Another conditional probability problem

[kenny1999]

Homework Statement

I've spent a few hours but unable to work out the solution of the following questions successfully.

The question is as follows:

I am having problems with (a)(iii) and (b)(ii)

Homework Equations

The Attempt at a Solution

(a)(iii) Although I can work out the solution at last, my original attempt was incorrect but I don't understand why.

My first attempt:
P(pass) = P(4 answer correct) + P(5 answer correct) + P(all answer correct)
= (3x3x1x1x1x1)/(6x6x6x6x6x6) + (3x1x1x1x1x1)/(6x6x6x6x6x6) + (1x1x1x1x1x1)/(6x6x6x6x6x6)

However, this is wrong. But I don't understand why...

(b)(iii) I can't work out the solution at all. Please help. Thank you.

 

[Ray Vickson]

Homework Statement

I've spent a few hours but unable to work out the solution of the following questions successfully.

The question is as follows:

I am having problems with (a)(iii) and (b)(ii)

Homework Equations

The Attempt at a Solution

(a)(iii) Although I can work out the solution at last, my original attempt was incorrect but I don't understand why.

My first attempt:
P(pass) = P(4 answer correct) + P(5 answer correct) + P(all answer correct)
= (3x3x1x1x1x1)/(6x6x6x6x6x6) + (3x1x1x1x1x1)/(6x6x6x6x6x6) + (1x1x1x1x1x1)/(6x6x6x6x6x6)

However, this is wrong. But I don't understand why...

(b)(iii) I can't work out the solution at all. Please help. Thank you.

Your "solution" for (a)(iii) neglect the fact that ANY 4 or more questions will suffice. Your expression 3^2 * 1^4/ 6^6 (for 4 correct answers) makes several errors. First, it assumes that questions 1 and 2 are answered incorrectly, while questions 3-6 are correct; but what we need is the probability that any \42 of the 6 questions is correctly answered. Secondly, it assumes that the probability of answering a question correctly is 1/6 and of answering incorrectly is 3/6---and that makes no sense at all. So, before showing us your incorrect attempt at (iii), you should first show us your attempts at (i) and (ii), so we can see if you are on the right track.

RGV

 

[kenny1999]

Your "solution" for (a)(iii) neglect the fact that ANY 4 or more questions will suffice. Your expression 3^2 * 1^4/ 6^6 (for 4 correct answers) makes several errors. First, it assumes that questions 1 and 2 are answered incorrectly, while questions 3-6 are correct; but what we need is the probability that any \42 of the 6 questions is correctly answered. Secondly, it assumes that the probability of answering a question correctly is 1/6 and of answering incorrectly is 3/6---and that makes no sense at all. So, before showing us your incorrect attempt at (iii), you should first show us your attempts at (i) and (ii), so we can see if you are on the right track.

RGV

OK, here a complete work by me shown below. Thanks for your time reading.

OK, I reformat my answer.

26.

(a) (i) P(gets 0 marks) = (3/4)(3/4)(3/4)(3/4)(3/4)(3/4)

(ii) P(all correct) = (1/4)(1/4)(1/4)(1/4)(1/4)(1/4)

(iii) P(pass the test) = P(4 answered correctly) + P(5 answered correctly) + P(6 answered correctly)


P(4 answered correctly) = (1/4)(1/4)(1/4)(1/4)(3/4)(3/4)x 6C4 ... (1)

P(5 answered correctly) = (1/4)(1/4)(1/4)(1/4)(1/4)(3/4)x 6C5 ... (2)

P(6 answered correctly) = (1/4)(1/4)(1/4)(1/4)(1/4)(1/4) ... (3)

P(pass the test) = (1)+(2)+(3)
=...


Am I correct now??


(b) (i) Let P: event that a student pass
S: event that a student has studied / prepared

P(P) = P(P and S) + P(P and S')
= P(P|S)P(S) + P(P|S')P(S')
= (1)(3/4) + (ans of (a)(iii) ) x (1/4)
=...



