Given:
The total number of marbles in the bag, T=15.
The number of white marble, W=2.
The number of blue marbles, B=5.
The number of red marbles, R=8.
The probability of choosing a white marble is,
[tex]P(White)=\frac{W}{T}=\frac{2}{15}[/tex]
The probability of choosing a blue marble is,
[tex]P(Blue)=\frac{B}{T}=\frac{5}{15}=\frac{1}{3}[/tex]
The probability of choosing a red marble is,
[tex]P(Red)=\frac{R}{T}=\frac{8}{15}[/tex]
The probability of choosing two reds in a row is a dependent event.
There are 8 red marbles in the bag. So, the probability of choosing a red in the first draw is,
[tex]P(\text{ Red in first draw)=}\frac{8}{15}[/tex]
Since one red ball is drawn from the bag, there are only 7 red balls remaining in the bag. Also, there are only 14 balls remaining in the bag. So, the probability of choosing a red in the second draw is,
[tex]P(\operatorname{Re}d\text{ in the second draw)=}\frac{7}{14}[/tex]
Now, the probability of choosing two reds in a draw is,
[tex]\begin{gathered} P(\text{Two reds in }a\text{ row)=}P(\operatorname{Re}d\text{ in first draw)}\times P(\operatorname{Re}d\text{ in second draw)} \\ =\frac{8}{15}\times\frac{7}{14} \\ =\frac{28}{105} \end{gathered}[/tex]
