Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Can someone please tell me where I can find some good coaching on probability? An online course would be the best. If that isn't available, I'll settle for whatever is good.

LOL.. I would like to request the same thing....
_________________

I'm trying to not just answer the problem but to explain how I came up with my answer. If I am incorrect or you have a better method please PM me your thoughts. Thanks!

probability of not hitting the target = pro(1st cannon not hitting)*pro(2nd not hitting)*pro(3rd not hitting) = .7*.6*.5 = .21 hence C
_________________

If A = the event that the 1st cannon hits the target B = the event that the 2nd cannon hits the target C = the event that the 3rd cannon hits the target

We need to find out P(A' * B' * C') where A' = the event that 1st cannon does not hit the target etc and '*' comes as we need the event A' AND B' AND C'.

As A, B, C are independent events, so is A', B', C' ; hence, P(A'*B'*C') = P(A')*P(B')*P(C') = (1-P(A))*(1-P(B))*(1-P(C)) = 0.7*0.6*0.5 = 0.21 (Ans)

Getting right is given for each canon. Then getting wrong is 1- probability of getting right for each canon. Probability of happening each case is AND operation whereas either case is OR. If AND then multiply all the probabilities. In this case probability of getting all wrong P is, (1-0.3)*(1-0.4)*(1-0.5) =0.21

I have seen so many of you telling the answer correctly. And kudos to you! I have to expose myself for thinking lamely. So, no one, of the participating pundits, seemed to have made a mistake of doing this \(1- (0.3*0.5*0.6)=.94\). What did I do? Or in other words, when would I do what I did, if ever? Yet, in other words, how should the problem sound so as to make my way of solving correct? I am no longer consider myself bad at probability, but this problem definitely prompts further training...

0.06 is the probability that all three canons hit the target. The other 0.94 includes the probability that none, one, and two cannons hit the target. Correct me if I am wrong.
_________________

0.06 is the probability that all three canons hit the target. The other 0.94 includes the probability that none, one, OR two cannons hit the target. Correct me if I am wrong.

Should be OR, otherwise it's correct.
_________________

Probability of A hitting the target = 0.3 (probability of miss = 1 - 0.3 = 0.7) Probability of B hitting the target = 0.4 (prob of miss = 0.6) Probability of C hitting/missing the target = 0.5