Solving Statistics Questions with Steps: Q1-4

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Statistic Question!

I have these four questions I'm having trouble solving...
Could anyone write the solutions, but with steps on how they did it so I can possibly learn! Thanks a lot

I'm pretty sure they use the binomial or poisson distribution formulas... can't get proper answers


1. In a large shipment of chips, 5% are defective. What is the probability that exactly two out of a sample of ten are defective?

2. On average, a system breaks down every 50hrs. Find the probability of more than two break-downs in a 24hr period.

3. Show that there are more families of 6 children split 4-2 than those with 3 boys and 3 girls.

4. People arrive at a bank at the rate of 60 per hour. Find the chance of getting 0,1,2 or 3 in the next minute.
 
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This is how I would to do it.
1. Let p=0.05 q=.95. Do a binomial of (10 choose 2)x p^2 x q^8
You need the number of ways to arrange 2 defects in 10 trials multiplied by the probabilty of each event, ie. two defects.
2. Need to convert the Lamba given to one that represents a breakdown per 24 hr period. Here Lamda is 1 per 50 hour period. That represents a lamda of 12/25 per 24 hr period. Now find the probability of no breakdowns, 1 breakdown, and 2 breakdowns in a 24 hr period and then subtract that from 1 leaves probability of more than 2.
3. No sure, but on first glance seems like showing that 6 choose 4 is larger than 6 choose 3. I would check on this though.
4. Same as 2. Convert the lambda to a mean for 1 minute and use Poisson distribution.
 
4. Same as 2. Convert the lambda to a mean for 1 minute and use Poisson distribution.

so would lambda be 1/60 and after I find the chance of getting 3 in the next minute... that's the chance of 3 not coming in the next minute? so I subtract that from 1 to get 3 coming in the next minute?

Thanks a lot for the help
 
Yeah use lamda as 1/60. Then it is just a case of finding probability for X=0 X=1 X=2 X=3. 1-P(x=3) is the chance of any number of people in the next minute other than 3 people, but they are not asking you for that. Just plug in x=i i=0,1,2,3 and youre done. You should have four different probabilities corresponding to each part.
 
THANKS!
I get it...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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