Question

For someone familiar with probability notation & jargon. Am I interpreting this probability equation correctly?
http://i.imgur.com/lWb4qXi.jpg

Answer 1

Say we have two coins, coin1 and coin2, and we flip them simultaneously 1,000 times.

data:
coin1 tossed 505 heads, and 495 tails
coin2 tossed 491 heads, and 509 tails
Tosses where coin1 = Heads and coin2 = Heads: 234
Tosses where coin1 = Tails and coin2 = Tails: 238
Tosses where coin1 = Heads and coin2 = Tails: 271
Tosses where coin1 = Tails and coin2 = Heads: 257

I think the greek E character in this case is called a "Set Membership Symbol"
And I think x and y are both sample spaces aka state spaces - jargon for vectors that
contain the different types of outcomes that can happen in an event, like {heads, tails}.

x (coin1): {heads, tails}
y (coin2): {heads, tails}

My question is this:
is the result of the equation, when used with my coin toss data:
P(coin1 heads, coin2 heads) + p(coin1 tails, coin2 tails)
I don't know if this is right, but I have assumed that elements of vector x and vector y are
both plugged into equation P(x,y), and that the index of each vector increases by 1 after
each iteration, if that makes sense. In other words, if my assumption is correct, then
this equation's result is the probability that events in vector x & vector y with the same
index occur simultaneously. If this is the case, I don't really understand the
applicability of such an equation, and unfortunately my book doesn't elaborate further. It
merely calls it a 'sum rule'.

Answer 2

@ParthKohli around by chance?

Answer 3

you would use that equation to (for example)
find the probability of getting heads, ignoring which coin:

P(heads) = P(coin1, heads) + P(coin2, heads)

Answer 4

or , for example, the probability of coin1
P(coin1) = P(coin1, heads) + P(coin1,tails)

Answer 5

That makes sense, thanks - so the sigma will iterate through all elements of the Y vector in this case?

Answer 6

Applying it, the odds that coin1 is heads, and coin2 is heads or tails
p(coin1_heads) = p(coin1_heads, coin2_tails) + p(coin1_heads, coin2_heads);
p(a, b) = nij (number times event a[i] and b[j] occur simultaneously) / n (total tosses)
234/1000 + 271/1000 = 505/1000, same probability as coin1 throwing heads in data. Some of the book's format had me confused, but I think I got it.