PROBABILITY....who can try to solve it?

This equates to having 6 red and 5 blue and trying to draw 2 red in 3 draws.

There are 4 possible successful draws:

all 3 red

red, red, blue

red, blue, red

blue, red, red

The probabilities of each would be:

6/11 * 5/10 * 4/9 = 120/990

6/11 * 5/10 * 5/9 = 150/990

6/11 * 5/10 * 5/9 = 150/990

5/11 * 6/10 * 5/9 = 150/990

Combined probability would be:

(120 + 150 + 150 + 150) / 990

570 / 990

57.575757...%

Solving this problem is not just trying to get all the events. It is like this:

There are 6 red and 7 blue balls. So, all in all there are 13 balls. 2 blue balls have been drawn already without replacement. That means there are 11 balls left, 5 of which are blue and the other 6 are red. Remember that the red balls are identical with each other and so as the blue balls, meaning you cant distinguish them.

The probability of getting a red ball in the next draw is the number of red balls over the total number of balls, that is, 6/11.

Since, you are not going to replace the ball, the next draw would give you a probability of 5/10(only 5 red balls left and 10 balls all in all).

In the third draw, the probability would be 4/9.

Since, we are asked to get the probability of getting at least 2 red balls that means we need to get the probability of getting 2 red balls and 3 red balls and add it.

The probability of getting 2 red balls in 3 draws is:

(6/11)(5/10)= 3/11

The probability of getting 3 red balls in 3 draws is:

(6/11)(5/10)(4/9) = 4/33

Adding this would give:

4/33+3/11 = 13/33 or 39.39%

you start with

4 red, 7 blue

sucessfull draws are

rrb

rbr

brr

rrr

rrb can be accomplished in 4*3*1 ways

rbr '' '' '' '' 4*1*3 ways

brr '' '' '' '' 1*4*3 ways

rrr '' '' '' '' 4*3*2 ways

in total there are 11*10*9 ways to draw balls.

netto the probability of at least 2 reds is then

(4*3*1 + 4*1*3 + 1*4*3 + 4*3*2) / 11 * 10 * 9

6 + 7 = 13

13 - 2 = 11

(number of outcomes favourable to event[getting at least 2 red balls in next 3 turns]) = 6

number of total outcomes = 11

answer: 6/11

6/11

So really this is about drawing from 6 red and 5 blue.

In the next 3 draws

P(RRR)=6/11*5/10*4/9=120/990

P(RRB)=6/11*5/10*5/9=150/990

P(RBR)=6/11*5/10*5/9=150/990

P(BRR)=5/11*6/10*5/9=150/990

TOTAL = 570/990

So the required probability is 57/99 or about 57.6%

You have 6 red and 5 blue to start.

6/11 + 6/11 - ( 6/11*5/10)

12/11 - 30/110

120/110 - 30/110

90/110

9/11

total=6+7=13.

.

.But -2 ........=13-2=11

..red numbers 6.......but 6-2=4

.

. p=4/11

So, now you have 6 red and 5 blues.

P(at least 2 red in 3 draws)

=Choose(6,2)*Choose(5,1)/Choose(11,3)

+Choose(6,3)/Choose(11,3)

= 57.57%