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In a shooting game, the probability for Jack to hit a target is 0.6. Suppose he makes 8 shots, find the probabilities that he can hit the target in more than 5 shots.

I find this question in an exercise and do not know how to solve it. I have tried my best but my answer is different from the one in the answer key. Can anyone suggest a clear solution to it?

My trial: (0.6^6)(0.4^2)+(0.6^7)(0.4)+0.6^8

But it is wrong...

(The answer is 0.3154, which is correct to 4 significant figures)

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To a few more, it is $0.31539456$. Luckily short binomial distribution calculation. Note that "more than $5$" means $6$ or $7$ or $8$. –  André Nicolas Apr 13 '13 at 14:15
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3 Answers 3

up vote 3 down vote accepted

The number of shots is n=8 and the probability of a 'successful' shot is p=0.6. This is the Binomial distribution.

The tail probability of a binomial distribution B(n,p) $X$, i.e. $P(X\geq x)$ is

$$\sum_{r\geq x}\binom n r (1-p)^{n-r}(p)^r$$

here your x is 6, n is 8, p=0.6, go figure...

EDIT: Okay you don't know what a binomial distribution is. Fine.

So you want 6 or more shots, that is 6,7 or 8.

How many ways can you have exactly 8 out of 8? well exactly 1 way. What is the probability of this? $0.6^8$

How many ways can you have exactly 7 out of 8? well, you can miss 1 of the 8. What is the probabilty of missing 1 out of 8? $0.4\times 0.6^7$ so it is $8\times 0.4\times 0.6^7$

How many ways can you have exactly 6 out of 8? well, you can miss 2 of the 8. This is not so easy. The formula for missing x out of n is $\binom n x$

This means $\dfrac{n\times(n-1)...\times 1}{[x\times (x-1)....1][(n-x)\times(n-x-1)...1]}$

In this case this is $\dfrac{8\times7...\times 1}{[6\times5... \times 1][2\times 1]}=\dfrac{8\times 7}{2\times 1}=28$

So there are 28 ways you can exactly miss 2 out of 8. The probability of each miss is $(0.6)^6\times 0.4^2$

Putting these all together, we have $0.6^8+8\times 0.6^7\times0.4+28\times (0.6)^6\times 0.4^2$ whatever that is.

If this still doesn't make sense, you should look at binomial distribution.

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Are there easier ways? I don't know binomial distributions :( –  redcap Apr 13 '13 at 14:14
@redcap okay, let me edit this... –  Lost1 Apr 13 '13 at 14:14
@Lost1, the formula is $\sum \binom nrp^r(1-p)^{n-r}$, right? –  lab bhattacharjee Apr 13 '13 at 14:15
@labbhattacharjee thanks! –  Lost1 Apr 13 '13 at 14:24
@Lost1 Thank you very much, you make it clear! –  redcap Apr 13 '13 at 14:29
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Using Binomial distribution, the required probability will be $$\sum_{6\le r\le 8}\binom 8r(0.6)^r(1-0.6)^{8-r}$$

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There are two ways to approach this problem:

  1. Find the probability asked for or ...
  2. Find the probability of what you don't want and subtract it from $1$. Since the problem has $8$ trials and is asking for more than $5$ trials being a successful hit, we only have to count the probabilities that Jack hit the target $6, 7$, or $8$ times. This is much easier than counting the probabilities that Jack can hit the target $1, 2, 3, 4, or 5$ times because this will take longer. Therefore, path $(1)$ is the way to go.

$(1)$ Finding the probability of hitting the target more than 5 times is very easy, as you have already completed half the job. Your answer:$$(0.6^6)(0.4^2)+(0.6^7)(0.4)+0.6^8$$ isn't necessarily incorrect, but it's just not enough. $$(0.6^6)(0.4^2)$$ is just showing that Jack his the target exactly 6 times and missed the other two times. But if we had a total of 8 shots, how do we know which 6 shots he made and which 2 shots he missed? Therefore you must multiply $(0.6^6)(0.4^2)$ by $8 \choose 2$, which is 56. Similarly, $$(0.6^7)(0.4)$$ is Jack making exactly 7 shots. So you must multiply this probability by $8 \choose 1$, which is 8. And lastly, $$0.6^8$$ is Jack hitting the target 8 times, never missing once. So we multiply this probability by $8 \choose 0$, which is 1.

Therefore our desired probability is just $$(0.6^6)(0.4^2){8 \choose 2}+(0.6^7)(0.4){8 \choose 2}+0.6^8{8 \choose 0} \approx 0.3154$$

HINT: Note that numbers I multiplied your probabilities with are the number of ways we can arrange the desired outcomes. For example, let Xs and Os denote Jack hitting and missing the target respectively. In how many ways can we arrange $$XXXXXOO$$, which is just the probability that Jack hits the target 6 times, and misses twice. This is $$\begin{align} \frac{8!}{2!6!} & = {8 \choose 2} \end{align} $$ Use this to figure out the respective probabilities: That Jack makes no more than 5 shots, than Jack makes exactly 3 shots, that Jack makes exactly $m$ shots, where $m$ is a positive integer such that $0\le m\le 8$. And all of the scenarios require that Jack only has 8 shots.

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I added some details to my answer to the question you put a bounty on. Hopefully this makes the method clearer. –  Bill Dubuque Jan 22 at 19:13
It does thanks. –  Vishwa Iyer Jan 22 at 23:59
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