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# George is aiming his arrows at the target shown

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Manager
Joined: 19 Aug 2016
Posts: 153
Location: India
GMAT 1: 640 Q47 V31
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George is aiming his arrows at the target shown [#permalink]

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02 May 2017, 23:52
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Difficulty:

35% (medium)

Question Stats:

74% (01:18) correct 26% (01:33) wrong based on 69 sessions

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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.

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

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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: 167
Concentration: Finance, Strategy
GMAT 1: 620 Q47 V28
GMAT 2: 640 Q48 V28
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Re: George is aiming his arrows at the target shown [#permalink]

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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!
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Manager
Joined: 17 Apr 2016
Posts: 100
Re: George is aiming his arrows at the target shown [#permalink]

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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]

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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.
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Re: George is aiming his arrows at the target shown [#permalink]

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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

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Re: George is aiming his arrows at the target shown [#permalink]

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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.

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Re: George is aiming his arrows at the target shown   [#permalink] 06 May 2017, 17:54
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