A phone store has items consisting of 5 models of iPhones, 4 models of Samsung, 8 models of LG, and 7 models of HTC. You're asked you to go and purchase 3 iPhones, 2 Samsungs, 2 LGs, and 4 HTCs.
i) In how many ways can you purchase these if you do not want to select more than one of any model?
ii) There are HTC and iPhone models that you really wanted. What is the probability that those two phone models were selected during your random selection?
I found for a) that there are 58800 ways to make this purchase after multiplying combinations of each phone type.
I'm trying to understand how I can do b). I know that there are 7 HTC models to choose from. I will pick 4 different HTC models without replacement, and order doesn't matter. So the total number of outcomes of picking 4 HTC models from 7 is 7C4 which is equal to 35. Since each HTC model has a 1/7 chance of being picked, would my probability of picking my favorite modem be 1/35?
And if I follow that logic for the iPhone, my probability would be 1/10. And so what I need to find is
P(Picking fav iPhone AND picking fav HTC) = 1/35 * 1/10 ?
Is my reasoning correct?
No. You have chosen four HTC phones out of seven available, so the chance any specific model is chosen is $4/7$