Quant - Probability + Combination + Permutation

If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4


i. if n is multiple of 2 then n+2 will be such that n* (n+2) will always be divisible by 8. Total such cases = number of even nos = 96/2 = 48
ii. In addition to i above, if n+1 is divisible by 8 then it will also suffice the requirement. Total such cases = 96/8 = 12

So, probability = (48+12)/96 = 5/8


2
Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob?

A. 24
B. 30
C. 60
D. 90
E. 120


i. (4+2+3+1)* 3! = 60
ii. There are 5! possible ways for the 5 participants to finish the race without any ties.
Logically, Meg will finish ahead of Bob in exactly half of these outcomes

5!/2 = 60

3.
A set consist of 2n-1 element. What is the number of subsets of this set which contain at most n-1 elements?

A. 2^(2n-2)
B. 2^(2n) - 2
C. 2^(2n) -1
D. 2^(2n)
E. 2^(2n-1)


for example, let n=2 then: 2n-1=3, n-1=1

the number of subsets: 3C1+1 (for empty subset) = 4

A. 2^(2n-2) ==> 4
B. 2^(2n) - 2 ==> 14
C. 2^(2n) -1 ==> 15
D. 2^(2n) ==> 16
E. 2^(2n-1) ==> 8

let n=3 then 2n-1=5, n-1=2

the number of subsets: 5C2+5C1+1 (for empty subset) = 10+5+1=16

A. 2^(2n-2) ==> 16
B. 2^(2n) - 2 ==> 62
C. 2^(2n) -1 ==> 63
D. 2^(2n) ==> 64
E. 2^(2n-1) ==> 32

4.
A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

E.

N=8C2*6C2*4C2*2C2/4!=105

4!=4P4 - we should exclude arrangement of 4 pairs in different order.

5.

Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

2r 3g 3b

a total of 8 balls.

[2/8*(3/8)*(3/8)*(3/8)*(3/8)*] *[5!/(2!*2!)] (we have 5!/2!2! b/c we have 2 green picks and 2 blue picks and 1 red pick but we don't need to worry about that one--> 5*4*3*2/2*2=30) --> 30(2*3^4/8^5) 81*2*30 -->
162*30= 4860/32768 -->~14%

6. Standard Deviation + Probability
Three fair coins are labeled with a zero (0) on one side and a one (1) on the other side. Jimmy flips all three coins at once and computes the sum of the numbers displayed. He does this over 1000 times, writing down the sums in a long list. What is the expected standard deviation of the sums on this list?
A) 1/2
B)3/4
C)sqrt3/2
D)sqrt5/2
E) 5/4

Approach 1:
variance of binomial distribution is n*p*(1-p), n = 3, p = 1/2

variance = 3*1/2*(1-1/2)= 3/4

stdev = sqrt(variance) = sqrt(3)/2

After many trials sample standard deviation is close to theoretical standard deviation = sqrt(3)/2.

C is the answer


Approach 2:
http://www.gmatclub.com/forum/7-t58616


7.
How many subordinates does Marcia Have?

1) there are between 200 and 500 lists she could make consisting of the names of at least 2 of her subordinates.
2) there are 28 ways that she could decide which 2 subordinates she will recommend promoting.

D

1. nc0 + nc1 + nc2 + nc3 + ..... + ncn = 2^n
Hence, nc2 + nc3 + ..... + ncn = 2^n - n - 1
If RHS of above equation is between 200 and 500 => n = 8 as 2^8 = 256 while 2^9 = 512

So. sufficient

2. nC2 describes this situation.

nC2=n(n-1)/2=28 ==> n(n-1)=56 ==> n=8


8.