GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Oct 2018, 00:32

### GMAT Club Daily Prep

#### 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

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

# George is aiming his arrows at the target shown

Author Message
TAGS:

### Hide Tags

Manager
Joined: 19 Aug 2016
Posts: 152
Location: India
GMAT 1: 640 Q47 V31
GPA: 3.82
George is aiming his arrows at the target shown  [#permalink]

### Show Tags

02 May 2017, 23:52
1
00:00

Difficulty:

35% (medium)

Question Stats:

69% (02:14) correct 31% (02:03) wrong based on 73 sessions

### HideShow timer Statistics

George is aiming his arrows at the target shown by his mentor. If the probability of hitting the target is 0.7, what is the approximate probability that he will hit the target only three times in his first five attempts?

A. 0.1
B. 0.3
C. 0.85
D. 0.5
E. 0.65

Could someone explain in a simplified manner how is combination being used here.

_________________

Consider giving me Kudos if you find my posts useful, challenging and helpful!

Senior Manager
Joined: 24 Apr 2016
Posts: 333
Re: George is aiming his arrows at the target shown  [#permalink]

### Show Tags

03 May 2017, 00:08
1
Given:

Probability of hitting the target is 0.7
Probability of Not hitting the target is 0.3

Now,probability that he will hit the target only three times in his first five attempts, this means, out of 5 attempts he will hit the target 3 times and miss the target 2 times ==> which can be represented as = 0.7 * 0.7 * 0.7 * 0.3 * 0.3

Now the three successful and unsuccessful attempts can happen in any sequence and which needs to be accounted for above.

So one of the sequence may be - Success,Failure,Success,Success,Failure
Another sequence may be - Failure, Success,Failure,Success,Success

So to account the different sequences we can considers that there are five places where three Success needs to be placed and this can be represented by 5C3 = 5!/3!*2! = 10

Therefore,

the approximate probability that he will hit the target only three times in his first five attempts

= 0.7 * 0.7 * 0.7 * 0.3 * 0.3 * 10

= 0.3087

Joined: 24 Mar 2015
Posts: 166
Concentration: Finance, Strategy
GMAT 1: 620 Q47 V28
GMAT 2: 640 Q48 V28
GMAT 3: 680 Q49 V35
WE: Engineering (Real Estate)
Re: George is aiming his arrows at the target shown  [#permalink]

### Show Tags

03 May 2017, 01:50
ashikaverma13 wrote:
George is aiming his arrows at the target shown by his mentor. If the probability of hitting the target is 0.7, what is the approximate probability that he will hit the target only three times in his first five attempts?

A. 0.1
B. 0.3
C. 0.85
D. 0.5
E. 0.65

Could someone explain in a simplified manner how is combination being used here.

How many ways can there be exactly 3 successful tries out of 5?

Consider $$\frac{5!}{(2!*3!)}=\frac{(5*4*3*2*1)}{(3*2)*2}=10$$

So there are 10 cases with 3 successful tries out of 5. Now we need to consider the probability of 1 combination

$$0.7*0.7*0.7*0.3*0.3=\frac{7^3*3^2}{100000}$$=0.03

$$0.03*10=0.3$$

Hope this helps!
_________________

Kindly click on the +1 Kudos if you found my post helpful. Thank you!

|
|
|
V

Manager
Joined: 17 Apr 2016
Posts: 89
Re: George is aiming his arrows at the target shown  [#permalink]

### Show Tags

04 May 2017, 19:36
quantumliner wrote:
Given:

Probability of hitting the target is 0.7
Probability of Not hitting the target is 0.3

Now,probability that he will hit the target only three times in his first five attempts, this means, out of 5 attempts he will hit the target 3 times and miss the target 2 times ==> which can be represented as = 0.7 * 0.7 * 0.7 * 0.3 * 0.3

Now the three successful and unsuccessful attempts can happen in any sequence and which needs to be accounted for above.

