Question: Chintu put some Black marbles and some White marbles into a jar. He then asked his brother Pintu to take out a marble. Pintu drew out a Black marble. Chintu asked Pintu to draw out another marble, and again he drew out a Black marble.
Pintu thought there must be more Black marbles than White marbles in the jar and asked Chintu, "I wonder what is the probability of me drawing a Black marble on a third try?"
Chintu replied, "Exactly 9/10 of what it was of drawing a Black marble on your first draw."
Can you help Pintu to determine how many marbles of each colour had been in the jar in the beginning? Give the minimal possible answer. Also, Pintu knew that there were at least seven marbles in the jar in the beginning.
Answer:
There were 8 Black marbles and 4 White marbles in the jar.
Let's assume that initially there are total N marbles. Also, B and W are the number of Black and White marbles respectively. Thus, N = B + W
The probability of drawing a Black marble on the first draw = B/(B+W)
Similarly, after drawing 2 Black marbles, the probability of drawing a Black marble on the third draw = (B-2)/(B+W-2)
Chintu said that the probability of drawing a Black marble on third draw (after drawing
Black marbles on first two draws) is exactly 9/10 of what it was of drawing a Black marble
on first draw.
(B - 2) / (B + W - 2) = (9 / 10) * B / (B + W)
10 * (B + W) * (B - 2) = 9 * B * (B + W - 2)
10B2 - 20B + 10BW - 20W = 9B2 + 9BW - 18B
B2 - 2B + BW - 20W = 0
We know that N = B + W, hence substitute W = N - B
B2 - 2B + B(N - B) - 20(N - B) = 0
B2 - 2B + BN - B2 - 20N + 20B = 0
18B + BN - 20N = 0
B(18 + N) = 20N
B = 20N / (18 + N)
Now, we know that the value of N is at least 7. Hence, using trial-n-error on the equation, the minimal value of N must be 12 so that B=8 and W=4.
Hence, initially there were 8 Black marbles and 4 White marbles in the jar.