<p>Suppose the probability that an American League team will win the World Series is 60%. What is the probability that a National League team will win the World Series in the 5th game? Express your answer to the nearest whole percent.</p>
<p>Any suggestions about how to go about it?</p>
<p>That’s an unclear question; unless the answer is 60%, it requires previous knowledge of the World Series, and even if the answer is 60%, the setup is misleading.</p>
<p>Probability that an AL team will win 4 games of the WS: 3/5
" " NL team " ": 2/5</p>
<p>prob. that the AL team will win any one game:
x^4 = 3/5
x = frthrt (3/5) (~ .88)</p>
<p>Probability that the NL team will win 4 games and lose 1 after the 5th game:</p>
<p>(frthrt (2/5))^4 * frthrt (3/5) = ~35%</p>
<p>World series is determined by first to four wins in a seven game series. If both NL and AL teams had equal probability (0.5) to win each game, AL (or NL) would win the world series 50% of the time. Given that an AL team would win the world series 60% of the time (in either 4, 5 , 6 or 7 games), how do I calculate an AL team’s probability to win a game? I do not think it is 60%.</p>
<p>It’s not.
The probability that one team will win the world series is really just their probability to win 4 games. The probability that the NL team will win the world series is 40%. Thus it’s x * x * x *x = .4
where x is the probability that the NL team will win individual 1 game. Thus, x = frthrt(2/5), x^4 * frthrt(3/5) = the probability you’re looking for (35%)</p>
<p>^Thanks.
10Char</p>
<p>^^ I think there is more to it than that. And this is WAY too hard to be on the SAT…</p>
<p>For example, you are assuming they win in 4 straight. But they could also win in 5, 6 or 7 games…the probabilities are all different and they all have to be included…The probability of winning the series is not just x^4, where x is the probability of winning 1 game.</p>
<p>Out of curiosity, where did you get this problem? I’ve seen it discussed recently, but I don’t remember where.</p>
<p>Wait, wait, wait: are you sure you didn’t mean to say that the American League’s probability of winning an INDIVIDUAL game is 60%??? That would make it a MUCH easier question (but still too hard for the SAT).</p>
<p>If that’s the question, then the answer is:</p>
<p>(4)(.6)(.4)^4=.061</p>
<p>This represents the fact that there is a .6 chance of losing, then (.4)^4 for the four games you win, and then multiply all that by 4 because the game you lose can land in any one of the first 4 positions.</p>