I need help with the following questions, i attempted to solve them, and i also put my answers here too.

When you are answering them, please show all your steps as i am kinda lost in this field.

1. suppose S = {1,2,3} and P({1,2}) = 1/3 and P({2,3}) = 2/3

compute P({1}), P({2}), P({3})

my answer

P({1}) + P({2}) + P({3}) = 1 <--- 1 being 100% and 0 being 0%

P({1}) + P({2}) = 1/3

P({2}) + P({3}) = 2/3

a) P({1}) + (1/3 - P({1})) + P({3}) = 1

P({3}) = 2/3

b) P({1}) + (2/3 - P({3})) + P({3}) = 1

P({1}) = 1/3

so then 1/3 + P({2}) + 2/3 = 1

so P({2}) = 0

therefore: P({1}) = 1/3, P({2}) = 0, P({2}) = 2/3

2. Suppose A1 watches the six o'clock news a 2/3 of the time and watches eleven o'clock news 1/2 of the time and watches both news 1/3 of the time. For a randomely selected day what is the probability that A1 watches only the six o'clock new?s, and whats the probability that A1 watches neither news?

my answer

chance for 6oclock = 2/3

chance for 11oclock news = 1/2 => complementary is 1/2

for only six o'clock news

has to watch the 6oclock news which is 2/3, and not watch the 11oclock news which is 1/2

2/3*1/2 = 1/3

so only watching 6 o'clock news is 1/3 chance...

for watching neither

avoid watching 6o'clock news which is 1/3 (complementary) and avoid watching eleven o'clock news 1/2 (complementary)

so its

1/3*1/2 = 1/6

so watching neither is 1/6 chance...