I need help

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Alpha2005

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A bag contains 8 blue, 12 red and 10 yellow identical balls. Two balls are drawn at random with replacement. Determine the probability that the balls are :
(a) of different colours
(b) of the same colour
(c) red and yellow
my idea
1.) p(different color) =p(b^r) u p(b^y) u p(r^y)
=8/75+4/45+2/15=74/225
2.)p(same colour) =p(b^b) u p( r^r) u p( y^y)
= 77/225
3.) p(r^y)= 22/30= 11/15
 
I agree with your answers for (a) and (b) if you are allowed to treat blue/red the same as red/blue. If not, you need to double your three fractions before adding.

For (c), your solution indicates one draw. You need to separate the colors into two draws: 1st draw red (probability = ?); 2nd draw yellow (probability = ?) then multiply those probabilities.
 
First hint: it's actually easier if you don't simplify fractions, since you are going to be adding them and need a common denominator.

Second hint: show another step or two of your work, so it's easier for others (or you) to check.

Take the first bit, your 8/75. What you did was [MATH]P(B\wedge R) = P(B)\cdot P(R) = \frac{8}{30}\cdot\frac{12}{30} = \frac{96}{900} =\frac{8}{75}[/MATH]. But that is really [MATH]P(B\text{, then }R)[/MATH]. To find [MATH]P(B\wedge R)[/MATH], you need [MATH]P((B\text{, then }R)\vee (R\text{, then }B))[/MATH].

That's what R.M. was pointing out.

Your answer for (3) is [MATH]P(R\vee Y)[/MATH], not [MATH]P(R\wedge Y)[/MATH], and on one draw, as R.M. said.
 
Alpha2005 said:
A bag contains 8 blue, 12 red and 10 yellow identical balls. Two balls are drawn at random with replacement. Determine the probability that the balls are :
(a) of different colours
(b) of the same colour
(c) red and yellow
my idea
1.) p(different color) =p(b^r) u p(b^y) u p(r^y)
=8/75+4/45+2/15=74/225
2.)p(same colour) =p(b^b) u p( r^r) u p( y^y)
= 77/225
3.) p(r^y)= 22/30= 11/15
Click to expand...
My first to you is: did you understand that with replacement means independence?
So \(\mathcal{P}(BR)=\mathcal{P}(B)\cdot\mathcal{P}(R)=\dfrac{96}{900}\)
Second question: are parts a) & b) complements? If not the same then different?
If we want to model this how to do it? Think ordered pairs. There are \(900\) pairs.
Of those pairs \(308\) have same colour: \(\mathcal{(RR)}.~ \mathcal{(BB)},~ \mathcal{(YY)} \)
So the probability of drawing two balls of same colour is \(\mathcal{P}\left[(BB)\cup(YY)\cup(RR)\right]=\dfrac{308}{900} \)
Edit: Prof Peterson posited as I was typing.
 
Dr.Peterson said:
First hint: it's actually easier if you don't simplify fractions, since you are going to be adding them and need a common denominator.

Second hint: show another step or two of your work, so it's easier for others (or you) to check.

Take the first bit, your 8/75. What you did was [MATH]P(B\wedge R) = P(B)\cdot P(R) = \frac{8}{30}\cdot\frac{12}{30} = \frac{96}{900} =\frac{8}{75}[/MATH]. But that is really [MATH]P(B\text{, then }R)[/MATH]. To find [MATH]P(B\wedge R)[/MATH], you need [MATH]P((B\text{, then }R)\vee (R\text{, then }B))[/MATH].

That's what R.M. was pointing out.

Your answer for (3) is [MATH]P(R\vee Y)[/MATH], not [MATH]P(R\wedge Y)[/MATH], and on one draw, as R.M. said.
Click to expand...
I appreciate this thank you
 
It wa
R.M. said:
I agree with your answers for (a) and (b) if you are allowed to treat blue/red the same as red/blue. If not, you need to double your three fractions before adding.

For (c), your solution indicates one draw. You need to separate the colors into two draws: 1st draw red (probability = ?); 2nd draw yellow (probability = ?) then multiply those probabilities.
Click to expand...
It is a mistake thank you
 
I think I might have gotten it
Please anyone who thinks it is not correct should please help to identify the mistake
Is this correct sir
1.) b----8 , r---12 y---10 , total no of balls ---30

No of ways the color can be arranged together In a group of two 3C2=3

B$R, B$Y, R$Y and each can also be inverted I.e red then blue for the first arrangement

Pr( different color) = p(b^r) u p(b^y. ) u p( r^y) u p(R^B) u p( y^b) u p( y^r)

= {p(b)*p(r) } + {p(b)*p(y) } + {p(r) *p(y)}. +. {p(r)*p(b)} + {p(y)*p(b) ) + {p(y) *p(r)

= {8/30* 12/30}*2 + {8/30*10/30}*2 +. {12/30 * 10/30}*2

=96/450 + 80/450 + 120/450

= 296/450

=148/225

2.) 2.)p(same colour) =p(b^b) u p( r^r) u p( y^y)

= (8/30)^2+(12/30)^2+(10/30) ^2

= 77/225


3.) p(r^y) = pr(r) *pr ( y)

12/30 * 10/30

120/900
 
Your answer for (3) is the probability of Red and then Yellow (that is, in that order), not Red and Yellow. Otherwise, looks good.
 
As pka pointed out, P(same color) + P(different colors) = 1. So P(different color) = 1 - P(same color) = 1 - 148/225.

You did not have to do what you did to find P(different colors)! However if you did it your way to make sure the sum is 1, then this is ok. However from experience it is not usually done for that reason.
 
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