Probability problem (ii): A ball is drawn at random from a box containing 6 red balls, 4 white balls, and...

[chijioke]

Here is my work.

Total number of balls = 6 + 4 + 5 =15

Number of red balls = 6

Number of white balls = 4

Number of blue balls = 5

Probability that a ball picked is red =Pr(R)

Probability that a ball picked is blue =Pr(B)

Probability that a ball picked is white =Pr(W)

Probability that a ball picked is not white =Prnt(W)


[a]Pr(R) =[math]\frac{\overset{2}{\cancel{6}}}{\underset{5}{\cancel{15}}}=\frac{2}{5}[/math](b)Pr(B) =[math]\frac{\overset{1}{\cancel{5}}}{\underset{3}{\cancel{15}}}=\frac{1}{3}[/math][c]Pr(W)=[math]\frac{4}{15}[/math][d]Prnt(W)= Pr(R) + Pr(B) = [math]\frac{2}{5}+\frac{1}{3} = \frac{6}{15}+\frac{5}{15} =\frac{11}{15}[/math]
The (d) can also be approached like this:

Prnt(W)= 1- Pr(W) =[math]1- \frac{4}{15} =\frac{11}{15}[/math]
So is my work correct? I am more interested in the (d) part. Thank you.

 

[Dr.Peterson]

View attachment 35918

Here is my work.

Total number of balls = 6 + 4 + 5 =15

Number of red balls = 6

Number of white balls = 4

Number of blue balls = 5

Probability that a ball picked is red =Pr(R)

Probability that a ball picked is blue =Pr(B)

Probability that a ball picked is white =Pr(W)

Probability that a ball picked is not white =Prnt(W)


[a]Pr(R) =[math]\frac{\overset{2}{\cancel{6}}}{\underset{5}{\cancel{15}}}=\frac{2}{5}[/math](b)Pr(B) =[math]\frac{\overset{1}{\cancel{5}}}{\underset{3}{\cancel{15}}}=\frac{1}{3}[/math][c]Pr(W)=[math]\frac{4}{15}[/math][d]Prnt(W)= Pr(R) + Pr(B) = [math]\frac{2}{5}+\frac{1}{3} = \frac{6}{15}+\frac{5}{15} =\frac{11}{15}[/math]
The (d) can also be approached like this:

Prnt(W)= 1- Pr(W) =[math]1- \frac{4}{15} =\frac{11}{15}[/math]
So is my work correct? I am more interested in the (d) part. Thank you.

Good work, especially on both ways of solving (d).

The only thing I would change is the notation you made up for "not". Rather than Prnt(W), it should at least be Pr(not W). There are several other notations, such as [imath]Pr(\neg W)[/imath], [imath]Pr(W')[/imath], [imath]Pr(W^c)[/imath], [imath]Pr(\overline{W})[/imath], and Pr(~W). Have you been taught any of these?

 

[chijioke]

Have you been taught any of these?

No.

 

[Dr.Peterson]

Have you been taught any of these?

No.

Then use Pr(not W), which is what I would use with a newcomer to probability.

Here is a good introduction that uses W' (and mentions two of the others).

 

[pka]

Why make things so complicated? This is just a trivial beginning probability question.
It is designed to teach you to solve by elementary means. There no need to simply any answer.
You are given a universe of [imath]15\text{ balls, of which }6\text{ are red, }5\text{ are blue, and }4\text{ are white .}[/imath]
At random select any one of those fifteen balls:
[imath]\text{the probability that ball is red equals }\dfrac{6}{15}\;\text{ the probability that ball is blue equals }\dfrac{5}{15}[/imath]

[imath]\text{the probability that ball is white equals }\dfrac{4}{15}\;\text{ the probability that ball is not white equals }\dfrac{5+6}{15}[/imath]

Do you understand how easy this question should have been for you?

 

[chijioke]

Why make things so complicated? This is just a trivial beginning probability question.
It is designed to teach you to solve by elementary means. There no need to simply any answer.
You are given a universe of [imath]15\text{ balls, of which }6\text{ are red, }5\text{ are blue, and }4\text{ are white .}[/imath]
At random select any one of those fifteen balls:
[imath]\text{the probability that ball is red equals }\dfrac{6}{15}\;\text{ the probability that ball is blue equals }\dfrac{5}{15}[/imath]

[imath]\text{the probability that ball is white equals }\dfrac{4}{15}\;\text{ the probability that ball is not white equals }\dfrac{5+6}{15}[/imath]

Do you understand how easy this question should have been for you?

Yes.