Jennifer writes the letters M-O-N-T-A-N-A on cards and then places the cards in a hat. What is the probability of picking an M?
There is a total of 7 letters in "MONTANA." Since only one letter is an M, the probability of picking an M is 1/7.
Jamal writes the letters K-A-N-S-A-S on cards and then places the cards in a hat. What is the probability of picking an S?
There is a total of 7 letters in "KANSAS." Since there are two letters that are S, the probability of picking an S is 2/7.
Jamal writes the letters K-A-N-S-A-S on cards and then places the cards in a hat. What is the probability of picking a vowel?
Out of the 7 letters in "KANSAS," there are two vowels: A and the second A. Therefore, there is a total of 2 vowels.
The probability of picking a vowel is then:
2 (number of vowels) / 7 (total number of letters) = 2/7
5. You and 4 friends go to a concert. In how many different ways can you sit in the assigned
seats?
20
24
48
120
Assuming that each seat is distinct (i.e. one seat is not the same as another), the number of ways you and your four friends can sit in the assigned seats is equal to the number of permutations of 5 objects, which is given by:
5! = 5 x 4 x 3 x 2 x 1 = 120
Therefore, the answer is 120.
14. How many ways can 3 students be chosen from a class of 20 to represent their class at a banquet?
6,840
3,420
1,140
2,280
The number of ways to choose 3 students from a class of 20 is given by the combination formula:
C(20, 3) = 20! / (3! * (20 - 3)!) = 20 x 19 x 18 / (3 x 2 x 1) = 1,140
Therefore, the answer is 1,140.
Simplify ¹⁰P⁴ (1 point)
210
360
5,040
151,200,
The formula for permutations is given by:
nPr = n! / (n - r)!
Using this formula with n = 10 and r = 4, we get:
10P4 = 10! / (10 - 4)! = (10 x 9 x 8 x 7) / (4 x 3 x 2 x 1) = 5,040
Therefore, the answer is 5,040.
The probability of a certain hockey player making a goal after hitting a slap shot is 1/5 How many successful slap shots would you expect her to make after 120 attempts?
5
20
24
60
The expected number of successful slap shots can be calculated by multiplying the probability of success (1/5) by the total number of attempts (120):
Expected number of successful slap shots = (1/5) x 120 = 24
Therefore, you would expect her to make 24 successful slap shots after 120 attempts.
The answer is option C (24).
A basket contains the following pieces of fruit: 3 apples, 2 oranges, 2 bananas, 2 pears, and 5 peaches. Jameson picks a fruit at random and does not replace it. Then Brittany picks a fruit at random. What is the probability that Jameson gets a banana and Brittany gets a pear?
The probability of Jameson selecting a banana is 2/14 (there are 2 bananas out of a total of 14 fruits), and the probability of Brittany selecting a pear after Jameson selects a fruit and does not replace it is 2/13 (there are now only 2 pears remaining out of a total of 13 fruits).
The probability of both events happening together (Jameson selecting a banana and Brittany selecting a pear) is the product of the individual probabilities:
Probability = (2/14) x (2/13) = 4/182 = 2/91
Therefore, the probability that Jameson gets a banana and Brittany gets a pear is 2/91.
You have five $1 bills, four $5 bills, six $10 bills, and three $20 bills in your wallet. You select a bill at random. Without replacing the bill, you choose a second bill. What is P($1, then $10)?
The probability of selecting a $1 bill on the first selection is 5/18 (there are 5 $1 bills out of a total of 18 bills).
Without replacement, the probability of selecting a $10 bill on the second selection is 6/17 (there are now 6 $10 bills out of a total of 17 bills remaining).
The probability of both events happening together (selecting a $1 bill then a $10 bill) is the product of the individual probabilities:
Probability = (5/18) x (6/17) = 5/51
Therefore, the probability of selecting a $1 bill on the first selection then a $10 bill on the second selection is 5/51.
A bag contains 5 green marbles, 8 red marbles, 11 orange marbles, 7 brown marbles, and 12 blue marbles. You choose a marble, replace it, and choose again. What is P (red, then blue)?
The probability of selecting a red marble on the first selection is 8/43 (there are 8 red marbles out of a total of 43 marbles).
After replacing the marble, the probability of selecting a blue marble on the second selection is 12/43 (there are now 12 blue marbles out of a total of 43 marbles).
The probability of both events happening together (selecting a red marble then a blue marble) is the product of the individual probabilities:
Probability = (8/43) x (12/43) = 96/1849
Therefore, the probability of selecting a red marble on the first selection then a blue marble on the second selection is 96/1849, which is approximately 0.052.
A bag contains 6 green marbles and 5 white marbles. You select a marble at random. What are the odds. in favor of picking a green marble?
The odds in favor of an event are defined as the ratio of the probability of the event occurring to the probability of the event not occurring.
In this case, the probability of picking a green marble is 6/11 (since there are 6 green marbles out of a total of 11 marbles), and the probability of not picking a green marble (i.e. picking a white marble) is 5/11.
Therefore, the odds in favor of picking a green marble are:
(6/11) / (5/11) = 6/5
Hence, the odds in favor of picking a green marble are 6 to 5.
A spinner is divided into 10 equal sections numbered from 1 to 10. You spin the spinner once, What is P (divisible by 2)?
Jamal writes the letters K-A-N-S-A-S on cards and then places the cards in a hat. What are the odds against picking an S?
The probability of picking an S can be calculated by counting the number of S's and dividing by the total number of letters:
Number of S's: 2
Total number of letters: 7
Probability of picking an S: 2/7
The odds against an event are defined as the ratio of the probability of the event not occurring to the probability of the event occurring:
Probability of not picking an S: 5/7
Odds against picking an S: (5/7) / (2/7) = 5/2
Therefore, the odds against picking an S are 5 to 2.
Jamal writes the letters K-A-N-S-A-S on cards and then places the cards in a hat. What are the odds against picking an S?
1:2
2:1
2:6
6:2
The probability of picking an S can be calculated by counting the number of S's and dividing by the total number of letters:
Number of S's: 2
Total number of letters: 7
Probability of picking an S: 2/7
The odds against an event are defined as the ratio of the probability of the event not occurring to the probability of the event occurring:
Probability of not picking an S: 5/7
Odds against picking an S: 5/2
Therefore, the odds against picking an S are 5 to 2. The answer is 5:2
Jamal writes the letters K-A-N-S-A-S on cards and then places the cards in a hat. What are the odds in favor of picking an S?
A. 1:2
B. 4:2
C. 2:6
D. 6:2
The probability of picking an S can be calculated by counting the number of S's and dividing by the total number of letters:
Number of S's: 2
Total number of letters: 7
Probability of picking an S: 2/7
The odds in favor of an event are defined as the ratio of the probability of the event occurring to the probability of the event not occurring:
Probability of picking an S: 2/7
Probability of not picking an S: 5/7
Odds in favor of picking an S: (2/7) / (5/7) = 2/5
Therefore, the odds in favor of picking an S are 2 to 5. The answer is A. 2:5.