In order to make the national tennis team, Matt

So assuming he plays Steve-Larry-Steve, his probability of going through is 1*(3/10)*1 = 3/10 --> shouldn't it be 1*(3/10)*1 + (7/10)*1*(3/10) = still 51/100 , just want to make sure it was missed and not that i am doing something wrong.

No. Assuming he plays Steve-Larry-Steve, his probability of going through is only 1*(3/10)*1 = 3/10
He must win two games in a row. That means he must win the game with Larry in any case.
Taking the best case scenario i.e. the case in which he will definitely roll over Steve i.e. his probability of winning over Steve is 1,
we get 1*(3/10)*1 (Win-Win-Win) - the only way in which he goes through if he plays Steve-Larry-Steve.

Think what happens if the probability of winning from Steve is less than 1, say 9/10.

His probability of going through will be (9/10)*(3/10) (Win-Win) + (1/10)*(3/10)*(9/10) (Lose-Win-Win)
= 270/1000 + 27/1000 = 297/1000 (less than 3/10)

Statement 1. - clearly not sufficient
Statement 2. - probability- Matt beats Larry = 3/10, lets says his chance of beating steve is 99%, i.e, 99/100, then
L-S-L = 3/10 x 99/100 x 3/10 = 891/10000 = no way, he cant make it -
S-L-S = 99/100 x 3/10 X 99/100 = 29403/100000 = oh no... ... sorry matt, try next year bro

Dear VeritasPrepKarishma, What is the methodology used to calculate the yellow highlight? I have no idea how you get there? Why whatever happens to third doesn't matter?

Matt beat Larry = 30%=0.3
Matt beat Steve = ?
Larry-Steve-Larry
(Win)(Win)(Lose)?
(Win)(Win)(Win)?

There are only two ways in which Matt can play the games:

Larry-Steve-Larry
Steve-Larry-Steve

According to data given in stmnt 2, we know that the probability that Matt beats Larry is 30%. We would like to find the maximum probability that Matt goes through to see if it is more than 51%. So in the best case scenario, Matt beats Steve with a probability of 1 i.e. definitely beats Steve. This will give us the maximum probability of Matt going through. If we find that it is not more than 51%, then we can be sure that whatever the actual probability of Matt beating Steve, the probability of Matt going through cannot be more than 51%.

If he plays Steve-Larry-Steve, his probability of going through is 1*(3/10)*1 = 3/10
He will win the first and third games and to go through, he must win the second one too since he needs to win two consecutive games (given in the question). So just winning 1st and 3rd are not enough.

If he plays Larry-Steve-Larry, his probability of going through is (3/10)*1*1 + (7/10)*1*(3/10) = 51/100
How do we get this? To win two consecutive games, he could win the first and second games or second and third games.
(3/10)*1*1 -> Wins first two games and goes through. Then the outcome of the third game is immaterial. Whether he wins or loses, he will still go through.
(7/10)*1*(3/10) -> Loses the first game with the probability of 7/10 since he plays it against Larry, but wins the other two games - one against Steve and the other against Larry.

Dear VeritasPrepKarishma, What if Matt loses to Larry on the third game, the probability will become (3/10)*1*(7/10)? Then, your probability in blue font still valid since you assuming that he wins the third game.

When I say the probability is (3/10)*1*1, it includes both Matt winning from Larry and Matt losing to Larry in the third game.

Winning + Losing
(3/10)*1*(3/10) + (3/10)*1*(7/10)
= (3/10)*1*(3/10 + 7/10)
= (3/10)*1*1

Basically, it doesn't matter what happens in the third game.

VeritasPrepKarishma

Do I always have the liberty to choose probability of 1 when Matt beats Steve??

I actually consider 40% & 50% probability of beating Steve (they both are better than 30%) and got two different result.Hence got the answer choic E.

No, not at all. The probability of Matt beating Steve could very well be 40%, 50%, 67%, or 100% (i.e. 1) etc.

But in this question we need to find if the probability of Matt winning is more than 51%. So let's try to find out the BEST POSSIBLE case for Matt. If that probability is more than 51%, then we know that we do not enough information to answer the question.

Say, we assume that probability of Matt beating Steve is 40%. Say, in that case, we get that probability of Matt making the team is 10% (just assumption).

Say, we assume that probability of Matt beating Steve is 50%. Say, in that case, we get that probability of Matt making the team is 15% (just assumption).

Say, we assume that probability of Matt beating Steve is 80%. Say, in that case, we get that probability of Matt making the team is 35% (just assumption).

Say, we assume that probability of Matt beating Steve is 100%. In that case, we get that probability of Matt making the team is 51% (we have calculated above).

Now, we see that 51% is the highest possible probability of Matt making the team. So can the probability of Matt making the team be more than 51%? No.

If, instead, at 100%, we would have got that the probability of Matt making the team is, say 55%, then we would have been unable to say whether the probability of Matt making the team is more than 51% or not. The answer would depend on the actual probability of Matt beating Steve.

But in our original question, no matter what the probability of Matt beating Steve, the probability of Matt making the team will NEVER be more than 51%.

Statement 1: Matt's probability of beating Steve are better than Matt's probability of beating Larry; is clearly not sufficient.

Statement 2: The probability that Matt beats Larry is 30%. Let us assume the best-case scenario.

Let us assume that the probability that Matt beats Steve is 100%. so the combined probability is (3/10*1*3/10 + 1*3/10*1) (Larry-Steve-Larry OR Steve-Larry-Steve) = 39%. Therefore, we can conclude that Matt's probability of making the team is NOT greater than 51%

Ans: B

B

1> P (beating s) > P(beating L) >> cool, but how much?? let, P beating S is 90%, therefore P losing to S is 10%. works for us. he is most probably winning the first game. Now, P of (beating L) could be anything from 0% to 89%. if it's 0% then Matt isn't winning the tournament- he needs to win 2 consec. games. But, if it's 89% then sure! he's going to win! Overall. Not suff.

2> P (beating Larry is 30%) -- this one's more straight forward. with this logic p (losing to L) is 70%. too high! he will eventually meet Larry after winning from Steve and there's a 70% chance he's going to lose against Larry. so yes, there you go. there's no way his chances of winning (the tournament- that needs 2 cosec. wins) are more than 51%

I know this is not methodical way to go about it, but for DS questions this way always help to visualise what's happening, without indulging into concepts.

To solve this problem realize that there are two cases 1) 2/3 P(of beating larry) * 1/3 P(of beating steve) =1 2) or vice versa

1) Is not sufficient because we don't know the probabilities of beating steve and larry. If the probability of beating larry was 1% and of beating steve 2%, the probability for matt of making in the national team would be 2/3(1/100) * 1 (2/100) < 51%. Viceversa same thing. If the probability instead were Beating steve=90% and beating larry =89% matt would have a probability higher than 51% to make it to the national team

2) This is sufficient because:

If the probability of beating larry is 30% the probability of not beating larry is 70%

Thus in the case Larry-steve-larry the probability of making to the champ = probab. of winning against larry *2/3 * prob of winning agains steve *1/3= 2/3(3/100) * 1/3 (X) =1 thus 20% * 1/3 (X) =1 where x is the probability of beating steve.

Now maximize the prob of winning against steve to 100% thus you can see that 20% of 33% is below 51%.

Thus the probability of making it to the national team will always be under 51%.

Mam, why we took probability of winning with steve as 1........how to deduce this ......it can be anything between 0 and 1

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