You roll a 4-sided die repeatedly. on your odd-numbered rolls (1st,3rd,5th, etc.) you are victorious if you get a 4. on your even-numbered rolls, you are victorious if you get a 3 or 4. you stop once you are victorious. let y be the number of times you roll.
find e[y].

E (Y) = 3

Step-by-step explanation:

If a 4-sided die is being rolled repeatedly; and the odd-numbered rolls (1st 3rd,5th, etc.)

The probability of odd number roll will be, p(T) =

However, on your even-numbered rolls, you are victorious if you get a 3 or 4. Also, the  probability of even number roll, p(U) =

In order to  calculate: E (Y); We can say Y to be the number of times you roll.

We know that;

E (Y) = E ( Y|T ) p(T) + E ( Y|U ) p(U)

Let us calculate E ( Y|T ) and E ( Y|U )

Y|T ≅ geometric =

Y|U ≅ geometric =

also; x ≅ geometric (p)

∴ E (x) =

⇒ = 4 ; also = 2

E (Y) = 4 × + 2 ×

        = 2+1

E (Y) = 3

yes   because as the functions are graphed they eventually will intersect unless otherwise

step-by-step explanation:

x=0

step-by-step explanation: