Question

Drawing a red marble from a bag of six red and 4 blue replacing it and then drawing a blue marble

Answer 1

Answer:

The probability of picking a red marble and a blue marble = 6/25

Step-by-step explanation:

Here, the question is INCOMPLETE.

What is the probability of this independent event. Drawing a red marble from a bag of 6 red and 4 blue marbles, replacing it, and then drawing a blue marble.

Here, total number of red marble in bag = 6

total number of blue marble in bag = 4

So, total marbles in the bag = 6 + 4 = 10 marbles

Now,if E : Any given event, then

$\textrm{P (Event E)} = \frac{\textrm{Total number of favorable outcomes}}{\textrm{Total Outcomes}}$

So, here $\textrm{P (Picking a Red marble)} = \frac{\textrm{Total number of red marble}}{\textrm{Total marble}} = \frac{6}{10} = \frac{3}{5}$

⇒ P(a Red ball)  = 3/5    ...... (1)

Now, if the picked marble is REPLACED, then total marbles in the bag  = 10

So, here $\textrm{P (a Blue marble)} = \frac{\textrm{Total number of blue marble}}{\textrm{Total marble}} = \frac{4}{10} = \frac{2}{5}$

⇒ P(a Blue marble)  = 2/5    ...... (2)

So, the probability of picking a red marble and a blue marble AFTER REPLACING

=P(a Red marble) x P(a Blue marble)  = 3/5 x 2/5  = 6 /25

Hence, the probability of picking a red marble and a blue marble = 6/25