Answer: [tex]\dfrac{2}{13}[/tex]
Step-by-step explanation:
Given : A bag contains 6 red balls, 4 green balls, and 3 blue balls.
Total balls = 6+4+3=13
Probability of drawing first ball as green :
[tex]P(G)=\dfrac{\text{Number of green balls}}{\text{Total balls}}\\\\=\dfrac{4}{13}[/tex]
If the first ball will be green, then the total balls left in bag = 13-1=12 and number of red balls remains the same.
Now, The conditional probability of drawing a red ball given that first ball was green :-
[tex]P(R|G)=\dfrac{\text{Number of red balls}}{\text{Total balls left}}\\\\=\dfrac{6}{12}=\dfrac{1}{2}[/tex]
Now, the probability that the first ball will be green and the second will be red will be :-
[tex]P(G\cap R)=P(G|R)\times P(G)\\\\=\dfrac{4}{13}\times\dfrac{1}{2}=\dfrac{2}{13}[/tex]
Hence, the required probability = [tex]\dfrac{2}{13}[/tex]
