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probability

kiroro22

New member
Joined
Jan 23, 2011
Messages
10
hi, I got stuck after studying the begining of probability. so could anyone solve these problems,please. many thanks


1.An urn contains five red balls and one white ball. A ball is drawn and then it and another ball of the same color are placed back in the urn. Finally, a second ball is drawn.

a.What is the probability that the second ball is white?
I am pretty sure how to solve this, but if anyone find any mistake or awkward bit please let me know,

R-red, W-white

if first one was R(5/6) then second will be two cases-----R(6/7)
- -----W(1/7)


if first one was W(1/6)then second will be two cases-----R(5/7)
-----W(2/7)

so P(W)=(5/6)(1/7)+(1/6)(2/7)=7/42

my answer is 7/42,,,,is it correct?


b.If the second ball is white, what is the probability that the first was red?

using P(A and B)/P(A)=P(B?A)

in this case P(R ?W)= P(R and W)/P(W)

so (5/6)(1/7)/(7/42)=5/7

my answer is 5/7 I am not sure whether it is right..








2.A team of 2 players enters a quiz show, in which they have to answer 'true-false' questions. Each player will, independently, give the correct answer with probability p. Before the show they are trying to decide between the following strategies.

a.take turns answering questions.

b.Both consider each question and then either give the common answer if they agree or, if they disagree, flip a fair coin to determine the answer.

which, if either, is the better strategy?



3.A year after completing a certain 'quit smoking' class, 48% of the women in the class and 37% of the men had not resumed smoking. These people all attended a success party to celebrate their achievement. If 62% of the original class were male, use apporpriate results in probability theory to determine;

a.the percentage of the original class who attended the party

b.the percentage of those attending the party who were women.





4.A machine consists of four components, labelled 1,2,3,4 arranged as shown.

A --> 1 -->B -->3 -->C

A --> 2 -->B -->4 -->C

Each component has probability p of failure, independently of the others. For the machine to function, A must be connected to C by a path of working components. What is the probability that the machine functions? what is the range of permissible values for p if we require the probability of machine failure to be smaller than 0.05?
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, kiroro22!

Here is some help . . .


1.An urn contains five red balls and one white ball.
A ball is drawn and then it and another ball of the same color are placed back in the urn.
Finally, a second ball is drawn.

a.What is the probability that the second ball is white?
. . My answer is 7/42 , , , is it correct?

b. If the second ball is white, what is the probability that the first was red?
. . My answer is 5/7 . . . I am not sure whether it is right..

Both answers are correct . . . Good work!




3. A year after completing a certain 'quit smoking' class, 48% of the women in the class and 37% of the men had not resumed smoking.
These people all attended a success party to celebrate their achievement.
If 62% of the original class were male, use apporpriate results in probability theory to determine;

a. the percentage of the original class who attended the party

b.the percentage of those attending the party who were women.

I found it easier to use some convenient numbers.


Suppose there were 10,000 people in the original class.

\(\displaystyle \text{Then: }\;\begin{Bmatrix} 62\%\text{ were male:}\;\; & \text{6200 men}\;\;\; \\ 38\%\text{ were female:} & \text{3800 women}\end{Bmatrix}\)


\(\displaystyle \text{Those who successfully quite smoking and attended the party:}\)

. . \(\displaystyle \begin{Bmatrix} 37\%\text{ of 6200} &=& 2294\text{ men}\;\;\; \\ 48\%\text{ of 3800} &=& \text{1824 women}\end{Bmatrix}\)


\(\displaystyle \text{A total of: }\:2294 + 1824 \:=\: 4118\text{ people attended the party:}\)

. . \(\displaystyle \text{{\bf(a)} The percentage is: }\:\frac{4\,\!118}{10,\!000} \;=\;41.18\%\)


\(\displaystyle \text{Of the 4118 people at the party, 1824 were women.}\)

. . \(\displaystyle \text{{\bf(b)} The percentage is: }\;\frac{1824}{4118} \;\approx\;44.29\%\)

 

kiroro22

New member
Joined
Jan 23, 2011
Messages
10
thank you very much indeed, no.3 was actually easy if I thought the way you did. very easy to follow your solution.

I really don't know how and where to start no.2,,,so can anyone solve this no.2??
 

kiroro22

New member
Joined
Jan 23, 2011
Messages
10
For Question 3, is there any way I could solve using appropriate results in probability theory?
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,091
kiroro22 said:
For Question 3, is there any way I could solve using appropriate results in probability theory?
Probability is not involved in this question.

You can start with

Let the original number of people in the class = P

Then

# of men = 0.62 * P

# of women = 0.38 * P

and so on....
 
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