Respuesta :
The probability if 3 marbles being red and 1 non red
= [tex](0.2 m) \times (\frac{0.2 m - 1}{m-1}) \times (\frac{0.2 m - 2}{m-2}) \times (\frac{0.8 m }{m-3})[/tex]
Step-by-step explanation:
Here, let us assume that the total number of the marbles in the bag = m
Now, as given :
The total percentage of red marbles in bag = 20% of m
So, total red marbles = 0.2 m
Marbles which are not red = m - 0.2 m = 0.8 m
The number of marbles chosen at random = 4
Now, probability of getting first red marble = [tex]\frac{\textrm{Total Red Marbles}}{\textrm{Total Marbles }} = \frac{0.2 m}{m} = 0.2[/tex]
Again, total marbles left = m - 1
Probability of getting second red marble = [tex]\frac{\textrm{Total Red Marbles}}{\textrm{Total Marbles }} = \frac{0.2 m - 1 }{m-1}[/tex]
Again, total marbles left = m - 2
Probability of getting third red marble = [tex]\frac{\textrm{Total Red Marbles}}{\textrm{Total Marbles }} = \frac{0.2 m - 2 }{m-2}[/tex]
Again, total marbles left = m - 3
Probability of getting fourth non - red marble = [tex]\frac{\textrm{Total non Red Marbles}}{\textrm{Total Marbles }} = (\frac{0.8 m }{m-3})[/tex]
So, the probability if 3 marbles being red and 1 non red
= [tex](0.2 m) \times (\frac{0.2 m - 1}{m-1}) \times (\frac{0.2 m - 2}{m-2}) \times (\frac{0.8 m }{m-3})[/tex]
