Question

A new battery's voltage may be acceptable (A) or unacceptable (U). A certain ash-light requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that 90% of all batteries have acceptable voltages. Let Y denote the number of batteries that must be tested.

(a) What is p(2), that is, P(Y = 2)?

(b) What is p(3)? [Hint: There are two different outcomes that result in Y = 3.]

(c) To have Y = 5, what must be true of the fifth battery selected? List the four outcomes for which Y = 5 and then determine p(5).

(d) Use the answers for parts (a)-(c) to obtain a general formula for p(y)

 

 

Answer 1

Solution

(a) What is p(2)

we have $P(A)=0.9\implies P(Y=2)=0.9\times 0.9=0.81$

Therefore. $P(Y=2)=0.81$

(b) What is p(3)

$\oplus\hspace{2mm} Y=3 : \Big\{(AUA),(UAA)\Big\}$

$\implies P(Y=3)=2\times 0.9\times 0.1\times 0.9=0.162$

Therefore.  $P(Y=3)=0.162$

(c) List the four outcomes for which Y = 5 and then determine p(5).

$\oplus\hspace{2mm} Y=5 : \Big\{(AUUUA),(UAUUA),(UUAUA),(UUUAA)\Big\}$

$\implies P(Y=5)=4\times0.9\times0.9\times0.1\times0.1\times0.1=0.0032$

Therefore.  $P(Y=5)=0.0032$

(d) general formula for p(y).

$\implies P(Y=y)=(y-1)(0.9)^2(0.1)^{Y-2}$

 

 

Therefore,

a).$P(Y=2)=0.81$

b). $P(Y=3)=0.162$

c). $P(Y=5)=0.0032$

d). $P(Y=y)=(y-1)(0.9)^2(0.1)^{Y-2}$