Asked by Dontworryabtit

a coin is tossed, and a started number cube is rolled. What is the probability that the coin shows heads and the number cube shows an even number
1/6
1
1/4
1/2
Answered by Dontworryabtit
a bag contains 3 blue marbles, 9 green marbles, and 11 yellow marbles. Twice you draw a marble and replace it. Find P(blue, then green.)
27/529
27/23
15/529
12/23
Answered by Dontworryabtit
the diagram shwos the contents of a jar of marbles. You select two marbles at random. One marble is drawn and not replaced. Then a secnd marble is drawn. What is the probability of select a red marbe and then another red marble?
3/64
1/4
1/20
1/30
Answered by Dontworryabtit
is the answer
3/64
1/4
1/20
1/30
Answered by Dontworryabtit
from a committee of 6 girls and 4 boys, a name is selected. Then another name is selected. What is the probablilty that both people drawn will be girls?
Answered by Dontworryabtit
write the number f permutations in factorial form. then simplify J K L M N O P
7!;720
8!:40,320
7!;49
7!;5,040
Answered by Dontworryabtit
how many 2 digit numbers can be formed using only the digits 2,3,5, and 6, if the digits are not be reapeated within a number?
11
12
10
2
Answered by Dontworryabtit
A juice cmpany decides to test five different brand of juice. The different brands have been labeled A, B, C, D, and E. The company decides to compare each brand with the other brands by pairing together different brands. How many different pairs will result by selecting two different brands at a time.
11
15
120
10
Answered by Dontworryabtit
There are candidates running for 3 seats on a committee. How many different election results are possible.
36
84
504
56
Answered by Dontworryabtit
There are 9 candidates running for 3 seats on a committee. How many different election results are possible.
36
84
504
56
Answered by Dontworryabtit
how many different ways there to rearrange the letters of the word EXAM, if you dont care if the result is a reconizable word
4
12
24
256
Answered by Bot
1/4
Answered by Bot
The probability of drawing a blue marble on any one draw is 3/23, because there are 3 blue marbles out of a total of 23 marbles. Similarly, the probability of drawing a green marble on any one draw is 9/23. Because the marbles are replaced after each draw, the probability of drawing a blue marble followed by a green marble is independent and can be found by multiplying the individual probabilities:

P(blue, then green) = P(blue) x P(green) = (3/23) x (9/23) = 27/529

Therefore, the answer is 27/529.
Answered by Bot
There are a total of 25 marbles in the jar. The probability of selecting a red marble on the first draw is 8/25, because there are 8 red marbles out of a total of 25 marbles. After the first red marble is drawn and not replaced, there are only 24 marbles left in the jar and 7 red marbles. Therefore, the probability of selecting another red marble on the second draw is 7/24.

To find the probability of selecting a red marble and then another red marble, we need to multiply the probabilities of the individual events:

P(red, then red) = P(red on first draw) x P(red on second draw | red on first draw)
= (8/25) x (7/24)
= 7/75

Therefore, the answer is 7/75.
Answered by Bot
No, the answer is not any of those options. The correct answer is 7/75 as shown in my previous response.
Answered by Bot
The probability of selecting a girl on the first draw is 6/10, because there are 6 girls out of a total of 10 people. After one person is selected and removed from the committee, there are only 9 people left, including 5 girls. Therefore, the probability of selecting another girl on the second draw, given that the first person selected was a girl, is 5/9.

To find the probability of both people drawn being girls, we need to multiply the probabilities of the individual events:

P(both girls) = P(first girl) x P(second girl | first girl)
= (6/10) x (5/9)
= 1/3

Therefore, the probability that both people drawn will be girls is 1/3 or approximately 0.3333 (rounded to four decimal places).
Answered by Bot
7!; 5,040 can be simplified as follows:

7! = 7 x 6 x 5 x 4 x 3 x 2 x 1
= 7 x 6 x 5 x 4 x 3 x 2
= 5,040

Therefore, the number of permutations is 5,040 and can be written as 7!.
Answered by Bot
There are four possible choices for the first digit (2, 3, 5, or 6), and three possible choices for the second digit (any digit except for the one already chosen for the first digit). Therefore, the total number of 2-digit numbers that can be formed using only the digits 2, 3, 5, and 6 without repetition is:

4 x 3 = 12

Therefore, the answer is 12.
Answered by Bot
To determine the number of pairs that can be formed when selecting two brands at a time from a set of five, we can use the formula for combinations:

nC2 = n!/[(n-2)! x 2!]

where n is the number of items in the set.

Substituting n = 5, we get:

5C2 = 5!/[(5-2)! x 2!] = (5 x 4)/(2 x 1) = 10

Therefore, there are 10 different pairs that can be formed by selecting two different brands at a time.

The answer is 10.
Answered by Bot
To determine the number of different election results possible for 3 seats on a committee, we can use the formula for permutations:

nPr = n!/(n-r)!

where n is the total number of candidates and r is the number of seats available.

Substituting n = the total number of candidates and r = 3, we get:

nP3 = n!/(n-3)!

For example, if there are 7 candidates running, then the number of different election results possible would be:

nP3 = 7!/(7-3)! = 7!/4! = 7 x 6 x 5 = 210

Therefore, we need to know the total number of candidates to determine the number of different election results possible.

The answer cannot be determined based on the information given.
Answered by Bot
To determine the number of different election results possible for 3 seats on a committee with 9 candidates, we can use the formula for permutations:

nP3 = n!/(n-3)!

Substituting n = 9, we get:

nP3 = 9!/(9-3)! = 9!/6! = 9 x 8 x 7 = 504

Therefore, there are 504 different election results possible.

The answer is 504.
Answered by Bot
The word EXAM has 4 letters. To find the number of different ways to rearrange the letters, we can use the formula for permutations of n objects taken r at a time:

nPn = n!

Substituting n = 4, we get:

4P4 = 4! = 4 x 3 x 2 x 1 = 24

Therefore, there are 24 different ways to rearrange the letters of the word EXAM, regardless of whether the resulting arrangement is a recognizable word or not.

The answer is 24.