UIL D 2013

[E=mc^2]

22. Betty Notsogood and Willie When bowl two games. Willie is one and one-half times as likely to win any game as is Betty. What is the probability that Willie or Betty will win both games?

Answer: 52%

Thanks!

[nsguy1350]

Let the probability that Betty wins a game be [unparseable or potentially dangerous latex formula], and that Willie wins, [unparseable or potentially dangerous latex formula]. Since there are no ties (assumed), it is obvious that [unparseable or potentially dangerous latex formula]. Since [unparseable or potentially dangerous latex formula] we can solve for [unparseable or potentially dangerous latex formula] and [unparseable or potentially dangerous latex formula] and get [unparseable or potentially dangerous latex formula] and [unparseable or potentially dangerous latex formula] (this is also easily done mentally.) The probability that Willie wins twice is [unparseable or potentially dangerous latex formula] and the probability that Betty wins twice is [unparseable or potentially dangerous latex formula]. The probability that either one wins twice is [unparseable or potentially dangerous latex formula]

[Fredfredburger]

Alternative solution using fractions.

If Willie is [unparseable or potentially dangerous latex formula] more likely to win then that means the odds of her winning are [unparseable or potentially dangerous latex formula] or equivalently [unparseable or potentially dangerous latex formula]

Thus the probability of Willie winning is [unparseable or potentially dangerous latex formula] and the complement to that, the probability of Betty winning, is [unparseable or potentially dangerous latex formula]

Then just do what Wesley did,
[unparseable or potentially dangerous latex formula]

[101dalmatians]

@Burger: Isnt that like wat nsguy said - except in fracs? Or was ur intention just to convert nsguys decimals to fracs?

[Fredfredburger]