Data Sufficiency

In order to make the national tennis team, Matt has to play a three-game series against Larry and Steve, and in doing so win two games in a row. He's given a choice, however: he can choose the order in which he plays against his opponents but cannot play the same opponent in consecutive games (so he could play Larry-Steve-Larry OR Steve-Larry-Steve). Assuming that Matt chooses the three-game sequence that maximizes his probability of making the national team, is his probability of making the team greater than 51%?

(1) Matt's probability of beating Steve are better than Matt's probability of beating Larry

(2) The probability that Matt beats Larry is 30%.

Let's start with the order of opponents.

Since I have to win two games in a row, I MUST win the middle game: if I don't, it's impossible for me to make the tennis team. Therefore, I need to play the INFERIOR opponent in the middle game: this strategy gives me not only the best chance of winning the essential game, but also TWO chances to beat the more difficult opponent. If this makes intuitive sense, you can skip the next portion, but if it doesn't, read on to see algebraically why this is true.

Suppose that my chances of beating the superior player are x, that my chances of beating the inferior player are y, and that y > x.

* Win first game, win second game, lose third game
* Lose first game, win second game, win third game
* Win first game, win second game, win third game

If I play the inferior player twice, my probabilities for each of these scenarios are:

y * x * (1 - y)
(1 - y) * x * y
y * x * y

So my chances of making the team are xy(1-y) + (1-y)xy + xy², or xy² + 2*(1-y)*xy.

If I play the superior player twice, my probabilities for each of these scenarios are:

x * y * (1 - x)
(1 - x) * y * x
x * y * x

So my chances of making the team are xy(1-x) + (1-x)xy + x²y, or x²y + 2*(1-x)*xy

Now let's compare the two probabilities:

xy² + 2*(1-y)*xy VS x²y + 2*(1-x)*xy

Dividing both sides by xy, we have

y + 2*(1-y) VS x + 2*(1-x)

or

2 - y VS 2 - x

Since y > x, my probability of winning is higher on the RIGHT HAND SIDE. Since this probability represents the scenario in which I play the superior player twice, that should be my strategy.

Now let's move to the statements!

S1 tells me the order in which I should play my opponents: I should play Larry first, then Steve, then Larry. This doesn't give me the chances of winning each match, however, so I don't know if my chances of making the team are greater than 51%. INSUFFICIENT!

S2 tells me that my chances of beating Larry are 30%. This seems insufficient, but let's ponder it for a bit. The question wants to know my MINIMUM probability of making the team, specifically whether that probability is at least 51%. So let's look at the BEST CASE scenario here, and see if that gets us to 51%.

My best case scenario is that Steve is a total schmuck and my chances of beating him are 100%. Let's assume that's true. If my chances of beating Steve are 100%, I'll play him in the middle game, guaranteeing me a win. Now I have TWO chances to beat Larry, and I only need to win one game. Hence my probability of making the team is my probability of beating Larry AT LEAST ONCE.

To beat Larry at least once, I only need not to lose both games. I lose both games 7/10 * 7/10, or 49% of the time. I therefore beat Larry at least once in the other 51% of games.

So even in the BEST CASE SCENARIO -- where my chances of beating Larry are 30% and my chances of beating Steve are 100% -- my probability of making the team is NOT greater than 51%.

Hence S2 is SUFFICIENT, and the answer is B.