Respuesta :
Answer:
y 3 4 5
___________________________________
P(Y) 0.1875 0.5 0.3125
Step-by-step explanation:
For this case we have the following statement : SOME DOGS ARE BROWN
And if we count the number of letters in the statement we have 16 possible letters. We can define the following random variable
Y= Length of word containing a selected letter.
For example the word DOGS have 4 of the total 16 letters, so we can define the probability using empirical approximation for each word like this:
[tex] P(SOME) = \frac{4}{16}= \frac{1}{4}=0.25[/tex]
And similar for the other words we have:
[tex] P(DOGS) = \frac{4}{16}= \frac{1}{4}=0.25[/tex]
[tex]P(ARE) = \frac{3}{16}= 0.1875[/tex]
[tex] P(BROWN)=\frac{5}{16}= 0.3125[/tex]
And we have the following distribution for the words in the statement with the random variable Y defined previously:
Word SOME DOGS ARE BROWN
________________________________________
y 4 4 3 5
P(Word) 0.25 0.25 0.1875 0.3125
And as we can see the possible values for Y are 3,4 and 5, so then we can define the probability distribution for Y like this:
y 3 4 5
___________________________________
P(Y) 0.1875 0.5 0.3125