Question

5. Suppose 85% of all batteries from a supplier have acceptable voltages. A certain type of flashlight requires two type-D batteries, and the flashlight will work only if both its batteries have acceptable voltages. Treat all batteries and flashlights as independent of each other. (a) In 20 randomly selected batteries, what is the probability that 19 or 20 have acceptable voltages? (b) Suppose 20 randomly selected batteries are placed into 10 flashlights. What is the probability that at least 9 of the flashlights function properly? Hint: Determine the probability that a single flashlight functions, then use that to find the desired probability. (e) You probabilities in parts a and b should differ slightly. Why do you think this is?

Answer 1

5.

(a)

Suppose, random variable X denotes number of batteries with acceptable voltage.

We define getting a battery with acceptable voltage as success. Then we can model Binomial distribution as follows.

:: X~Bin(n = 20, p=0.85)

:: P(X = 1) = (n.)»" (1 – p)*-* = (20) (0.85)" (0.15)20-1

So, required probability is given by

.:. P(X = 19)+ P(X = 20) = (*)(0.5"(0.15)+(20) (0.5) (0.15).

  

= 0.1367983 +0.0387595 = 0.1755578

(b)

A flashlight works if both of the batteries attached to that particular flashlight work.

So, probability that a flashlight works = 0.85 * 0.85 = 0.7225

Suppose, random variable Y denotes number of flashlights functioning properly.

We define getting a flashlight functioning properly as success. Then we can model Binomial distribution as follows.

HY ~ Bin(n = 10, p = 0.7225)

:: P(Y = y) = (") p*(1 – p)"-= () (0.7225)*(0.2775)10-y

So, required probability is given by

PIY = 9)+P(Y = 10) = (. )(0.7225)'(0.2775)'+(10) (0.7225):9 (0.2775)

= 0.1488688 +0.0387595 = 0.1876283

(c)

The probabilities in (a) and (b) differs as these two are calculated in two different instances.

In (a) we allowed 0 or 1 battery to be defective whereas in (b) we allowed 0 (in case of 10 flashlights working properly), 1 or 2 (in case of 9 flashlights working properly) batteries to be defective.