Problem Solving - 25

Two canoe riders must be selected from each of two groups of campers. One group consists of three men and one woman, and the other group consists of two women and one man. What is the probability that two men and two women will be selected?

(A) 1/6
(B) 1/4
(C) 2/7
(D) 1/3
(E) 1/2


Answer: E

1st Group: 3 Men, 1 Woman
2nd Group: 1 Man, 2 Women

Total number of ways of selection: 4C2 * 3C2 = 18

No of ways of selecting two men and two women:

1. 2 men and 0 woman from 1st group and 0 men and 2 women from 2nd group = 3C2 = 3
2. 1 man and 1 woman from 1st group and 1 man and 1 woman from 2nd group = 3C1* 2C1 = 6 Thus Total ways from the above = 3 + 6 = 9

Therefore Probability = 9/18 = 1/2

Problem Solving - 24

On his drive to work, Bob listens to one of three radio stations, A,B, or C. He first turns to A. If A is playing a song he likes, he listens to it; if not, he turns to B. If B is playing a song he likes, he listens to it; if not, he turns to C. If C is playing a song he likes, he listens to it; if not, he turns off the radio. For each station , the probability is 0.30 that any given moment the station is playing a song Leo likes. On his drive to work, what is the probability that Bob will hear a song he likes?

A) 0.027
B) 0.090
C) 0.417
D) 0.657
E) 0.900

Answer: D

Probability Leo listens to A = (Probability that A is playing a song of Leo's choice ) = 0.3

Probability Leo listens to B = (Probability that A is not playing a song of Leo's choice ) * (Probability that B is playing a song of Leo's choice ) = (1-0.3) * 0.3 = 0.7*0.3

Probobility Leo listens to C = (Probability that A is not playing a song of Leo's choice ) * (Probability that B is not playing a song of Leo's choice ) * (Probability that C is playing a song of Leo's choice ) = 0.7 * 0.7 * 0.3

Total Probability = 0.3 + 0.7*0.3 + 0.7*0.7*0.3 = 0.657

Manhattan Challenge Problem of the week! - 05/28/07

Baseball's World Series matches 2 teams against each other in a best-of-seven series. The first team to win four games wins the series and no subsequent games are played. If you have no special information about either of the teams, what is the probability that the World Series will consist of fewer than 7 games?

(A) 12.5%
(B) 25%
(C) 31.25%
(D) 68.75%
(E) 75%

Answer: D

OE - In order to determine the probability that the World Series will last fewer than 7 games, we can first determine the probability that the World Series WILL last exactly 7 games and then subtract this value from 1.

In order for the World Series to last exactly 7 games, the first 6 games of the series must results in 3 wins and 3 losses for each team.

Let's analyze one way this could happen:

Game 1 -----Game 2 -----Game 3 -----Game 4 -----Game 5 -----Game 6
T1 Wins-----T1 Wins-----T1 Wins-----T1 Loses-----T1 Loses-----T1 Loses

There are many other ways this could happen. Using the permutation formula, there are 6!/(3!)(3!) = 20 ways for the two teams to split the first 6 games (3 wins for each).

There are then 2 possible outcomes to break the tie in Game 7. Thus, there are a total of 20 × 2 = 40 ways for the World Series to last the full 7 games.

The probability that any one of these 40 ways occurs can be calculated from the fact that the probability of a team winning a game equals the probability of a team losing a game = 1/2.

Given that 7 distinct events must happen in any 7 game series, and that each of these events has a probability of 1/2, the probability that any one particular 7 game series occurs is

(1/2) ^ 7 = 1/ 128

Since there are 40 possible different 7 game series, the probability that the World Series will last exactly 7 games is:

40 * 1/128 = 40/ 128 = .3125 = 31.255%

Thus the probability that the World Series will last fewer than 7 games is:

100% - 31.25% = 68.75%.

The correct answer is D.

DS Question - 18

Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

1) The probability that the ball will both be white and have an even number painted on it is 0.

2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2.

Answer: E

The question is asking for P(W) or P(E)
=>P(W) or P(E) = P(W) + P(E) - P(W and E)

From statement (1) --- it is given that P(W and E) = 0 --- insufficient as we do not know individual probabilities.
From statement (2) --- it is given that P(W) - P(E) = 0.2 --- insufficient

Combining both statement (1) and (2) it is still insufficient as P(W) + P(E) cannot be calculated....P(W U E) = P(W) + P(E) - P( W and E)

NOTE : P(A and B) = P(A) * P(B) --- This is true only iff both A & B are Independent. Otherwise, it would be -- P(A and B) = P (A) * P(B/A) = P(B) * P(A/B).
From the question we do not know that if both A and B are independent.

Problem Solving - 15

During a behavioral experiment in a psychology class, each student is asked to compute his or her lucky number by raising 7 to the power of the student's favorite day of the week (numbered 1 through 7 for Monday through Sunday respectively), multiplying the result by 3, and adding this to the doubled age of the student in years, rounded to the nearest year. If a class consists of 28 students, what is the probability that the median lucky number in the class will be a non-integer?

(A) 0%
(B) 10%
(C) 20%
(D) 30%
(E) 40%

Answer - A
Since any power of 7 is odd, the product of this power and 3 will always be odd. Adding this odd number to the doubled age of the student (an even number, since it is the product of 2 and some integer) will always yield an odd integer. Therefore, all lucky numbers in the class will be odd.

The results of the experiment will yield a set of 28 odd integers, whose median will be the average of the 14th and 15th greatest integers in the set. Since both of these integers will be odd, their sum will always be even and their average will always be an integer. Therefore, the probability that the median lucky number will be a non-integer is 0%.

Problem solving - 13

If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?

A) 25%
B) 50%
C) 62.5%
D) 72.5%
E) 75%

Answer - C
n is even - anytime n is even, it is divisible by 8
total nos using sequence theorm 96 = 2 + (#-1) 2,
hence # = 48
n is odd - again 48 no
but in 1-8, only one combination is divisible by 8, when n is 7,15, 22... hence 12 cases
probalility = possible outcomes / total outcomes = 48+12 / 96 = 60/96 = .625 = 62.5%

OR

we need to check for what values of n, n(n+1)(n+2) is divisible 8....
n(n+2) is divisible by 8 for all values of even numbers and there are 48 even nos in 1 to 96....
now the remaining part (n+1) is divisible by 8 for 12 odd numbers....such as 7, 15, 23, 31, 39.... to find this u can divide 96 by 8....
so totally 48 + 12 = 60 numbers are there between 1 to 96 for which n(n+1)(n+2) is divisible 8....
now when calculate the % it would be ( 60 / 96 ) * 100 = 62.5 %....

Problem solving - 11

A number is selected at random from first 30 natural numbers. What is the probability that the number is a multiple of either 3 or 13?

(A) 17/30
(B) 2/5
(C) 7/15
(D) 4/15
(E) 11/30

Answer -- B
The first 30 natural nos are 1,2,3.....28,29,30.
There are 10 multiples of 3 in the above range and there are 2 multiples of 13 in the same range.
Hence there are total 12 i.e (10+2) nos which can be either a multiple of 3 or 13.
Total numbers = 30.
So the probability is 12/30 = 2/5.

Manhattan Challenge Problem of the Week ! - dec 18

Chicago Rain

If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?

(A) 1/32

(B) 2/25

(C) 5/16

(D) 8/25

(E) 3/4

Answer: Correct ans is C. For OE click on the link below.
Manhattan Challenge Problem