Probability with different populations and functions

jackcb
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Homework Statement



In an organsataion employees work either in marketing or sales. 40% are male. 70% of the men work in sales. 60% female. 50% of the women work in marketing.

What is the probability that a randomly selected employee is either male and/or works in marketing?

Homework Equations


The Attempt at a Solution



P(male) = 0.4
P(employee works in marketing) = 0.42
P(male working in marketing) = 0.4 x 0.3 = 0.12

I know it's simple, but I'm very confused now (and stupid!). I would be really grateful if someone could help with this, please?
 
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It's been a while since I did probablity, but I'm pretty sure this is not Calculus, and that question doesn't make much sense to me, I guess I lost touch with basic maths...
 
It may help to think of numbers of employees rather than probabilities. Suppose the total number of employees is N. So the number of male employees is 0.4*N eg. Then if you add the number of females who work in marketing to the total number of males and divide by N, you will have it.
 
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Thank you.

So it there are 100 employees, 40 will be male (of which 12 work in marketing), and there will be 30 women working in marketing too. So I add 40 and 30, getting 70 and the 70/100 = 0.7. Is that right, please?
 
Or just say .4*N are male and .6*.5*N are female and in marketing. Total .7*N. Probability .7*N/N=.7.
 
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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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