dave13 wrote:
Bunuel wrote:
There are three red, three blue, and three green balls in an opaque bag. Alan draws three balls at random without replacement, then passes the bag to Bridget, who does the same, as does Charlie. What is the probability that Charlie will draw three balls of the same color?
A. 1/56
B. 1/28
C. 1/14
D. 1/4
E. 1/3
VeritasKarishmai used method (1- all non favorable results) and focused only on Charlie.
Case 1. Charlie draws two of the same color and one of another
\(\frac{3}{9}*\frac{2}{8}*\frac{3}{7} = \frac{1}{28} \)
so since i can have 6 such events: (2R +1G, 2R+1B, 2B+1G, 2B+1R, 2G+1R, 2G+1B ) multiply 1/28 by 6 = \(\frac{3}{14}\)
Case 2: Charlie draws 3 balls of different colors (R *G * B)
\(\frac{3}{9}*\frac{3}{8}*\frac{3}{7} = \frac{3}{56}\\
\)
So total number of unfavorable outcomes for Charlie is \(\frac{3}{14}+\frac{3}{56} = \frac{15}{56}\)
so i get this answer \(1- \frac{15}{56} = \frac{41}{56 }\)
can you pls let me know where am I making mistake ?
Yes, whether Alan and Bridget pick 6 balls and Charlie is left with 3 or Charlie picks 3 balls and Alan and Bridget are left with 6 is the same. Hence, focus on Charlie is fine.
But it is easier to calculate the direct probability.
In the first pick, Charlie can pick any colour ball. Probability = 1
In the second pick, Charlie must pick the same colour as picked previously. Probability = 2/8
In the third pick, Charlie must pick again the same colour as picked 2 times before. Probability = 1/7
Total Probability = 1*(2/8)*(1/7) = 1/28
Answer (B)
If you do go the (1 - Non favourable) way,
Charlie pick 2 balls of same colour: First pick any colour, second pick same colour as before, third pick a different colour. Say RRB. We multiply by 3!/2! to get arrangements of RRB.
Probability = 1*(2/8)*(6/7) * 3!/2! = 18/28
Charlie picks all 3 balls of diff colours: First pick any colour, second pick another colour, third pick the third colour.
Probability = 1*(6/8)*(3/7) = 9/28
Total Probability = 27/28
Required probability = 1 - 27/28 = 1/28