WT 4th page 191 - first example...

[nimish.tiwari]

Amir, I too am pondering on this since yesterday. I think your approach is correct but with a minor correction.

P(med1 or med2) = P(med1) + P(med2)

P(med1)= 1/14
P(med2) = P(not getting selected for 1st medicine) AND P(getting selected for 2nd med)

P(not getting selected for 1st med) = 13/14
P(getting selected for 2nd) = 1/13

So, P(med2) = 13/14 * 1/13 = 1/14

Hence, P(med1 or med2) = 1/14 + 1/14 = 1/7.

The explanation in Strategy Guide did miss this part I guess.
Can someone from MGMAT team confirm on my understanding of the problem here and the solution mentioned above? Or is that I have got the approach wrong but luckily right with the end answer?

Thanks!!

[Ben Ku]

Hi Nimish,

There's nothing really wrong with your approach. It works. The point the Strategy Guide wanted to make was that it's simpler if you "take advantage of the symmetry"; in other words, even though the question states Progaine first, then Ropecia, the order here doesn't matter because we're ONLY concerned with Donald.

We can basically rephrase the question as: "In a group of 14 patients, if one person will receive Progaine and another person will receive Ropecia, what is the probability that Donald will receive either drug?"

So to solve this problem, you can find P(D receives P) OR P(D receives R). We ADD the probabilities.

(1/14) + (1/14) = 1/7.

Hope that helps.

[gmatalongthewatchtower]

Manhattan GMAT Staff,

I am completely lost. I think this is a good example of conditional probability. I am not sure why we are ignoring it.

Here's my take:

Let P = Event that Prograine is administered.
R = Event that Ropecia is administered.
O = Event that placebo is administered {We dont need this event!}

Now P(P) = 1/14 (Agree) ......(iii)
Given that P has happened,

P(R/P) = 1/13. This should not be P(R).

The ask is to find out P(R OR P)

P(R OR P) = P(R) + P(P) - P(R AND P)

Now, I know P(P), I can calculate P(R and P) = P(R/P).P(P);

P(R) is unknown. Now, to calculate P(R):
One way to calculate P(R) would be; P(R) = P(P and R)/P(P/R);
Now P(P/R) = 1/13; and {Think about this event}
P(P and R) = 1/13x1/14......(ii)
Therefore, P(R) = 1/14..........(i)

Using i,ii and iii,
P(R OR P) = (1/14) + (1/14) - (1/13x1/14) = (2/14) - {(1/13)x(1/14)} = 25/(13x14)

My answer isn't right ? I am not sure what's wrong with this method ? Can anyone please help me ?

-Voodoo

[gmatalongthewatchtower]

can anyone pls reply ?

thanks

[tim]

we get to these questions when we can. responding to your own post pushes yours to the back of the queue..

this problem can certainly be solved using conditional probability, but it is not efficient to do so, and you're using the formulas incorrectly. what we want is P(P) and P(R/~P), and we want the probability that one of those happens:

P(P) = 1/14
P(R/~P) = 13/14 * 1/13 = 1/14
(remember, you have to factor in the 13/14 chance that he doesn't get P and multiply that by the probability that he then gets R)

adding these together (we want one OR the other), we get 1/14 + 1/14 = 1/7