A bag has red, black and white stones.
There are 2 red stones.
Twice as many black as red, that means 4 black stones.
The remaining two-fifths of the stones are white.
That is,
If three-fifth of the stones are 6.
Then, two fifth of the stones are,
[tex]\begin{gathered} \frac{2}{3}\times6=2\times2 \\ =4 \end{gathered}[/tex]Therefore, there are 4 white stones.
The probability that the first stone is red and the second stone is white is,
Using the formula,
[tex]^nC_r=\frac{n!}{(n-r)!r!}[/tex]Here, n is the total number of items and r is the number of selected items.
[tex]\begin{gathered} \frac{^2C_1}{^{10}C_1}\times\frac{^4C_1}{^9C_1}=\frac{\frac{2!}{(2-1)!1!}}{\frac{10!}{(10-1)!1!}}\times\frac{\frac{4!}{(4-1)!1!}}{\frac{9!}{(9-1)!1!}}\Rightarrow\frac{\frac{2\times1}{1}}{\frac{10\times9!}{9!}}\times\frac{\frac{4\times3!}{3!}}{\frac{9\times8!}{8!}} \\ =\frac{2}{10}\times\frac{4}{9} \\ =\frac{1}{5}\times\frac{4}{9} \\ =\frac{4}{45} \end{gathered}[/tex]Hence, the correct option is B.
The probability that the first stone is red and the second stone is white is,
[tex]\frac{4}{45}[/tex]