I have a problem with probability I have a question which I struggle slightly to get my head around. 
A box contains 25 mobile phones.


*

*10 phones have WiFi but no Camera

*9 phones have a Camera but no WiFi

*6 phones have both camera and WiFi


Chose 5 phones at random (without replacement) from the box.
What is the probability that


*

*a) None of the 5 phones has both WiFi and a Camera

*b) All 5 phones have a Camera

*c) At least 2 of the phones has WiFi

*d) Exactly 2 phones have WiFi but no Camera, and the other 4 phones have a Camera.


I am studying for a repeat exam and this is a question that has come up on a past exam paper. It is part two to a three part question. I fully understand part one and three. Just this one catches me out. I can't ask the lecturer for any assistance as I am repeating the maths module outside of college assistance.
Should I be using the nCr function on my calculator? If so, how should I be using it? Say for part 'a)', would I be using it like 6C5?
Appreciate any help that will push me in the right direction :) 
 A: Some hints:
For the first question, you choose five from those that don't have both WiFi and a Camera.
For the second question, you choose five from those that have at least a Camera (they can also have WiFi, but don't have to).
For the third question, it may be easier to calculate the number of combinations that have exactly zero phones or exactly one phone with WiFi.  Calculating the zero-phones case is similar to the first question.  Calculating the one-phone case you'd first choose the phone that has WiFi, then choose the four phones that don't from what's left.  Once you get these two counts, subtract them from the total number of combinations $_{25}C_5$ because you want the number that have at least two phones with WiFi (which is everything that you haven't counted).
For the fourth question, choose the two phones that have WiFi but no camera, and then choose three (I'm guessing the $6$ is a typo!) from what's left that have a Camera (they may also have WiFi but don't have to).
In each case, you get the total number of combinations that have the desired properties.  Then divide by the total number of combinations to get the probability.
EDIT based on your comment:  There are two main ways to solve the third question.  You calculated the number of combinations with two WiFi/no Camera phones (call this $N_2$), and did so correctly: $_{10}C_2 \cdot _{15}C_3$.  If you go this route, you'll also need to calculate the same for three, four, and five WiFi/no Camera phones.  This is counting the possibilities directly (you do not subtract from the total number $N_T = _{25}C_{5}$).
The way I suggested saves you a few calculations.  Since
$$N_T = N_0 + N_1 + N_2 + N_3 + N_4 + N_5,$$
we see that
$$N_{\geq 2} = N_2 + N_3 + N_4 + N_5 = N_T - N_0 - N_1.$$
This is the complement set: everything that's not what you want.  If you subtract everything that you don't want from everything, you get what you want.  In this case, you only need to calculate the number of possibilities for zero ($N_0$) and one ($N_1$) WiFi/no Camera phones, and subtract from the total. 
A: Probability in this case is the ratio $$\frac {\text{number of favourable cases}} {\text{number of all possible cases} } $$
Let's take the first question which asks the probability of drawing 5 phones which doesn't have both Wifi and camera. 
So they must have only Wifi or only camera (or neither but that's not possible in this problem). So they can be chosen from the 10 phones which have only Wifi or from the 9 which have only camera or any combination of them indifferently. 
So the favourable cases are the number of possible way I have to choose 5 elements from 19 while all the cases are the number of way I have to choose 5 elements from 25.
