no of red balls = 6
no of blue balls = 4
Total no of balls = 10
Question => what is the probability that 2 balls selected are red balls?
The probability that both balls are red = (Probability that the first ball is red)*(Probability that second ball is red)
Probability that the first ball is red = $$\frac{no\ of\ red\ balls}{total\ no\ of\ balls}=\frac{6}{10}$$
$$Probability\ that\ the\ second\ ball\ is\ red=\frac{no\ of\ red\ balls\ remaining}{Total\ no\ of\ balls\ }$$
$$=\frac{6-1}{10-1}=\frac{5}{9}$$
$$Probablity\ that\ both\ balls\ are\ red=\frac{6}{10}\cdot\frac{6}{9}=\frac{1}{3}$$
This is the result of the probability that the 2 balls selected are red balls (The main question asked; but note that the resulting answer 1/3 cannot be found among the provided options)
However, the Probability of selecting blue balls=
$$\frac{4}{10}=\cdot\frac{3}{9}=\frac{2}{5}\cdot\frac{1}{3}=\frac{2}{15}$$
This is the result of the probability that the 2 balls selected are blue balls, and thus, it results in one of the provided options, which is OPTION E.
