prob

[dsaqwert]

there are 10 cancer patients in a hospital. Doctors offer two type of curing method. method A is risky and has %80 chance to succeed and method B is the routine one and has %40 chance of success. %20 of the patients prefer the risky method and the remaining undergo the routine way. If 4 out of 10 patients cured after the treatments, what is the probability that all of the risky method choosers are cured.

 

 

1/3 but not OA, a modified question i've come up with

 

[Priyadarshi]

My approach will be :

 

Out of 10 patients only 2 will opt for the risky method , assuming all risky method choosers are cured out of the four 2 are risky method chooser and two are routine method .

 

So probbility of the first person is :2/10*8/10 ( prob of choosing risky method and prob of getting cured )

 

simillary the second person is 2/10 * 8/10

 

Third and fourth person woul have chosen routine method so : 8/10*4/10

 

so total prob that all the risky method people are cured is :

 

(2/10*8/10)^2* (8/10*4/10)^2

 

= 8^2/5^8 ..hmm differs a lot from ur answer

[dsaqwert]

hi priyadarshi, practically that number you found seems to be too small to be the answer

[mohican]

can it be 29/210 ?

[dsaqwert]

can it be 29/210 ?

 

as i said there is no OA, what is your reasoning

[arjmen]

Possible ways:

2C0*(0.2^2)*8C4*(0.4^4)*(0.6^4) + 2C1*(0.2*0.8)*8C3*(0.4^3)*(0.6^5) + 2C2*(0.8^2)*8C2*(0.4^2)*(0.6^6)

 

Favourable event:

2C2*(0.8^2)*8C2*(0.4^2)*(0.6^6)

 

After some calculation, I get 72/125

 

Not totally confident about the answer though.

[dsaqwert]

hmm, arjmen your approach seems true, except that given numbers hint the answer should not be greater than 0.5

 

my reasoning is as follows;

 

a cured person would have experienced either the risky or the routine treatment,

P(risky given that cured)=P(risk taker)*P(risky cure) / [P(risk taker)*P(risky cure)+P(routine follower)*P(routine cure)]

P(risk taker given that cured)=0.2*0.8/(0.2*0.8+0.8*0.4)=1/3

and P(routine follower given that cured)=1- P(risk taker given that cured)=2/3

 

now, for 4 cured person in total there are three possible groupings,

(Ri,Ri,Ro,Ro), (Ri,Ro,Ro,Ro), (Ro,Ro,Ro,Ro)

for P[(Ri,Ri,Ro,Ro)]=4c2*(1/3)^2*(2/3)^2=24/81

for P[(Ri,Ro,Ro,Ro)]=4c1*(1/3)^1*(2/3)^3=32/81

for P[(Ro,Ro,Ro,Ro)]=4c4*(2/3)^4=16/81

 

P(2 risk taker out of 4 cured)=P[(Ri,Ri,Ro,Ro)] / {P[(Ri,Ri,Ro,Ro)]+P[(Ri,Ro,Ro,Ro)]+P[(Ro,Ro,Ro,Ro)]}

P(2 risk taker out of 4 cured)=24/(24+32+16)=1/3

[800ForSure]

The Question stem confuses me..

 

1) If 4 out of 10 patients cured [???after the treatment?????], what is the probability that all of the risky method choosers are cured...[ or out of 4 cured, 2 are those who selected the risky method]

 

Total = 2C0*8C4 + 2C1*8C3 + 2C2*8C2 = 114

Favorable = ( 2C2*8C2)

 

Probability = 28/114 = 14/57

 

2) If 4 out of 10 patients cured,what is the probability that both the risky method choosers are cured successfully,

 

Probability that both are selected, 14/57

Probability that both selected are successfully cured [ (14/57)* (0.8^2) ]

 

.......My concern is that the patient selected will either be cured successfully or unsuccessfully so If you calculat all the possible cases for a Number of selected patients the sum of total successful/unsuccessful combo is going to be 1 for e.g.

 

2 patients are selected who chose the risky method and other two who didn't...

 

POSSIBLE Ways..

 

2C2*8C2 [

(0.8)^2(0.4)^2 + 2(0.8)^2(0.4)(0.6)+(0.8)^2(0.6)^2+

(0.2)^2(0.4)^2 + 2(0.2)^2(0.4)(0.6)+(0.2)^2(0.6)^2+

2(0.8)(0.2)(0.4)^2 + 4(0.8)(0.2)(0.6)(0.4)+2(0.8)(0.2)(0.6)^2

]

 

Which is equal to 2C2*8C2 [1] = 2C2*8C2

 

so as far as total selection is concerned, the success rate will not have any effect on TOTAL.

 

I might be unable to explain it here but if you've specific questions/doubts,i'll try to address that....PART3 is going to follow

:crazy: :crazy:

[800ForSure]

In addition to my earlier response.. If the Question Stem says

 

3) Out of 10 patients cured, 4 are cured successfully...what is the probability, in that case my answer will be

 

total ways

W1 = 2C0*(0.2)^2*8C4*(0.4)^4(0.6)^4

W2 = 2C1*(0.8)*8C3*(0.4)^3*(0.6)^6

W3 = 2C2*(0.8)^2*8C2*(0.4)^2*(0.6)^6

 

Total T = W1+W2+W3

 

Fav = W3

 

Probability = W3/T :p

[arjmen]

Hmm, I don't see why there is a need to include P(risk taker) etc.

 

We know that there are 10 patients, that 2 particular patients opt for the risky procedure (and the rest for the standard procedure) and that 4 of the 10 patients will be cured.

 

So we have 2 groups, one with 2 candidates and the other with 8 candidates. We need to find the probability of both candidates from the 1st group being elected given their probability of success and the probability of success of the candidates in the 2nd group.

 

800ForSure,

Aren't you missing a '0.2' term in W2?

W2 = 2C1*(0.8)*(0.2)*8C3*(0.4)^3*(0.6)^6

[800ForSure]

agreed with you Arjmen..i missed it. but then we'll also have to redue (0.6)^6 to (0.6)^5

so

 

W2 = 2C1*(0.8)*(0.2)*8C3*(0.4)^3*(0.6)^5

[arjmen]

Now our answers match, 800ForSure (well, one of your answers and all of mine):D

[dsaqwert]

800ForSure, sorry for not being clear, saying 4 out of 10 patients cured means they are successfully cured.

[FZanjani]

dsaqwert -

 

given (as per your q stem) that 2 people have chosen th erisky treatment and 8 have chosen the non-risky one then there are possibly only three scenarios -

 

___________________Treat - A _|_Treat - B

No. of patients

successfully cured _______0_____|____4___

 

_______________________1_____|____3___

_______________________2_____|____2___

 

and in only one scenario do both the people who chose risky treatment survive - so the proobability should be 1/3

 

 

This is more of a question than a definitive answer from my side.