Can someone help me out with this Maths Question?

I think its from Edexcel June 2003 Maths A 1387 (Non Calc):

A bag contains 3 black beads, 5 red beads, and 2 green beads. Gianna takes a bead at random from the bag, records it colour and replaces it. She does this two more times.

Work out the probability that, of the three beads Gianna takes, exactly two are the same.

Thanks to anyone who helps!

Reply 1

icy_shadow2006

i did this paper a few days ago,lol

if it helps u u can draw the tree diagram but can't be assed to do it on the comp. so these are the calculations;

3/10 x 2/9 = 1/15

5/10 x 4/9 = 2/9

2/10 x 1/9 = 1/45

1/15 + 2/9 + 1/45 = 14/45

ans = 14/45

do u understand that?

Reply 2

icy_shadow2006

lol, u helped me, and i helped u!!!

Reply 3

Could you explain it a bit more please? Thanks.

Reply 4

icy_shadow2006

if u think about it her first choice could be anything
the probablity that it would be black is 3/10 because there are 3 black beads, 10 becuase there are 10 in total.

Then the chance that her second choice would also be black is 2/9, 2 because one black bead is gone, and 9 because as 1 bead gone there is a total of 9. U times these two together to find the total chance of getting 2 black beads in a row.

Then u do the same to the other colours and add them all together

does that make things clearer or not?

Reply 5

But Gianna replaces the beads.

awwww

i didn't see that, sorry,
then u just do the same thing but keep to the origonal numbers
eg. 3/10 x 3/10
do that for eac colour and then add the total 2gether
sorry about that

Reply 8

Is there a short way of doing this? It's a non-calc paper.

Reply 9

icy_shadow2006

don't think so, but it will carry a few marks, probably 3 or 4

Reply 10

Ok, thanks.

Anyone else who knows a short way?

Reply 11

kidda_puttar

There is no short way unfortunately!

If you find it a bit hard, draw a tree diagram (google it)....it'll explain to you exactly what you need to do and make the question easier.

Reply 12

Chewwy

you can attack it frome either way, both are equally long and painful:

(3((2x2x3)+(2x2x5)+(3x3x2)+(3x3x5)+(5x5x2)+(5x5x3)))/1000 = 1 - 6(2x5x3/1000) - 2x2x2/1000 - 5x5x5/1000 - 3x3x3/1000 = 660/1000

Reply 13

Newton

There are six possible combinations in which what you are asked for can happen

BRR, BGG
RBB, RGG
GBB, GGG

But either of these combinations can be achieved in three ways

BRR=RBR=RRB
BGG=GBG=GGB

So you want the probability that the outcome is either of these combinations

=>P(EXACTLY TWO ARE SAME)=P(BRR) OR P(BGG) OR...

=>P(2S)=P(BRR)+P(BGG)+...

P(B)=(3/10)

P(R)=(5/10)

P(G)=(2/10)

=>P(2S)=

(3/10)((5/10)^2)+(3/10)((2/10)^2)
+
(5/10)((3/10)^2)+(5/10)((2/10)^2)
+
(2/10)((3/10)^2)+(2/10)((5/10)^2)

=

(87/1000)+(13/200)+(17/250)

=

(11/50)

Although since there are three ways in which each combination can be achieved

P(2S)=3*(11/50)

=>P(2S)=(33/50)

Newton.

Reply 14

davros

i got 33/50 (same answer as chewwy because 660/1000 = 66/100 = 33/50)

Remember if you're subtracting the case when 3 are all the same you CAN have GGG because the balls are replaced after each draw!

Reply 15

El Stevo

davros

i got 33/50 (same answer as chewwy because 660/1000 = 66/100 = 33/50)

Remember if you're subtracting the case when 3 are all the same you CAN have GGG because the balls are replaced after each draw!

GGG is not exactly two greens, as Newton noticed.

Reply 16

davros

no you misunderstand me!

You can either work out exactly 2 of a colour and 1 of a different one

OR

work out prob(2 same 1 diff) = 1 - p(all 3 same) - p(all 3 different)

because there are 3 mutually exclusive situations.

Reply 17

El Stevo

two reds and a none-red = 3(5/10)^2(5/10)
two blacks and a none-black = 3(3/10)^2(7/10)
to greens and a none-green = 3(2/10)^2)(8/10)

= 3/8 + 189/1000 + 12/125
= 33/50