probability that at most two white balls are drawn?

Question

a bag contains a red ball, a white ball, and a green ball. a ball is drawn, the color noted, and the ball replaced. this experiment is repeated 3 times.

 

find the probability of the event E: at most 2 white balls are drawn?

Answer 1

Hi Aubrey,

 

When we are considering probability with replacement then we follow these steps:

    1.  Determine the probability of each event independently (as a separate event).

    2.  Determine the final probability of all the events happening by multiplying together the probabilities obtained in step 1.

 

Let's define our three separate events as follows where at most there are two white balls drawn in three separate drawings:

    1.  P(1):  probability of NOT drawing a white ball.

    2.  P(2):  probability of drawing the 1st white ball.

    3.  P(3):  probability of drawing the 2nd white ball.

 

It does NOT matter in which order the above three separate events occur as long as they do occur.  We will simply multiply their probabilities together in whatever order they occur.

 

Let's calculate the probability for each of the above three separate events:

 

    1.  P(1) = desired outcomes / possible outcomes

         P(1) = 2/3 (no white ball drawn)

    2.  P(2) = desired outcomes / possible outcomes
         P(2) = 1/3 (1st white ball drawn)

    3.  P(3) = desired outcomes / possible outcomes

         P(3) = 1/3 (2nd white ball drawn)

 

Now we'll determine the final probability, P(E), of ALL three events happening:

    P(E) = P(1) x P(2) x P(3)

    P(E) = 2/3 x 1/3 x 1/3

    P(E) = 2/27

 

Thanks for submitting this problem and glad to help.

 

God bless, Jordan.