Probability

Hi, I have a few questions on an excel file that I have to complete regarding probability. Here are the questions:
1) Two cards are drawn in succession from a well shuffled pack of cards (without replacement).
Find the probability that:
i) both cards are red
ii)both cards are the same colour
iii) the cards are different colours

2) A box contains 7 red balls and 11 black calls. Two balls are taken out one after the other without replacement. What is the probability that both balls are black?

3) Of two air-conditioners X and Y in a computer lab, X has a probability 0.04 of breaking down on any one day and Y has a probability 0.02. On the first day os summer when both air conditioners are called into use, what is the probability that:
i) Neither will break down
ii) Exactly one will break down
iii) Both will break down

Here are my answers:
1i) P(1st red) = 1/2, P(2nd red) = 25/51 so P(both red) = 1/2 *25/51 = 25/102

ii) 25/102 (same concept as above)

iii) P(1st red) = 1/2, P(2nd black) = 26/51 so P(different colours) = 1/2 * 26/51 = 13/51

2) p(1st ball black) = 11/18, P(2nd ball black) = 10/17 so P(both balls black) = 11/18 * 10/17 = 55/153

3i) P(neither break) = (1-0.04) * (1/0.02) = 0.96*0.98 = 588/625

ii) P(one breaks) = 0.04 * (1-0.02) + 0.02 * (1-0.04) = 0.04*0.98 +0.02*0.96 = 0.0392+0.0192 = 73/1250

iii) P(both) = 0.04 * 0.02 = 1/1250

My problem is that with the above answers I get 71% for the 3 questions, I would like to know where and what I'm doing wrong.

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3) Of two air-conditioners X and Y in a computer lab,
X has a probability 0.04 of breaking down on any one day and Y has a probability 0.02.
On the first day os summer when both air conditioners are called into use, what is the probability that:

i) Neither will break down

ii) Exactly one will break down

iii) Both will break down


Here are my answers:

\(\displaystyle i)\; P(\text{neither break}) \;=\; (1-0.04)\cdot(1/0.02) \;=\; (0.96)(0.98) \;=\;\frac{588}{625}\)

\(\displaystyle ii)\; P(\text{one breaks}) \;=\; (0.04)(1-0.02) \;+\; (0.02)(1-0.04) \;=\; (0.04)(0.98) \;+\;(0.02)(0.96) \;=\; 0.0392\;+\;0.0192 \;=\; \frac{73}{1250}\)

\(\displaystyle iii)\; P(\text{both break}) \;=\; (0.04)(0.02) \;=\;\frac{1}{1250}\)

My problem is that with the above answers I get 71% for the 3 questions, . How did you get 71%?

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