Complete Question:
There are 10 marbles. What is the probability of drawing 2 blue marbles if the first marble is NOT placed back in the bag before the second draw.
2 yellow, 3 pink, 5 blue
Answer:
[tex]P(B_1\ and\ B_2) = \frac{2}{9}[/tex]
Step-by-step explanation:
Given
[tex]Blue = 5[/tex]
[tex]Pink = 3[/tex]
[tex]Yellow = 2[/tex]
[tex]Total = 10[/tex]
When the first blue marble is selected, the probability is:
[tex]P(B_1) = \frac{n(Blue)}{Total}[/tex]
[tex]P(B_1) = \frac{5}{10}[/tex]
Since the first marble is not returned, we have the following marbles left
[tex]Blue = 4[/tex]
[tex]Pink = 3[/tex]
[tex]Yellow = 2[/tex]
[tex]Total = 9[/tex]
The probability of picking a second blue marble is:
[tex]P(B_2) = \frac{n(Blue)}{Total}[/tex]
[tex]P(B_1) = \frac{4}{9}[/tex]
So, the required probability is:
[tex]P(B_1\ and\ B_2) = P(B_1) * P(B_2)[/tex]
[tex]P(B_1\ and\ B_2) = \frac{5}{10} * \frac{4}{9}[/tex]
[tex]P(B_1\ and\ B_2) = \frac{5*4}{10*9}[/tex]
[tex]P(B_1\ and\ B_2) = \frac{20}{90}[/tex]
[tex]P(B_1\ and\ B_2) = \frac{2}{9}[/tex]
