. The spinner is divided into equal parts. Use a tree diagram to find the probability that the spinner will land on a consonant the first time and a vowel the second time if the spinner is spun twice.
3 answers
I'm sorry, I cannot create visual content as I am a text-based program.
The spinner is divided into equal parts. Use a tree diagram to find the probability that the spinner will land on a consonant the first time and a vowel the second time if the spinner is spun twice.
A circle is divided equally into three sections.· One of the sections is labeled with an upper E.
· One of the sections is labeled with an upper U.
· One of the sections is labeled with an upper N.
· An arrow originating from the center of the circle is pointing at the section labeled with upper U. (1 point)
two-thirds
two-ninths
three-fourths
start fraction 5 over 9 end fraction
A circle is divided equally into three sections.· One of the sections is labeled with an upper E.
· One of the sections is labeled with an upper U.
· One of the sections is labeled with an upper N.
· An arrow originating from the center of the circle is pointing at the section labeled with upper U. (1 point)
two-thirds
two-ninths
three-fourths
start fraction 5 over 9 end fraction
Here is the tree diagram for the given scenario:
```
C V
/ \ / \
E U E U
/|\ | /|\ |
C V C C C V C V
```
Starting from the top, the first branch represents the first spin and the second branch represents the second spin. The letters C and V represent consonants and vowels, respectively.
From the diagram, we can see that there are four possible outcomes for the two spins that have a consonant on the first spin and a vowel on the second spin: EC, UC, VC, and EV. So, the probability of this occurring is:
P(consonant first, vowel second) = P(EC) + P(UC) + P(VC) + P(EV)
P(EC) = 1/3 * 2/3 = 2/9 (probability of landing on a consonant on the first spin is 2/3, and on the second spin is 1/3)
P(UC) = 1/3 * 1/3 = 1/9 (probability of landing on a vowel on the first spin is 1/3, and on the second spin is 1/3)
P(VC) = 1/3 * 1/3 = 1/9 (probability of landing on a vowel on the first spin is 1/3, and on the second spin is 1/3)
P(EV) = 1/3 * 1/3 = 1/9 (probability of landing on a consonant on the first spin is 2/3, and on the second spin is 1/3)
Therefore,
P(consonant first, vowel second) = 2/9 + 1/9 + 1/9 + 1/9 = 5/9
Thus, the probability that the spinner will land on a consonant the first time and a vowel the second time is 5/9.
```
C V
/ \ / \
E U E U
/|\ | /|\ |
C V C C C V C V
```
Starting from the top, the first branch represents the first spin and the second branch represents the second spin. The letters C and V represent consonants and vowels, respectively.
From the diagram, we can see that there are four possible outcomes for the two spins that have a consonant on the first spin and a vowel on the second spin: EC, UC, VC, and EV. So, the probability of this occurring is:
P(consonant first, vowel second) = P(EC) + P(UC) + P(VC) + P(EV)
P(EC) = 1/3 * 2/3 = 2/9 (probability of landing on a consonant on the first spin is 2/3, and on the second spin is 1/3)
P(UC) = 1/3 * 1/3 = 1/9 (probability of landing on a vowel on the first spin is 1/3, and on the second spin is 1/3)
P(VC) = 1/3 * 1/3 = 1/9 (probability of landing on a vowel on the first spin is 1/3, and on the second spin is 1/3)
P(EV) = 1/3 * 1/3 = 1/9 (probability of landing on a consonant on the first spin is 2/3, and on the second spin is 1/3)
Therefore,
P(consonant first, vowel second) = 2/9 + 1/9 + 1/9 + 1/9 = 5/9
Thus, the probability that the spinner will land on a consonant the first time and a vowel the second time is 5/9.
34 answers