Shooting game - a probability question

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)

Solution 1:

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.

Solution 2:

Using Binomial distribution, the required probability will be $$\sum_{6\le r\le 8}\binom 8r(0.6)^r(1-0.6)^{8-r}$$