After P(P) is evaluted, then

P(both students pass) = P(P) x P(P)
= ...

is it right??

However, there are 40 students, and two students are selected. Will the probability of pass increase or decrease when the first student is selected? Do we have to think like drawing two red balls from a set of 40 balls (for example) without replacement? The probability of drawing first red balls must be different from probability of drawing the second red balls... I am quite confused.



(b)(ii) P(S|P) = P(S and P) / P(P)
= P(P|S)P(S) / P(P)
= (1)(3/4) / P(P)
=... (1)

P(S'|P) = 1 - P(S|P)
= ... (2)

so the required probability

P(at least one of the student doesn't prepare given both pass)

= P(both don't prepare given both pass) + P(1st student prepare, 2nd student no prepare) + P(1st student no prepare, 2nd prepare)

= (ans 2) x (ans 2) + (ans 1) x (ans 2) + (ans 2) x (ans 1)




Am I right? The answer is not important to me because I am not a student. But I really need to understand the concept quick

 

[kenny1999]

Your "solution" for (a)(iii) neglect the fact that ANY 4 or more questions will suffice. Your expression 3^2 * 1^4/ 6^6 (for 4 correct answers) makes several errors. First, it assumes that questions 1 and 2 are answered incorrectly, while questions 3-6 are correct; but what we need is the probability that any \42 of the 6 questions is correctly answered. Secondly, it assumes that the probability of answering a question correctly is 1/6 and of answering incorrectly is 3/6---and that makes no sense at all. So, before showing us your incorrect attempt at (iii), you should first show us your attempts at (i) and (ii), so we can see if you are on the right track.

RGV

OK, here a complete work by me shown below. Thanks for your time reading.

 

[Ray Vickson]

OK, here a complete work by me shown below. Thanks for your time reading.

This is looking a lot better; your solution to (a) looks OK, and part of your solution to (b) is good, but you don't seem to finish (b).

Let E = {both pass}, and in your group of two students, let
S0 = {0 students study}, S1 = {1 student studies}, S2 = {2 students study}.
You want P(S0 or S1|E). This is
$$P(S_0 \cup S_1 | E ) = \frac{ P((S_0 \cup S_1)\cap E )}{P(E)} = \frac{ P(S_0 \cap E ) + P(S_1 \cap E)}{P(E)},$$
because the events S0 and S1 are disjoint. Note that
$$P(E) = P(E|S_0 ) P(S_0) + P(E|S_1) P(S_1) + P(E|S_2) P(S_2),$$
and if $p_a$ is the probability you computed in (a), then
$$P(E|S_0) = p_a^2, \; P(E|S_1) = 1 \cdot p_a, \; P(E|S_2) = 1,$$
as you said. Since you know $P(S_0), \, P(S_1) \text{ and } P(S_2),$ you can just put it all together to answer (b).

RGV

 

[kenny1999]

This is looking a lot better; your solution to (a) looks OK, and part of your solution to (b) is good, but you don't seem to finish (b).

Let E = {both pass}, and in your group of two students, let
S0 = {0 students study}, S1 = {1 student studies}, S2 = {2 students study}.
You want P(S0 or S1|E). This is
$$P(S_0 \cup S_1 | E ) = \frac{ P((S_0 \cup S_1)\cap E )}{P(E)} = \frac{ P(S_0 \cap E ) + P(S_1 \cap E)}{P(E)},$$
because the events S0 and S1 are disjoint. Note that
$$P(E) = P(E|S_0 ) P(S_0) + P(E|S_1) P(S_1) + P(E|S_2) P(S_2),$$
and if $p_a$ is the probability you computed in (a), then
$$P(E|S_0) = p_a^2, \; P(E|S_1) = 1 \cdot p_a, \; P(E|S_2) = 1,$$
as you said. Since you know $P(S_0), \, P(S_1) \text{ and } P(S_2),$ you can just put it all together to answer (b).

RGV

Sorry any alternative method? I don't quite understand after reading carefuly