Binomial Distribution; Using a mean to suggest probability of success.

and thanks for taking the time to look at my question. This was for a homework task, on binomial distribution, which i thought i understood.

A consignment of china mugs is packed for transportation in boxes of six. After the journey, a random sample of 100 boxes was inspected and the number of broken mugs was counted, with the following results: Number of boxes - Number of broken mugs in a box
25-0, 41-1, 25-2, 7-3, 2-4, 0-5, 0-6.

Let X be the random variable for the number of broken mugs in a box.

a) Calculate the mean number of broken mugs in a box for the sample of 100 boxes. The sum of the broken mugs = 120. 120/100 = 1.20. Mean number of broken mugs in a box for the sample of 100 boxes = 1.20.

The next section of the question confuses me:

b)Explain why a Binomial distribution X ~ Bin(6, p) could be used to model X, and show that using the mean calculated in (a) as the expected value suggests that the probability of success p, is equal to 0.2 where ‘success’ is a mug being broken. You should address each of the assumptions of a Binomial model in your answer.

I understand that the probability of success = 0.2 as 1.2/6 = 0.2, but i don't understand how to place this in the binom function in excel to show this. Any insights would be very helpful. =BINOMDIST(1.2;6;0.2;1) = 0.655,
=BINOMDIST(6;100;0.012;1) = 0.99
Why do i get the feeling that i'm overlooking something blindingly obvious?

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The data giving the number of boxes that have a particular number of broken mugs does not make sense. According to what is written, $25$ boxes have $0$ broken mugs, but then you write $5$ boxes and $6$ boxes have $0$ broken mugs. Please correct the errors. – heropup May 19 '14 at 20:55
Sorry about that, I had the last two numbers back-to-front. fixed it now. Thanks. – MapReduceFilter May 19 '14 at 20:59
Still flawed. "7-4" and "2-4" makes no sense. – heropup May 19 '14 at 20:59
I really don't know why i'm incapable of typing :'(. Hopefully fixed now. – MapReduceFilter May 19 '14 at 21:01

Your sample mean of the number of broken mugs in a box is $$\bar x = \frac{25 \cdot 0 + 41 \cdot 1 + 25 \cdot 2 + 7 \cdot 3 + 2 \cdot 4 + 0 \cdot 5 + 0 \cdot 6}{100} = 1.2.$$ Since the expected value of a binomial distribution with parameters $n$ and $p$ is ${\rm E}[X] = np$, if we use the estimate $\bar x = 6\hat p$, then $\hat p = \frac{1.2}{6} = 0.2$, as claimed.