A certain sports league has ten teams in its Western Conference and eight teams in its Eastern Conference. At the end of the season, two teams, one from each conference, play in The Big Game. Two teams in each conference are located in Texas. If each team in each conference has an equal probability of making it to The Big Game, and if each team in The Big Game has an equal probability of winning that game, what is the probability that a team from Texas will win The Big Game?

a) 3/20

b) 2/9

c) 9/40

d) 27/80

e) 5/16

What would be the probability of P{East winning the Super bowl}. Is it 1/2? Or 8/18 because of equally likely probabilities?

Just curious. Here's why I am asking:

Here's what I did initially:

P(T=texas wins) = P(Texas Wins/East) P(East) + P (Texas Wins/West) P(West)

= 10/18 * 2/10 + 8/18 * 2/8

= 2/9

If I change P(Texas Wins/East) = P(Texas Wins/East) = 1/2, ( the question says that "If each team in each conference has an equal probability of making it to The Big Game, and if each team in The Big Game has an equal probability of winning that game," --- I am not sure whether this is an accurate translation...)

then P(texas wins) = 1/2 * 2/10 + 1/2 * 2/8 = 9/40... Any thoughts? Please help....