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George is aiming his arrows at the target shown [#permalink]
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02 May 2017, 22:52
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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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Re: George is aiming his arrows at the target shown [#permalink]
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02 May 2017, 23:08
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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
Answer is B. 0.3



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Re: George is aiming his arrows at the target shown [#permalink]
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03 May 2017, 00:50
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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Re: George is aiming his arrows at the target shown [#permalink]
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04 May 2017, 18:36
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
Answer is B. 0.3 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?



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Re: George is aiming his arrows at the target shown [#permalink]
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04 May 2017, 20:06
adityapareshshah wrote: 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
Answer is B. 0.3 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? adityapareshshahWhen 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, 20:21
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 The answer is (B)
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Re: George is aiming his arrows at the target shown [#permalink]
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06 May 2017, 16:54
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(HHHMM) = 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 HHHMM, which is a combination problem because order doesn’t matter. 5C3 = 5!/[3!(53)!] = 5!/(3! x 2!) = (5 x 4)/2 = 10 ways Thus, P(3H and 2M) ≈ 0.03 x 10 = 0.3. Answer: B
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