Hassan has a bag which contains 20 coloured discs. He randomly chose three discs from the bag and found that two were red and one was white.
a) Based on this sample, how many of each coloured disc are in the bag?
b) Hassan returned the three discs to the bag and then randomly seleceted another three. This time he selected two whites and one red. How many of each coloured disc do you think are in the bag?
---> Please help me on how to solve these questions... Thank you
To solve these questions, we can use a probability approach called the "binomial distribution."
In each situation, we are randomly choosing three discs from the bag without replacement, meaning that the probability of each selection changes with each draw.
a) Based on the first sample, where two discs are red and one is white, let's assume there are X red discs and Y white discs in the bag. We need to find the values of X and Y that satisfy this condition.
To begin, let's consider the probability of drawing a red disc. In the first draw, the probability is X/(X+Y), as there are X red discs and X+Y total discs in the bag. Similarly, in the second draw, the probability is (X-1)/(X+Y-1), and in the third draw, it is (X-2)/(X+Y-2).
Likewise, the probability of drawing a white disc in the first draw is Y/(X+Y), the second draw is (Y-1)/(X+Y-1), and the third draw is (Y-2)/(X+Y-2).
Since we have two red and one white disc in the sample, we can multiply the probabilities of the respective draws:
(X/(X+Y)) * ((X-1)/(X+Y-1)) * (Y/(X+Y-2)) = 2/3
To solve this equation, we need to find values of X and Y that satisfy it. We can start by trying different values until we find one that satisfies the equation.
b) For the second sample, where two discs are white and one is red, we can follow a similar approach.
(X/(X+Y)) * ((X-1)/(X+Y-1)) * (Y/(X+Y-2)) = 1/3
Again, we need to find values of X and Y that satisfy this equation by trying different combinations.
Keep in mind that the exact values for X and Y may not be unique, as there could be various combinations that satisfy the equation.