请教大家两道数学题，谢谢

songchangwen

Q7:

For a certain race, 3 teams were allowed to enter 3 members each.A team earned 6 – n points whenever one of its members finished in nth place, where 1 ≤ n ≤ 5.There were no ties, disqualifications, orwithdrawals.If no team earned more than 6 points, what is the least possible score a team could have earned?

A.0

B.1

C.2

D.3

E.4

十四

Q2:

If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p >1/2
?
(1)More than1/2
of the 10 employees are women.
(2)The probability that both representatives selected will be men is less than1/10
.


A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D. EACH statement ALONE is sufficient.

E. Statements (1) and (2) TOGETHER are NOT sufficient.

sdcar2010

For a certain race, 3 teams were allowed to enter 3 members each.A team earned 6 – n points whenever one of its members finished in nth place, where 1 ≤ n ≤ 5.There were no ties, disqualifications, orwithdrawals.If no team earned more than 6 points, what is the least possible score a team could have earned?

A.0

B.1

C.2

D.3

E.4


- - - - - - -
The only points each team member can get are 5, 4, 3, 2, 1. If no team earned more than 6 points, then let's team A gets 6 (5 and 1) and team B gets 6 (4 and 2), what is left is 3 points only for team C.  Hence, the least possible score a team could have earned is 3, if no team got more than 6 points.

sdcar2010

If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p >1/2

(1)More than1/2 of the 10 employees are women.
(2)The probability that both representatives selected will be men is less than1/10

(1) More than half of the 10 emplyees are women.  Okay, let's say 6 employees are women.  For the first pick, p(w1) = 6/10.  For the second pick, p(w2) = 5/9.  Then the probability that both representatives selected will be women = p(w1) * p(w2) = 1/3.  Not sufficient.

(2) The probability that both representatives selected will be men is less than1/10 .
Let's say there are x number of men among the ten candidates.  Using the same logic above, the probability that both representatives selected will be men = p(m1) * p(m2) = (x/10) * ((x-1)/9)
Then as the above equation is less than 1/10: (x/10) * ((x-1)/9) < 1/10, which is to say that x * (x-1) < 9.  The biggest number that fulfill this requirement is x = 3.  So at least there are 7 women among the ten candidates. Then you can calculate the probability that both representatives selected will be women = p(w1) * p(w2) =  (7/10) * (6/9) = 21/45.  Not sufficient, either.