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:

The probability that a target will be shot two out of two [#permalink]
17 Mar 2013, 17:53

3

This post received KUDOS

4

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

58% (01:57) correct
42% (01:18) wrong based on 132 sessions

The probability that B can shoot a target two out of two times is 0.25. What is the probability that the target will be missed by B immediately after such two shots?

Re: The probability that a target will be shot two out of two [#permalink]
17 Mar 2013, 22:23

Probablity of B can shoot a target 2/2 is 0.25 i.e. the probablity that he will shot both the times is 1/4. And the probality that he will score either 1 or 0 (miss 1 or both) is 3/4. Hence the probablity of miss the target should be 3/4 i.e. 0.75. Answer is C.

Re: The probability that a target will be shot two out of two [#permalink]
18 Mar 2013, 00:46

3

This post received KUDOS

The probability of 2 "times" is 0,25 \(x^2=0,25\) \(x=0,5\) x= probability of missing/hitting the target

The probability that he will miss the next shot is 0,5. "The streak" is not relevant.If I throw a coin the probability of getting one "side" is the same regardless of what I have obtained in the previous shots.

Correct me if I am wrong. _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: The probability that a target will be shot two out of two [#permalink]
18 Mar 2013, 05:26

Zarrolou wrote:

The probability of 2 "times" is 0,25 \(x^2=0,25\) \(x=0,5\) x= probability of missing/hitting the target

The probability that he will miss the next shot is 0,5. "The streak" is not relevant.If I throw a coin the probability of getting one "side" is the same regardless of what I have obtained in the previous shots.

Correct me if I am wrong.

OA is B and the above explanation is correct. _________________

Re: The probability that a target will be shot two out of two [#permalink]
24 Mar 2013, 00:42

Suppose there are total 'x' shots. B shoots 2 out of 2 times. means out of x shots (x>2) B shots 2 out of 2 ----> B shots at least 2 out of x. therefore, 2/x = 0.25 (given)

the target will be missed by B immediately after such two shots : this means he can shot just twice ...the third shot cannot happen which means he missed (x-2) shots. thus, the probabilty of missing just after 2 shots is (x-2)/x. (x-2)/x = 1 - 2/x = 1 - 0.25 = 0.75

Re: The probability that a target will be shot two out of two [#permalink]
24 Mar 2013, 02:13

1

This post received KUDOS

Expert's post

Perhaps wrote:

Suppose there are total 'x' shots. B shoots 2 out of 2 times. means out of x shots (x>2) B shots 2 out of 2 ----> B shots at least 2 out of x. therefore, 2/x = 0.25 (given)

the target will be missed by B immediately after such two shots : this means he can shot just twice ...the third shot cannot happen which means he missed (x-2) shots. thus, the probabilty of missing just after 2 shots is (x-2)/x. (x-2)/x = 1 - 2/x = 1 - 0.25 = 0.75

Answer : C

Please correct me.

Frankly I don't understand your reasoning above. Anyway:

The probability that B can shoot a target two out of two times is 0.25 --> if the probability of success is p, then the probability of two successes in a row is p*p, we are told that p*p=0.25 --> p=0.5.

Now, the third shoot is independent from the first two, thus the probability of success with third try is p=0.5 and the probability of missing in third try is 1-p=0.5.

Re: The probability that a target will be shot two out of two [#permalink]
24 Mar 2013, 04:38

Quote:

Frankly I don't understand your reasoning above. Anyway:

The probability that B can shoot a target two out of two times is 0.25 --> if the probability of success is p, then the probability of two successes in a row is p*p, we are told that p*p=0.25 --> p=0.5.

Now, the third shoot is independent from the first two, thus the probability of success with third try is p=0.5 and the probability of missing in third try is 1-p=0.5.

Answer: B.

hey bunuel ....i still cant understand the method:( why v r considering p=0.25 ... while in ques its clearly mentioned that success for two shots is 0.25 ....

Re: The probability that a target will be shot two out of two [#permalink]
24 Mar 2013, 04:40

Expert's post

Perhaps wrote:

Quote:

Frankly I don't understand your reasoning above. Anyway:

The probability that B can shoot a target two out of two times is 0.25 --> if the probability of success is p, then the probability of two successes in a row is p*p, we are told that p*p=0.25 --> p=0.5.

Now, the third shoot is independent from the first two, thus the probability of success with third try is p=0.5 and the probability of missing in third try is 1-p=0.5.

Answer: B.

hey bunuel ....i still cant understand the method:( why v r considering p=0.25 ... while in ques its clearly mentioned that success for two shots is 0.25 ....

Re: The probability that a target will be shot two out of two [#permalink]
02 Jul 2014, 05:35

There are only 2 possibilietes: hit or no hit. Hence for the first shot the probability of hitting the target is 1/2. For the second shot also 1/2. Thus, the probability of hitting both is 1/2*1/2 = 1/4. Not hitting the next target is again 1/2. Hence B.

Re: The probability that a target will be shot two out of two [#permalink]
15 Jul 2014, 22:23

Bunuel wrote:

Frankly I don't understand your reasoning above. Anyway:

The probability that B can shoot a target two out of two times is 0.25 --> if the probability of success is p, then the probability of two successes in a row is p*p, we are told that p*p=0.25 --> p=0.5.

Now, the third shoot is independent from the first two, thus the probability of success with third try is p=0.5 and the probability of missing in third try is 1-p=0.5.

Answer: B.

Hello Bunuel,

How is the third shot independent of the first two shots ? The question states that "What is the probability that the target will be missed by B immediately after such two shots". I assumed this to be a conditional probability. That is, we have to consider only those failures which follow two successes.

Re: The probability that a target will be shot two out of two [#permalink]
16 Jul 2014, 01:32

Expert's post

1

This post was BOOKMARKED

parul1591 wrote:

Bunuel wrote:

Frankly I don't understand your reasoning above. Anyway:

The probability that B can shoot a target two out of two times is 0.25 --> if the probability of success is p, then the probability of two successes in a row is p*p, we are told that p*p=0.25 --> p=0.5.

Now, the third shoot is independent from the first two, thus the probability of success with third try is p=0.5 and the probability of missing in third try is 1-p=0.5.

Answer: B.

Hello Bunuel,

How is the third shot independent of the first two shots ? The question states that "What is the probability that the target will be missed by B immediately after such two shots". I assumed this to be a conditional probability. That is, we have to consider only those failures which follow two successes.

Appreciate your thoughts on this ! Thank you

Agree that the wording is not perfect. But "missed by B immediately after such two shots" means that these two shoots have already been made. _________________

Re: The probability that a target will be shot two out of two [#permalink]
03 Nov 2014, 03:52

probability of at least 1 miss in the 2 hits is 0.75. (1-0.25). so (miss,hit)(hit,miss) (miss miss) is 0.75. in these 2 out of three is miss in the first attempt(which replicates for 3rd attempt). hence (2/3)*0.75=0.5.

Regds Siva

gmatclubot

Re: The probability that a target will be shot two out of two
[#permalink]
03 Nov 2014, 03:52

Low GPA MBA Acceptance Rate Analysis Many applicants worry about applying to business school if they have a low GPA. I analyzed the low GPA MBA acceptance rate at...

In out-of-the-way places of the heart, Where your thoughts never think to wander, This beginning has been quietly forming, Waiting until you were ready to emerge. For a long...