Part (a)

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There are two ways in which you can draw a red ball on the second draw : (i) You draw a red ball in Draw 1 and then draw another red ball on Draw 2 (without putting the first one back!) OR (ii) You draw a white ball in Draw 1 and then draw a red ball on Draw 2 (again without putting the first one back!)

Since there are 4 red balls and 2 white balls in the box, the probability of drawing a red ball in Draw 1 is 1/4 (one out of 4 red balls) and similarly probability of drawing a white ball in Draw 1 is 1/2.

Now, if you dont put the first ball back, we have reduced the number of RED or WHITE balls in the box. So if in Draw 1, you drew a RED ball and did not put it back, you will have only 3 red balls in the box remaining, so the prob. of drawing a red ball in Draw 2 now becomes 1/3 (see that the prob. increases!). Similarly if in Draw 1 you drew a white ball and did not put it back, the prob. of drawing a red ball in Draw 2 remains at 1/4 because you did NOT reduce the number of red balls in the box. It was 4 before and its still 4.

Probablity in Case(i) : 1/4*1/3 = 1/12.[1/4 from Draw1 and 1/3 from Draw 2]

Probablity in Case(ii) : 1/2*1/4 = 1/8.

The total probability of drawing a red ball on Draw 2 is the SUM of the probablities of the cases outlined above.

Total = 1/12 + 1/8 = 5/24.

Part (b)

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The only difference in this part is that you dont put the drawn ball back. So the probability will not change. Since before any Draw we will have 4 red and 2 white balls in the box, the probabilities will not be affected. Prob. of drawing a red ball will always be 1/4 and prob. of drawing white ball always 1/2. So proceeding as before :

Case(i) : 1/4*1/4 = 1/16

Case(ii) : 1/2*1/4 = 1/8

Total = 1/16 + 1/8 = 3/16.