# participent question

#### kingman

Dear Sir,
I really need help in the below question.
Thanks
Kingman

In a test, questions are picked randomly. The probability that a participant gets an easy question is .6 while the probability of getting a difficult question is .4.

The probability of any participant giving the correct answer to an easy question is .8 while the probability
of giving the correct answer to a difficult question is .3.

Find the probability that the participant give the correct answer twice in a row .

I have 2 approaches to the question but wonder how to prove that the solution are the same using Venn diagram. Let

E: an event that participant picked an easy question.
D: an event that participant picked a difficult question.
C: an event that the participant give the correct answer,
W: an event that the participant give the wrong answer.

Solution 1:
Probability = (.6*.8+.4*.3)2 =.36 (using product rule)
Solution 2:
Using Tree diagram,
Probability = .6*.8 * (.6*.8+.4*.3) + .4*.3 * (.6*.8+.4*.3) =.36

#### CaptainBlack

MHF Hall of Fame
Dear Sir,
I really need help in the below question.
Thanks
Kingman

In a test, questions are picked randomly. The probability that a participant gets an easy question is .6 while the probability of getting a difficult question is .4.

The probability of any participant giving the correct answer to an easy question is .8 while the probability
of giving the correct answer to a difficult question is .3.

Find the probability that the participant give the correct answer twice in a row .

I have 2 approaches to the question but wonder how to prove that the solution are the same using Venn diagram. Let

E: an event that participant picked an easy question.
D: an event that participant picked a difficult question.
C: an event that the participant give the correct answer,
W: an event that the participant give the wrong answer.

Solution 1:
Probability = (.6*.8+.4*.3)2 =.36 (using product rule)
Solution 2:
Using Tree diagram,
Probability = .6*.8 * (.6*.8+.4*.3) + .4*.3 * (.6*.8+.4*.3) =.36
Why do you need to use a Ven Diagram for this?

I would use a unit square divided into parts in the proportions given in the question, then the required probability will be the area corresponding to a correct answer. This is virtually imposible to describe without a diagram.

CB

kingman

#### kingman

use set operations

What I mean is to use set operation like union, intersection and P( A) ,P(A') P(c/E) set notations to represent the probabilities found in the two solutions and prove the probabilities of the events are equal.
Thanks
Kingman

#### undefined

MHF Hall of Honor
What I mean is to use set operation like union, intersection and P( A) ,P(A') P(c/E) set notations to represent the probabilities found in the two solutions and prove the probabilities of the events are equal.
Thanks
Kingman
Small note: If you have a keyboard like mine, you can find the key for | right above the Enter/Return key, as the shifted character above backslash (\).

$$\displaystyle P(E) = 0.6$$

$$\displaystyle P(D) = 0.4$$

$$\displaystyle P(C|E) = 0.8$$

$$\displaystyle P(C|D) = 0.3$$

So what you're asking for is

$$\displaystyle P(C) = P(C \cap (E \cup D))$$

$$\displaystyle = P((C \cap E) \cup (C \cap D))$$

$$\displaystyle = P(C \cap E) + P(C \cap D) - P((C \cap E) \cap (C \cap D))$$

$$\displaystyle = P(C \cap E) + P(C \cap D)$$

$$\displaystyle = P(E)\cdot P(C|E) + P(D)\cdot P(C|D)$$

$$\displaystyle = (0.6)(0.8) + (0.4)(0.3)$$

$$\displaystyle = \frac{3}{5}$$

Then the probability of this happening two times in a row is

$$\displaystyle P(X) = \left(\frac{3}{5}\right)^2 = \frac{9}{25} = 0.36$$

kingman

#### kingman

Dear Undefined,
Thanks very much for the answer .
Can you show me how to prove the term after the minus sign of the below expression is zero as you have written in your answer:

P(C)=

and can you show me how to prove the second solution as given in the question:
Using Tree diagram,
Probability = .6*.8 * (.6*.8+.4*.3) + .4*.3 *(.6*.8+.4*.3) =.36

* (P(E). P(C|E) +P(D).P(C|D))

also I wonder how can write the union and set symbol
in this message window?

Thank you very much
Kingman

#### undefined

MHF Hall of Honor
Dear Undefined,
Thanks very much for the answer .
Can you show me how to prove the term after the minus sign of the below expression is zero as you have written in your answer:

P(C)=
E and D are mutually exclusive events. The probability of them happening at the same time is zero.

and can you show me how to prove the second solution as given in the question:
Using Tree diagram,
Probability = .6*.8 * (.6*.8+.4*.3) + .4*.3 *(.6*.8+.4*.3) =.36

* (P(E). P(C|E) +P(D).P(C|D))
I'm not sure what you're asking. Didn't you have to draw the tree diagram in order to write down the expression? Or did you just copy the expression from somewhere? I originally assumed you had discovered the solution yourself, but I guess you got it from the back of the book or some such.

Try drawing the tree diagram and post again if you have trouble.

But one critique: I think it's overkill to draw a whole diagram to go along with the expression. Since the participant's answering the first question and second question are independent events, it's enough to find the probability of answering one question correctly, then square the result.

also I wonder how can write the union and set symbol
in this message window?

Thank you very much
Kingman
See the LaTeX tutorial thread, which is within the LaTeX Help subforum. The LaTeX command for union is \cup and for intersection is \cap.

There are also unicode symbols but I don't know the codes to enter them with a keyboard. They look like this and can be copy and pasted: ∪ ∩

kingman

#### kingman

need help

Dear Undefined,
I have drawn the tree diagram and saved its image in gif. format and I need help to know how to
send or paste the file in message window to you .
Thanks
Kingman

#### undefined

MHF Hall of Honor
Dear Undefined,
I have drawn the tree diagram and saved its image in gif. format and I need help to know how to
send or paste the file in message window to you .
Thanks
Kingman
When you are posting a reply, look down a little where it says "Additional Options" and then "Attach Files" where there is a button "Manage Attachments." Then you can choose "Browse" to find the file on your computer and then "Upload." Then it should appear on this forum!

#### kingman

Thanks

Dear Undefined,
Thanks for the guidance and I think I have manage to attach the tree diagram into this message box.
I would appreciate very much if you can show me how to write the required probability from the tree diagram and show that it is actually looks like

(P(E). P(C|E) +P(D).P(C|D))* (P(E). P(C|E) +P(D).P(C|D))

Thanks
Kingman

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#### undefined

MHF Hall of Honor
Dear Undefined,
Thanks for the guidance and I think I have manage to attach the tree diagram into this message box.
I would appreciate very much if you can show me how to write the required probability from the tree diagram and show that it is actually looks like

(P(E). P(C|E) +P(D).P(C|D))* (P(E). P(C|E) +P(D).P(C|D))

Thanks
Kingman
Okay, so your diagram accurately matches the expression given in the first post, that is,

.6*.8 * (.6*.8+.4*.3) + .4*.3 * (.6*.8+.4*.3) =.36

I've re-attached your image with a red rectangle on it; you can draw just what's in the red rectangle to be able to write

P(X) = .6*.8+.4*.3

where X is the event of answering a question correctly.

Then, without drawing any diagrams, you can reason that since answering the first question correctly is independent from answering the second question correctly, we may simply write that the desired probability is

P(X)*P(X)

This is because the situation is just like rolling a die two times in a row, and asking what is the probability of rolling a 3 two times in a row. Rolling a three once has probability (1/6). Doing it twice in a row has probability (1/6)(1/6) = (1/36).

Edit: Forgot to attach image. Also, some minor typos/etc.

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kingman