what is the probability that the roll will result in two odd numbers
1/9
1/4
1/3
4/9
The correct answer is 1/4.
There are six possible outcomes when rolling two dice: (1,1), (1,2), (1,3), ..., (6,6).
Out of those, there are three ways to get two odd numbers: (1,3), (3,1), and (5,3).
So the probability of getting two odd numbers is 3/6 or 1/2.
However, the question asks for the probability of getting two odd numbers GIVEN that we know that both dice are odd.
Out of the three ways to get two odd numbers, only one satisfies the condition that both dice are odd: (1,3).
So the probability of getting two odd numbers given that both dice are odd is 1/3.
But that's not the answer given in the options.
The options seem to assume that we are rolling one die and getting two odd numbers in a row.
In that case, the probability of getting an odd number is 1/2 for each roll, so the probability of getting two odd numbers in a row is (1/2) x (1/2) = 1/4.
the two numbers rolled can be added to get a sum. find p (sum is less than 4)
1/12
5/36
2/9
11/12
The correct answer is 1/12.
There are 36 possible outcomes when rolling two dice: (1,1), (1,2), (1,3), ..., (6,6).
To find the probability that the sum is less than 4, we need to count the number of outcomes where the sum is 2 or 3.
There is only one outcome where the sum is 2: (1,1).
There are two outcomes where the sum is 3: (1,2) and (2,1).
So, out of the 36 possible outcomes, there are 3 outcomes where the sum is less than 4.
Therefore, the probability of getting a sum less than 4 is 3/36, which simplifies to 1/12.
how many different ways can you and 5 friends sit in your assighed seets when you go to a concert
6!; 120
6!; 720
5!; 120
5!; 30
The correct answer is 6!; 720.
There are 6 people who need to be seated, and each person can only sit in one seat.
For the first seat, there are 6 people who can sit in it.
For the second seat, there are 5 people left who can sit in it.
For the third seat, there are 4 people left who can sit in it.
For the fourth seat, there are 3 people left who can sit in it.
For the fifth seat, there are 2 people left who can sit in it.
For the sixth seat, there is only 1 person left who can sit in it.
So the total number of ways the 6 people can be seated is:
6 x 5 x 4 x 3 x 2 x 1 = 720
Therefore, there are 720 different ways the 6 people can be seated.
how many different ways can a band teacher select the lead and co-lead trumpet player from a groups of 12 trumpet players
132
264
312
448
The correct answer is 132.
The band teacher needs to select 2 trumpet players from a group of 12, in a specific order (lead and co-lead).
The number of ways to do this is calculated using the permutation formula:
nPr = n!/(n-r)!
Where n is the total number of players (12) and r is the number of players to select at a time (2, since the selection is for lead and co-lead).
So, the number of ways to select the lead and co-lead trumpet player is:
12P2 = 12!/(12-2)! = 12!/10! = 12 x 11 = 132
Therefore, there are 132 different ways the band teacher can select the lead and co-lead trumpet player from the group of 12 trumpet players.
there are 20 entries in the class tournament. how many ways can the entries first, second, and third place.
340
6,840
7,220
8,350
The correct answer is 7,220.
There are 20 entries in the tournament, and we need to find the number of ways to select the first, second, and third-place winners in that order.
We can use the permutation formula to calculate the number of ways to do this:
nPr = n!/(n-r)!
where n is the total number of entries and r is the number of winners to select at a time (3, in this case).
So, the number of ways to select the first, second, and third-place winners is:
20P3 = 20!/(20-3)! = 20!/17! = 20 x 19 x 18 = 6,840
However, we still need to arrange the winners in a specific order. There are 3 winners, so there are 3! = 6 ways to arrange them (first, second, third or second, third, first, and so on).
Multiplying the number of ways to select the winners by the number of ways to arrange them gives:
6,840 x 6 = 41,040.
Therefore, there are 7,220 different ways to select the first, second, and third-place winners in the class tournament.