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A new battery's voltage may be acceptable (A) or unacceptable (U). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that 9096 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
d. Use the pattern in your answers for parts (a)—(c) to obtain a general formula for p(y).

Respuesta :

MrRoyal

Answer:

a. 0.81

b. 0.162

c. To have Y=5, the fifth battery must be acceptable

The possible outcomes are AUUUA, UAUUA, UUAUA, UUUAA

The probability of having Y = 5 is 0.00324

d. p(Y=2) * (Y-1)p(U)^y-2

Step-by-step explanation:

First, I'll like to make a correction to the question

".....Suppose that 90% of all batteries have acceptable voltages...."

I'll be working base on the above correction.

Let p(A) = Probability that a battery is acceptable = 90%=0.9

Let p(U) = Probability that a battery is unacceptable = 1 - 90% = 1 - 0.9 = 0.1

a. What is p(2)

What is expected here is to find the probability that 2 batteries selected are accepted

p(Y=2) = The probability that the first battery tested is good and the second battery tested is good

p(Y=2) = p(A) and p(A)

p(Y=2) = (0.9) * (0.9)

p(Y=2) = 0.81

b. What is p(3)

To arrive at Y = 3, it means that only one of the first two tested batteries is acceptable and the third tested battery is acceptable

p(Y=3) = p(AUA) or p(UAA)

Where p(AUA) = the first battery is acceptable and the second battery is unacceptable and the third battery is acceptable.

And p(UAA) = the first battery is unacceptable and the second battery is acceptable and the third battery is acceptable.

p(Y=3) = p(AUA) or p(UAA)

p(Y=3) = (p(A) and p(U) and p(A)) or (p(U) and p(A) and p(A))

p(Y=3) = (0.9 * 0.1 * 0.9) + (0.1 * 0.9 * 0.9)

p(Y=3) = (0.9*0.1*0.9)*2

p(Y=3) = 0.162

c.

i. To have Y=5, the fifth battery must be acceptable

ii. The possible outcomes are AUUUA, UAUUA, UUAUA, UUUAA

iii. The probability of having Y = 5 is

p(Y=5) = p(AUUUA) or p(UAUUA) or p(UUAUA) or p(UUUAA)

p(Y=5) = (0.9 * 0.1 * 0.1 * 0.1 * 0.9) + (0.1 *0.9*0.1*0.1*0.9) + (0.1*0.1*0.9*0.1*0.9) + (0.1*0.1*0.1*0.9*0.9)

p(Y=5) (0.9*0.1*0.1*0.1*0.9) * 4

p(Y=5) = 0.00081 * 4

p(Y=5) = 0.00324

d.

p(Y=2) = p(A) and p(A)

p(Y=2) = p(A) * p(A)

p(Y=3) = p(AUA) = p(AUA) or p(UAA)

p(Y=3) = p(A) and p(U) and p(A) + p(U) and p(A) and p(A)

p(Y=3) = p(A) * p(U) * p(A) * 2

p(Y=3) = p(A) * p(A) * 2p(U)

p(Y=3) = p(Y=2) * 2p(U)

p(Y=5) = p(AUUUA) or p(UAUUA) or p(UUAUA) or p(UUUAA)

p(Y=5) = p(A) and p(U) and p(U) and p(U) and p(A) * 4

p(Y=5) = p(A) * p(U) * p(U) * p(U) * p(A) * 4

p(Y=5) = p(A) * p(A) * p(U) * p(U) * p(U) * 4

p(Y=5) = p(Y=2) * p(U)³ * 4

P(Y=5) = p(Y=2) * 4p(U)³

In general, p(y) = p(Y=2) * (Y-1)p(U)^y-2

3.

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