A bag holds tiles with the letters A, E, I, O, and U. Which list of letter combinations is missing from the table that shows the possible outcomes for randomly choosing two letters out of the bag without replacement?
AE AI AO AU
EA EI EO EU
IA IE IO IU
OA OE OI OU
A. AA, EE, II, OO
B. UA, UE, UI, UO
C. UU, UA, UE, UI
D. IE, UO, UA, AI
A. AA, EE, II, OO
The letters for the word math are placed in a bag. Two letters are selected with replacement. Which statements are true about the tree diagram that could represent the possible outcomes? Select all that apply.
A. There would be 4 tree bases.
B. Each base would have 3 tree branches.
C. The diagram would show a total of 12 possible outcomes.
D. There would be 4 combinations that have double letters.
A. There would be 4 tree bases.
B. Each base would have 4 tree branches.
C. The diagram would show a total of 16 possible outcomes.
D. There would be 2 combinations that have double letters (MM and TT).
(Note: With replacement means that the letter is put back into the bag before the second letter is drawn, so it is possible to get the same letter twice in a row.)
huh
With replacement means that the letter is put back into the bag before the second letter is drawn. Here's how we can find the total number of outcomes and the number of outcomes with double letters:
- For the first letter, there are 4 choices: M, A, T, or H.
- For the second letter, there are again 4 choices, since we are putting the first letter back into the bag.
- The total number of outcomes is the number of ways we can choose two letters from 4 choices, with replacement. This is equal to 4 x 4 = 16.
- To find the number of outcomes with double letters, we can count them directly: they are MM, AA, TT, and HH. There are 2 of them (MM and TT).
So, the statements that are true about the tree diagram are:
A. There would be 4 tree bases.
B. Each base would have 4 tree branches.
C. The diagram would show a total of 16 possible outcomes.
D. There would be 2 combinations that have double letters (MM and TT).
A spinner has three equal sections labeled 1, 2, and 3.
Which organized list shows the possible outcomes for spinning the spinner twice?
A.
1,1; 1,2; 1,3
2,1; 2,2; 2,3
3,1; 3,2; 3,3
B.
1,2; 1,3
2,1; 2,3
3,1; 3,2
C.
1,1; 1,2; 1,3
2,2; 2,3
3,3
D.
1,1; 1,2; 1,3
A. 1,1; 1,2; 1,3
2,1; 2,2; 2,3
3,1; 3,2; 3,3
(Spinning the spinner twice means that we are choosing one section, recording the result, and then choosing again from the same sections.)
So, we can get the following outcomes:
1st spin 2nd spin
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
These outcomes correspond to the list in option A.
A spinner has three equal sections labeled 1, 2, and 3.
What is the probability of getting two numbers that have a sum of 5 on the spinner?
A. two-ninths
B. one-third
C. one-sixth
D. four-ninths
We can find the probability of getting two numbers that have a sum of 5 by first listing all the possible outcomes when spinning the spinner twice, as we did in the previous question:
1st spin 2nd spin
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
There are two outcomes that have a sum of 5: 2+3 and 3+2. So, the probability of getting two numbers that have a sum of 5 is:
P(sum is 5) = number of outcomes with sum of 5 / total number of outcomes
= 2 / 9
Therefore, the answer is A. two-ninths.
A bag contains 4 white, 3 blue, and 5 red marbles.
Find the probability of choosing 3 blue marbles in a row if the marbles are replaced.
A. two-elevenths
B. one over two hundred twenty
C. Fraction 1 over 27 end fraction
D. one over sixty-four
The probability of choosing a blue marble on one draw is 3/12, since there are 3 blue marbles out of 12 total marbles. Since the marbles are replaced, the probability of drawing a blue marble on one draw is the same for each draw and the draws are independent. Therefore, the probability of choosing 3 blue marbles in a row is:
(3/12) x (3/12) x (3/12) = (27/1728) = 1/64
Therefore, the answer is D. one over sixty-four.
A bag contains 4 white, 3 blue, and 5 red marbles.
Find the probability of choosing a blue marble, then a red marble if the marbles are not replaced.
A. five over forty-four
B. fifteen over thirty-five
C. two-thirds
The probability of choosing a blue marble on the first draw is 3/12, since there are 3 blue marbles out of 12 total marbles. If a blue marble has been selected, there will be 11 marbles left, including 5 red marbles. Therefore, the probability of choosing a red marble on the second draw given that a blue marble was selected on the first draw is 5/11. The probability of both events occurring is the product of their probabilities:
(3/12) x (5/11) = 15/132 = 5/44
Therefore, the answer is A. five over forty-four.
A bag contains 4 white, 3 blue, and 5 red marbles.
Find the probability of choosing 2 white marbles in a row if the marbles are not replaced.
A. The fraction states 1 over 11.
B. one-ninth
C. two-thirds
D. Start Fraction 1 over 16 End Fraction
The probability of choosing a white marble on the first draw is 4/12, since there are 4 white marbles out of 12 total marbles. If a white marble has been selected, there will be 11 marbles left, including 3 white marbles. Therefore, the probability of choosing another white marble on the second draw given that a white marble was selected on the first draw is 3/11. The probability of both events occurring is the product of their probabilities:
(4/12) x (3/11) = 1/11
Therefore, the answer is A. The fraction states 1 over 11.