Anew 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 () to obtain a general formula for p(y).



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.

points are handed out whenever you post answers.

brainliest answers are handed out by the person asking the question, so whether your answer is picked depends entirely on whether the asker bothers to pick

if you're asking about your rank, it depends on having a certain number of points and a certain number of brainliest answers. for example, at the time of this writing, i have 44338 points out of 15000 needed for the next (nearly triple the requirement! ) but i only have 24 out of 50 required brainliest answers. this number decays, meaning after a certain amount of time (i think about a month), the total number of brainliest answers you've earned doesn't count toward this total. (i've been stuck at the same rank for a long

the lesson to learn here is that you shouldn't worry about points or rank. chances are, no matter how good an answer you may provide, your efforts will go mostly unnoticed unless you focus on the newest/most active questions that get posted.

what is the answer to this