Probability of Two Marbles Drawn: Black & Diff Cols

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The problem involves drawing two marbles from a bag containing 3 black, 2 green, and 1 white marble, with replacement. The probability of drawing two black marbles is calculated as 1/4. For the probability of drawing two marbles of different colors, two methods yield the same result of 22/36. Both attempts at calculating the probability of different colors are confirmed as correct. The discussion emphasizes the accuracy of the calculations and the clarity of the approaches used.
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Homework Statement


A bag contains 3 black marbles, 2 green marbles and 1 white marble. Two marbles are drawn at random from the bag WITH replacement. Find the probabilities that the two marbles are

(a)black
(b)of different colours


Homework Equations





The Attempt at a Solution



(a)P(both black) = 3/6 x 3/6 = 1/4

(b) I have two attempts for this.

Attempt A
P(of different colours) = P(B,G) + P(B,W) + P(G,W) + P(G,B) + P(W,B) + P(W,G)
= (3/6)(2/6) + (3/6)(1/6) + (2/6)(1/6) + (2/6)(3/6) + (1/6)(3/6) + (1/6)(2/6)
= 22/36

Attempt B
P(of different colours) = 1 - P(both black) - P(both green) - P(both white)
= 1 - (3/6)(3/6) - (2/6)(2/6) - (1/6)(1/6)
= 22/36

Any problems for any of the attempt above?
 
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hi kenny1999! :wink:

looks fine :smile:

(B is probably neater than A)
 
tiny-tim said:
hi kenny1999! :wink:

looks fine :smile:

(B is probably neater than A)

but are they both correct?
 
yes of course!

(didn't i say so? :smile:)
 

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