Sorry for the poorly worded title but I'm not sure what to put for it. I have a question relating the the binomial distribution for which I have provided two solutions for but I am not sure which is correct. Any help would be much appreciated.
"For all newly sold smartphones, roughly about 55% run on Android, 35% run on iOS, and 10% run on Windows Phone. A mobile phone shop sells phones with all these operating systems. Their pool of customers is large enough to be assumed infinite. Calculate the probability that of the first 10 customers, 3 want to buy an Android phone, and 1 wants to buy a Windows phone."
First solution:
Let X = number wanting to buy Android
Let Y = number wanting to buy Windows
$P(X = 3 \cap Y = 1)=P(X=3)\cdot P(Y=1) = \frac{10!}{3!7!}(0.55)^{7}(0.45)^3\cdot\frac{10!}{1!7!}(0.1)^1(0.9)^{9} = 0.029$
Second solution:
Probability of wanting Android or Windows $= 0.55 + 0.1 = 0.65$
Let Z = number wanting Android or Windows
$P(Z=4)=\frac{10!}{4!6!}(0.65)^4(0.35)^6 = 0.069$
There are 16 combinations of wanting Android or Windows, 4 of which consist of 3 Android and 1 Windows, therefore probability $=\frac{1}{4}\cdot P(Z=4) = 0.017$
In addition to this, I have another question: "Calculate the probability that of the first 12 customers at least 2 want to buy an Android phone and at least 2 want to buy an iOS phone." but I think it better to look at this after understanding the former question.
Thanks!
Edit:
I thought I might as well provide my attempt a solution for the latter question to save time.
I thought that I would take the compliment and find the combinations that did not satisfy the criteria those being in the form x-y with x being number of Android and y being number of iOS:
\begin{array}{cccc} 0-0 & 1-0 \\ 0-1 & 1-1 \\ 0-2 & 1-2 & 2-0 & 2-1 \\ 0-3 & 1-3 & 3-0 & 3-1 \\ 0-4 & 1-4 & 4-0 & 4-1 \\ 0-5 & 1-5 & 5-0 & 5-1 \\ 0-6 & 1-6 & 6-0 & 6-1 \\ 0-7 & 1-7 & 7-0 & 7-1 \\ 0-8 & 1-8 & 8-0 & 8-1 \\ 0-9 & 1-9 & 9-0 & 9-1 \\ 0-10 & 1-10 & 10-0 & 10-1 \\ 0-11 & 1-11 & 11-0 & 11-1 \\ 0-12 & 1-12 & 12-0 & 12-1 \\ \end{array}
Then $P(X\geq 2 \cap Y \geq 2) = P(X=0 \cap Y=Any)$
$+P(X=1 \cap Y=Any)$
$+P(X\geq2 \cap Y=0)$
$+P(X\geq 2 \cap Y=1)$
$= P(X=0)P(Y=Any)$
$+P(X=1)P(Y=Any)$
$+P(X\geq 2)P(Y=0)$
$+P(X\geq 2)P(Y=1)$
(Where $Y=Any$ represents $0 \leq Y \leq 12$)
Your first method is correct. Your second method is incorrect.