So one of the sequence may be - Success,Failure,Success,Success,Failure
Another sequence may be - Failure, Success,Failure,Success,Success

So to account the different sequences we can considers that there are five places where three Success needs to be placed and this can be represented by 5C3 = 5!/3!*2! = 10

Therefore,

the approximate probability that he will hit the target only three times in his first five attempts

= 0.7 * 0.7 * 0.7 * 0.3 * 0.3 * 10

= 0.3087

Hi quantumliner,

I have a little silly doubt. Need to get concepts right.
We are considering the sequence for occurrence of Success and hence multiplying the probability by 10, but we aren't considering the failure sequence here.
What am I missing?
Senior Manager
Joined: 24 Apr 2016
Posts: 333
Re: George is aiming his arrows at the target shown  [#permalink]

### Show Tags

04 May 2017, 21:06
quantumliner wrote:
Given:

Probability of hitting the target is 0.7
Probability of Not hitting the target is 0.3

Now,probability that he will hit the target only three times in his first five attempts, this means, out of 5 attempts he will hit the target 3 times and miss the target 2 times ==> which can be represented as = 0.7 * 0.7 * 0.7 * 0.3 * 0.3

Now the three successful and unsuccessful attempts can happen in any sequence and which needs to be accounted for above.

So one of the sequence may be - Success,Failure,Success,Success,Failure
Another sequence may be - Failure, Success,Failure,Success,Success

So to account the different sequences we can considers that there are five places where three Success needs to be placed and this can be represented by 5C3 = 5!/3!*2! = 10

Therefore,

the approximate probability that he will hit the target only three times in his first five attempts

= 0.7 * 0.7 * 0.7 * 0.3 * 0.3 * 10

= 0.3087

Hi quantumliner,

I have a little silly doubt. Need to get concepts right.
We are considering the sequence for occurrence of Success and hence multiplying the probability by 10, but we aren't considering the failure sequence here.
What am I missing?

When I said we need to consider the sequence of success, it also includes the sequence of failure.

For example, lets take the sequence - "Success,Failure,Success,Success,Failure". This sequence is a unique sequence for both success and failure.

In another sequence "Failure, Success,Failure,Success,Success", This again is a unique sequence for both success and failure.

In simple words, we are asked to arrange 3 success and 2 failure in a row of 5 seats. Hopefully this helps.
Intern
Status: Not Applying
Joined: 27 Apr 2009
Posts: 47
Location: India
Schools: HBS '14 (A)
GMAT 1: 760 Q51 V41
Re: George is aiming his arrows at the target shown  [#permalink]

### Show Tags

04 May 2017, 21:21
A probability question based on P&C.
The question needs P(Hit, Hit, Hit, Not Hit, Not Hit).
Before you jump to find what this value is, realise that this is only one of the possible outcomes. The outcomes can be arranged in other possible ways. So, you need to find the number of arrangements of these outcomes. Also, you need to find the Probability of not hitting the target.

P(hitting) = 0.7
P(not hitting) = 1 - P(hitting) = 1 - 0.7 = 0.3

Number of arrangements of (H,H,H, NH, NH) = 5!/(3!*2!) = 10

P(3 hits and 2 fails) = 10 *(0.7*0.7*0.7*0.3*0.3) = 10*0.30 = 0.3

_________________

http://www.wiziuscareers.com

Experts in Helping you get through a top-50 school with Low GMAT

Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3846
Location: United States (CA)
Re: George is aiming his arrows at the target shown  [#permalink]

### Show Tags

06 May 2017, 17:54
ashikaverma13 wrote:
George is aiming his arrows at the target shown by his mentor. If the probability of hitting the target is 0.7, what is the approximate probability that he will hit the target only three times in his first five attempts?

A. 0.1
B. 0.3
C. 0.85
D. 0.5
E. 0.65

We need to determine the probability that George hits the target 3 times and misses twice, in 5 attempts.

Since the probability of a hit is 0.7, the probability of a miss is 0.3. Thus, we need to determine:

P(H-H-H-M-M) = 0.7 x 0.7 x 0.7 x 0.3 x 0.3 ≈ 0.03

However, we have to account for the number of ways we can arrange H-H-H-M-M, which is a combination problem because order doesn’t matter.

5C3 = 5!/[3!(5-3)!] = 5!/(3! x 2!) = (5 x 4)/2 = 10 ways

Thus, P(3H and 2M) ≈ 0.03 x 10 = 0.3.

_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Re: George is aiming his arrows at the target shown &nbs [#permalink] 06 May 2017, 17:54
Display posts from previous: Sort by