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

User Coquin
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1 Answer

4 votes

Answer:

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

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}} = (6)/(10) = (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}} = (4)/(10) = (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

User Eric Speelman
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