Question

10 marbles what is the probability of drawing 2 blue marbles if the first marble is placed back in the bag before the second draw

Answer 1

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:

$P(B_1\ and\ B_2) = \frac{2}{9}$

Step-by-step explanation:

Given

$Blue = 5$

$Pink = 3$

$Yellow = 2$

$Total = 10$

When the first blue marble is selected, the probability is:

$P(B_1) = \frac{n(Blue)}{Total}$

$P(B_1) = \frac{5}{10}$

Since the first marble is not returned, we have the following marbles left

$Blue = 4$

$Pink = 3$

$Yellow = 2$

$Total = 9$

The probability of picking a second blue marble is:

$P(B_2) = \frac{n(Blue)}{Total}$

$P(B_1) = \frac{4}{9}$

So, the required probability is:

$P(B_1\ and\ B_2) = P(B_1) * P(B_2)$

$P(B_1\ and\ B_2) = \frac{5}{10} * \frac{4}{9}$

$P(B_1\ and\ B_2) = \frac{5*4}{10*9}$

$P(B_1\ and\ B_2) = \frac{20}{90}$

$P(B_1\ and\ B_2) = \frac{2}{9}$