• Shuffle
    Toggle On
    Toggle Off
  • Alphabetize
    Toggle On
    Toggle Off
  • Front First
    Toggle On
    Toggle Off
  • Both Sides
    Toggle On
    Toggle Off
  • Read
    Toggle On
    Toggle Off
Reading...
Front

Card Range To Study

through

image

Play button

image

Play button

image

Progress

1/5

Click to flip

Use LEFT and RIGHT arrow keys to navigate between flashcards;

Use UP and DOWN arrow keys to flip the card;

H to show hint;

A reads text to speech;

5 Cards in this Set

  • Front
  • Back
A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the porbability that the sum of the three numbers on the balls selected from the box will be odd?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4
Since there are 50 odd and 50 even balls in the box, the probability of selecting an odd ball at random is 50/100=1/2; the probability of selecting an even ball is the same.
For the sum of the three numbers on the selected balls to be odd, either
1. all balls must be odd.
2. two balls are even and one is odd.

OOO
1/2*1/2*1/2=1/8
OEE
1/2*1/2*1/2=1/8
EOE
1/2*1/2*1/2=1/8
EEO
1/2*1/2*1/2=1/8

Adding all four probabilities give 4/8=1/2 as the probability that the sum of the three numbers will be odd.
A cmpany that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one of two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter)

A. 4
B. 5
C. 6
D. 12
E. 24
Since the problem asks for the minimum number of colors needed, start with the lowest answer choice available. Calculate each sucessive option until finding the minimum number of colors that can represent at least 12 distribution centers.

#of
colors by one by two total
number color colors
4 4 4!/2!2!=6 6+4=10
5 5 5!/2!3!=10 5+10=15

The correct answer is B
Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 1/2, and 5/8, respectively, what is th eprobability that Xavier and Yvonne, but not Zelda, will solve the problem?

A. 11/8
B. 7/8
C. 9/64
D. 5/64
E. 3/64
Since the individuals' probabilities are independent, they can be multiplied to figure out the combined probability. The probability of Xavier's sucess is given as 1/4, adn the probability of Yvonne's success is given as 1/2. Since the probability of Zelda's success is given as 5/8, the the probability of her NOT solving the problem is 1-5/8=3/8. Ths, the combined probability is 1/4*1/2*3/8=3/64.

E
There are 8 teams in a certain league and each team play each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

A. 15
B. 16
C. 28
D. 56
E. 64
Since no team needs to play itself, each team needs to play 7 other teams. In addition, each game needs to be counted only once, rather than once for each team that plays that game. Since two teams play each game, 8*7/2=28 games are needed.

C
A certain club has 10 members, including Harry. One of the 10 members is to be chosen at random to be the present, one of the remaining 9 members is to ebe chosen at random to be the secretary, and one of the remaining 8 members is be chosen at random to be the treasurer. What is the probability that Harry will be either the member chosen to be the secetary or the member chosen to be the treasurer?

A. 1/720
B. 1/80
C. 1/10
D. 1/9
E. 1/5
Possible outcomes

P S T TOTAL
9/10 1/9 - 1/10
9/10 8/9 1/8 1/10

1/10+1/10=1/5

E