Statistics and Discrete distributions

amai
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Homework Statement


The probabilities of blood types O, A, B and AB are 0.46, 0.39, 0.12, 0.03 respectively.
If a clinic is seeking either type O or B from six random individuals, what is the probability that at least 2 people have the desired blood type?

Homework Equations





The Attempt at a Solution



I'm not really sure where to start. From the questions, I don't think it's a binomial distribution or a poisson distribution. All i have so far is...

Where X=people with the desired blood type

P(X≥ 2)=1-P(0)-(P(1)

Any suggestions?
 
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Since the probability of being O is 0.46 and the probability of being B is 0.12, what is the probability of being either O or B?

What is the probability of not being either O or B, or in other words, what is the probability of being A or AB?
 
amai said:

Homework Statement


The probabilities of blood types O, A, B and AB are 0.46, 0.39, 0.12, 0.03 respectively.
If a clinic is seeking either type O or B from six random individuals, what is the probability that at least 2 people have the desired blood type?

Homework Equations





The Attempt at a Solution



I'm not really sure where to start. From the questions, I don't think it's a binomial distribution

You are conducting six independent trials, in which success in a trial is defined as "this person has type O or B blood". That gives you 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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