Lottery DS

[rahulvsd]

A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?

1) Players must match at least two numbers to win.

(2) X = 4

[spoiler]OA:C[/spoiler]

[pemdas]

edit: 10! is right
criteria for winning is described in statement(1), possible outcomes =10!
st(1) possible two number sets = C(10,2)=10!/8!*2! Our probability is one winning set (note: complex calc. is required if we are bound to NOT distinct digits). P=1/(10!/8!*2!) We also need to find probability for AT LEAST condition meaning #units in set >=2.
C(10,3)=
C(10,4)
C(10,5)...
It's clear without solving that we will succeed in numbers and find Probability of all 1/36 + 1/C(10,3) + 1/C(10,4) ..., But we won't be able to define X and limit our result to it, therefore Not Sufficient

st(2) no winning condition is given Not Sufficient;

Combined st(1&2): must be Sufficient, as we are provided with X and probability condition for winning.
c

A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?

1) Players must match at least two numbers to win.

(2) X = 4

[spoiler]OA:C[/spoiler]

[user123321]

A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?

1) Players must match at least two numbers to win.

(2) X = 4

[spoiler]OA:C[/spoiler]

1) let x = 2
then probability of winning the lottery = 1/10C2 = 1/45
let x = 3
then probability of winning the lottery = (1+8)/10C3 = 3/40

so with change of x probability is changing. hence insufficient.

2) with just knowing x = 4, we cannot get the probability because we need a condition for obtaining the probability.

combining both we get probability = (1+7+8c2)/10C4

hence sufficient => C

user123321

[pemdas]

@user123321, thanks for posting, i missed somewhere to include 0 and C=10! above
applicable to st(1) and st(2) with C(10,n) ...

[saketk]

A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?

1) Players must match at least two numbers to win.

(2) X = 4

[spoiler]OA:C[/spoiler]

Can you mention the source as well?

This question looks incomplete to me. How many times a player can play this game?

question says - X numbers are selected all at once from a lot containing distinct numbers from 0-9.

Since the digits are distinct then it will never be possible for the player to pick 2 matching number UNLESS he plays multiple times. Even if I pick all numbers at once then also I will have all distinct numbers 0,1,2,3..9.

This is not a right question IMO.

Again, please mention the source.

[GmatMathPro]

There is some ambiguity in the question about how exactly the lottery mechanism works. I assume the intent is that it works like a typical lottery system. That is, the player picks X numbers and the lottery people pick X numbers and then prizes are awarded based on the extent to which the player's numbers match the lottery numbers.

Statement 1) Players must match at least two numbers to win. It should be intuitively clear that this is not sufficient by looking at extremes. If X=2, matching 2 numbers will be fairly unlikely. if X=6, the chance of matching at least two numbers is 100% because the lottery people would only have a pool of 4 numbers that could come up that wouldn't match any of the player's numbers, so at least 2 of them MUST match with the player's numbers.

Statement 2) This is insufficient because we don't know how many numbers a player has to match to win. Although, I don't think this is very good form on the part of the question writers. i think they should introduce the fact that the player can still win even if he doesn't match ALL the numbers in the question stem, rather than in statement 1.

Statements 1&2. This tells us they are drawing 4 numbers and the player must match at least two of them to win. It should be obvious that the probability of this is calculable and unique. SUFFICIENT. Just for fun, though:

Whatever four numbers the player chooses.....

There are 6 numbers that don't match any of the player's numbers, so there are 6C4=15 ways of not matching any of his numbers. Or they could match exactly 1 of his numbers in 4*6C3=80 ways. (4 choices for which number of the player's to match, and 20 choices for which 3 non-matching numbers to choose). There are 10C4=210 possible drawings, so the player will fail to win on 80+15=95 of the 210 possible drawings. 95/210=19/42, so he will win with probability 1-19/42 = 23/42.

OR we can calculate it directly....

When the numbers are drawn, there is 4C4=1 way to get an exact match.

To match exactly 3, there are 4C3=4 ways to choose 3 of the player's numbers to match, and 6 ways to pick a non-matching number. So 6*4=24 ways to match exactly 3 numbers.

To match exactly 2, there are 4C2=6 ways to choose 2 of the player's numbers to match, and 6C2=15 ways to choose the 2 remaining non-matching numbers. So 6*15=90 ways to match exactly 2 numbers.

(1+24+90)/10C4 = 23/42 chance of winning

[rahulvsd]

Source is Grockit. Apologies for not mentioning it with the question.