probability tree (1 Viewer)

[h_g]

two white balls and one black ball are placed in a bag. Two balls are selected at random. Find the probability of choosing:

a) 2 white balls
b) 1 white and 1 black ball

 

[Axio]

W-W
W-W
W-B
W-B
B-W
B-W

a. 2/6 = 1/3
b. 4/6 = 2/3

 

[h_g]

how did u get for b) 2/3? shouldn't it be 1/3?

 

[Axio]

how did u get for b) 2/3? shouldn't it be 1/3?

I think the question is asking for the probability of picking a white and a black ball overall, not a white ball then a black ball... is that what you were getting at?

 

[h_g]

yea, can u post a pic of the tree diagram u drew.

 

[Axio]

yea, can u post a pic of the tree diagram u drew.

Just draw:

- a point
- then coming off of that point is three branches: W1, W2, B
- then coming off of those is two more branches each (for a total of six pathways), with either W1, W2 or B coming off of those (the one you have already used in a pathway doesn't get repeated)

 

[Papercutter]

Are the two balls picked with replacement?

 

[enigma_1]

Are the two balls picked with replacement?

^ We need to know this, OP

Is this a stupid question from MIF or something..

 

[Flop21]

With replacement:

a) 2/3 x 2/3 = 4/9

b) 2/3 x 1/3 + 1/3 x 2/3 = 4/9

The tree diagram will just have B and W in the first branch, then B and W coming off the first branches. Then if you want you put in the fractions on the branches. So 'B' on the first branch will be 1/3 then the B coming off that will also be 1/3 if the ball is replaced. So to work out 2 black balls for example, you would times these two fractions. As you always times along the branch, but add separate branches if need be. Like I did in b).

However, if it is without replacement:

a) 2/3 x 1/2 = 1/3

b) 1/3 x 2/2 + 2/3 x 1/2 = 2/3


If you need any more help let me know. And if I've done it wrong let me know also lol.