Respuesta :
Answer:
a) 3.5
b) 3.33
c) [tex]6 - 6\left(\begin{array}{ccc}5\\6\end{array}\right)^{10}[/tex]
Step-by-step explanation:
As given,
A fair die is rolled 10 times
a)
Expected value of Sum of the number in 10 rolls = [tex]\frac{1}{6}(1+2+3+4+5+6) = \frac{21}{6}[/tex]
= 3.5
∴ we get
Expected value of Sum of the number in 10 rolls = 3.5
b)
Ley Y : number of multiples of 3
Y be Binomial
Y - B(n = 10, p = [tex]\frac{2}{6}[/tex] )
Now,
Expected value = E(Y) = np = 10×[tex]\frac{2}{6}[/tex] = 3.33
c)
Let m = total number of faces in a die
⇒m = 6
As die is roll 10 times
⇒n = 10
Now,
Let Y = number of different faces appears
Now,
Expected value, E(Y) = m - m[tex]\left(\begin{array}{ccc}m-1\\m\end{array}\right)^{n}[/tex]
= [tex]6 - 6\left(\begin{array}{ccc}6-1\\6\end{array}\right)^{10} = 6 - 6\left(\begin{array}{ccc}5\\6\end{array}\right)^{10}[/tex]
In this exercise we have to use probability knowledge to calculate the times the numbers of a roll, in this way we have to:
a) [tex]3.5[/tex]
b) [tex]3.33[/tex]
c) [tex]6-6\left[\begin{array}{cc}5\\6\2\end{array}\right] ^{10}[/tex]
Given that a fair die is rolled 10 times:
a) Expected value of Sum of the number in 10 rolls :
[tex]\frac{1}{6} (1+2+3+4+5+6)= \frac{21}{6} = 3.5[/tex]
b) Using the formula of the binomial that is represented by:
[tex]Y - B(n = 10, p = \frac{2}{6})[/tex]
Expected value: [tex]E(Y) = np = (10)(\frac{2}{6} ) = 3.33[/tex]
c) Looking at the 10 times the roll does, we have that it will be. Let m = total number of faces in a die that represents m = 6, as die is roll 10 times that represents n = 10. Now, Expected value:
[tex]= E(Y) = m - m \left[\begin{array}{cc}m-1\\m\\ \end{array}\right] \\= 6-6\left[\begin{array}{cc}5\\6\2\end{array}\right] ^{10}[/tex]
See more about probability at brainly.com/question/795909
