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

Question

in a box, there are 6 red balls and 7 blue balls.Two blue balls are drawn in the first and second draw. assuming the balls are not returned, what is the probability of getting at least 2 red balls in the next three draws?

Answer 1

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...%

Answer 2

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%

Answer 3

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

Answer 4

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

Answer 5

6/11

Answer 6

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%

Answer 7

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

Answer 8

total=6+7=13.
.
.But -2 ........=13-2=11

..red numbers 6.......but 6-2=4
.
. p=4/11

Answer 9

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%