You can donate to a company: $10$ dollars , $20$ dollars or nothing.
In a mall there are $70$% young people and $30$ % old people.
$50$% from the old people aren't donating anything.
$20$% from the young people there are donating $20$ dollars.
$60$% from the people there are donating $10$ dollars.
The probability that an old one donates $20$ dollars is $3/7$ times smaller than the probability that a young one donates $20$ dollars.
If we choose randomly, one by one, $10$ people from that mall (old and young), what is the probability that exactly $7$ people donated $10$ dollars and the rest donated $20$ dollars?
I calculated and found that the probability for a person to donate $20$ dollars is $0.2$, and the probabilty for donating $10$ dollars is $0.6$ and for nothing is $0.2$.
My problem is that i don't know if i can use the binomial formula here such that: $p=\binom {10}{7}0.6^70.2^3=0.02687$ or that's wrong and i need a multinomial.
Thanks.
The probability for a person to donate \$20 is not equal to 0.2.
The probability for an old person is $\frac37\cdot0.2 = \frac{3}{35}$, and the probability for a young person is 0.2. Therefore the average probability is:
$$\frac{3}{35}\cdot0.3+0.2\cdot0.7=\frac{9}{350}+\frac{14}{100}=\frac{9}{350}+\frac{49}{350}=\frac{56}{350}=\frac{4}{25}=0.16$$
Other than that, you have the right probability.