I'm dumb I know, I need Probability help

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I did this problem but am not too sure it's right, have a look

Of all households making less than $25,000 annually, 45% of them use coupons weekly when shopping for groceries. Suppose 10 households making less than $25,000 annually are selected at random. What is the variance of the number of households that use coupons weekly?

I got 4.5

But I have no clue how to do this one, the book doesn't explain anything

Suppose 30% of all M&Ms are brown. If 7 M&Ms are randomly selected, what is the probability that at most 1 is brown?
 
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These are both binomial distribution problems. The probability that a selected house hold uses coupons weekly is 0.45 and the probability that is does not is 0.55. You should have a formula for the variance of a sample from a binomial distribution.

If 30% of all M&Ms are brown, then 70% are not- the chance that any randomly selected M&M is not brown is 0.7. What is the probability that all 10 are not brown? What is the probability that exactly one is brown?
Again, this is a binomial distribution.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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