5 red, 7 blue, 8 green. 2 drawn consecutively......suppose each trial...

Question

5 red, 7 blue, 8 green marbles drawn consecutively and without replacement.  Suppose each trial consists of a single marble being drawn and then placed back into bag.  If repeated, find the probability of drawing at least a red marble once in two trials?

Answer 1

Chance of drawing no red marble on the first and second trials is 15/20 or 3/4, because the marble is replaced, and there are 15 non-red marbles. So the chance of both trials producing no red marbles is 3/4*3/4=9/16. Therefore, if we subtract this from 1 we get 7/16 which is the chance of drawing at least one red marble in two trials. 7/16=0.4375 or 43.75%.

Answer 2

1. The chance to draw a red marble on the first trial is 5/20. ··· (1)

2. The chances to draw a red marble in two trials are:

(i). Chance the first trial draws a red, then the second trial draws a red: (5/20){(5-1)/(20-1)}

(ii). Chance the first trial draws a red, then the second trial fails a red: (5/20){(20-5)/(20-1)}

(iii). Chance the first trial fails a red, then the second trial draws a red: {(20-5)/20}{5/(20-1)}

Thus, the chance to draw at least one red in two trials is: (i)+(ii)+(iii)   That is:

(5/20){(5-1)/(20-1)} + (5/20){(20-5)/(20-1)} + {(20-5)/20}{5/(20-1)}

={5(5-1)+5(20-5)+5(20-5)} / 20(20-1)=(20+75+75) / 20x19 = 170/380 = 17/38 ··· (2)

CK: Chance each trial fails to draw a red: {(15-1)/20}{(20-5)/(20-1)}=210/380=21/38 ··· (3)

So, (2)+(3)=(17/38)+(21/38)=38/38=1   CKD.

The answer: The probability of at least one being red is 17/38. (=approx.44.7%).