Probability Question: A Question is tossed 4times..

a) Whats is the probability of tossing atleast three "heads" if a fair coin is tossed four times:

My attempt: atleast 3heads = heads 3times or heads four times
so: 4!/3!=4
4 x (1/16)= 1/4

probability for 4 heads = 1/16
ans: 1/4 + 1/16 = 5/16

2) Eight cards, numbered 4 to 11, are placed in a hat and two of the cards are drawn at random.
A) What is the probability that the sum of the numbers will be equal to 10?
B) What is the probability that the sum of the numbers will be greater than 14?

P(4 heads) = 1/16 (1/2 ^ 4)
P(3 heads) = 4 * 1/16 (each arrangement of 3 h, 1 t = 1/16 and there are 4 of them)

Total = 5/16


2 A)
There are 8 C 2 choices of pairs of cards = 8 * 7 / 2! = 28
Sum = 10:
4 + 6 is the only one to add to 10, so P(sum = 10) = 1/28

2 B)
Sum > 14 from:
4 + 11
5 + 10 or 5 + 11
6 + 9, 10, 11
7 + 8, 9, 10, 11
8 + 9, 10, 11
9 + 10, 11
10 + 11
Count those up and divide by 28
16/28 = 8/14 = 4/7

Check:
4 + (5 - 10) <= 14 (6 cases)
5 + (6 - 9) <= 14 (4 cases)
6 + (7 - 8) <= 14 (2 cases)
7 + anything > 14
same for the rest
6 + 4 + 2 = 12
and 12 + 16 = 28, so that's everything.