We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
42
4
avatar

A box contains R red balls, B blue balls, and no other balls. One ball is removed and set aside, and then a second ball is removed. On each draw, each ball in the box is equally likely to be removed. The probability that both of these balls are red is 2/7 . The probability that exactly one of these balls is red is 1/2 . Determine the pair (R, B).

 Nov 9, 2019
 #1
avatar+2499 
+1
 
 

I tried to solve it, but it was too hard. This was my attempt, can someone tell me what I did wrong?

 

The probability that one red ball is selected:

\(\frac{r}{r+b}\).

With \(r\) being the number of red balls over \((r+b)\) , the total number of balls.

 

The probability that one blue ball is selected is similar:

\(\frac{b}{r+b}\).

 

 

Ok, so the problem states: "The probability that both of these balls are red is 2/7"

Lets make an equation for this:

 

\(\frac{r}{r+b}*\frac{r}{r+b}=\frac{2}{7}\)

 

The problem also states: "The probability that exactly one of these balls is red is 1/2"

Lets make an equation for this:

 

\(\frac{r}{r+b}*\frac{b}{r+b}=\frac{1}{2}\)

 

 

 

So now we have a system of equations:

\(\frac{r}{r+b}*\frac{r}{r+b}=\frac{2}{7}\)

\(\frac{r}{r+b}*​​\frac{b}{r+b}=\frac{1}{2}\) 

 

We solve:(Simplify fraction multiplications)

\(\frac{r^2}{(r+b)^2}=\frac{2}{7}\)

\(\frac{rb}{(r+b)^2}=\frac{1}{2}\)

 

Now cross multiply:

\(7r^2=2(r+b)^2\)

\(2rb=(r+b)^2\)

 

Simplify:

\(\frac{7r^2}{2}=(r+b)^2\)

\(2rb=(r+b)^2\)

 

After this I got stuck, I determined the ratio between r and b, which is \(7r=4b\).

 

HelP!

 
 Nov 9, 2019
 #2
avatar
+1

Well, If you have R red balls out of a total (R + B), then when you draw the first one(assuming it is R), then don't you have left (R - 1) / (R + B - 1) for the 2nd draw?. Use "concrete numbers" to illustrate the point: Suppose you have 10 Red ball and 15 Blue balls. The first draw will be: 10 / [10+15]=10/25. The 2nd draw will have: 10 - 1/25 -1 =9/24. Then the probability of 2 Red ball will be: 10/25 x 9/24.

 Nov 9, 2019
 #3
avatar+2499 
+1

oohhhhh, ok

 

guest's way of determining probability + my strategy of solving using equations = your answer

CalculatorUser  Nov 9, 2019
 #4
avatar
+1

CU: Try these numbers:

 

a=4,    b=3>>>>>>>    4/7 * 3/6 =12/42 = 2/7
a=12,  b= 10>>>>>>   12/22 * 11/21 =132 / 462 = 2/7

 Nov 9, 2019

10 Online Users

avatar
avatar