Question

Two urns each contain green balls and blue balls. Urn I contains 4 green balls and 6 blue balls, and Urn II contains 6 green balls and 2 blue balls. A ball is drawn at random from each urn. What is the probability that both balls are blue?

A. 2/51
B. 3/20
C.1/10
D.4/153

Answer 1

Urn 1 contains a total of 4 + 6 = 10 balls. 6 of the 10 balls are blue. What is the probability of drawing a blue ball:
\[P(blue\ from\ urn\ 1)=\frac{number\ of\ blue\ balls}{total\ number\ of\ balls}=?\]

Answer 2

10

Answer 3

1/10 right?

Answer 4

Since it's a independent event, thus \[P(blue, urn1)=\frac{6}{10}\] and \[P(blue, urn2)=\frac{2}{8}\], then the \[P(blue, urn1\ and\ blue,urn2)=\frac{6}{10}.\frac{2}{8}\]

Answer 5

4/153

Answer 6

Not really. We are just looking at urn 1, that has 6 blue balls and a total number of balls = 4 + 6 = 10
So the probability of drawing a blue ball from urn 1 is 6/10.
By the same reasoning the probability of drawing a blue ball from urn 2 is 2/8.
The two events are independent therefore their intersection is given by
\[\frac{6\times 2}{10\times 8}= you\ can\ calculate\]
Hint: Remember to simplify your answer.

Answer 7

i got it

Answer 8

Good! What answer did you get?

Answer 9

i multiplied 6x2= 12 and 10x8=80 1280?

Answer 10

\[\frac{6\times 2}{10\times 8}=\frac{12}{80}=\frac{3}{20}\]

Answer 11

oh you divide 12/80? ok

Answer 12

thanks :)

Answer 13

You're welcome :)