Question

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 \(80 \%\) of all batteries have acceptable voltages, and let \(y\) denote the number of batteries that must be tested. a. What are the possible values of \(y\) ? b. What is \(p(2)=P(y=2) ?\) c. What is \(p(3) ?\) (Hint: There are two different outcomes that result in \(y=3\).)

Short answer

a. The possible values for y are any integer 2 or greater. b. The value of P(y=2), the probability that exactly two batteries need to be tested to find two acceptable ones, is 0.64. c. The value of P(y=3), the probability that three batteries need to be tested to find two acceptable ones, is 0.288.

Step-by-step solution

Determine possible values for y

Since two acceptable batteries are needed, the minimum number of batteries that must be tested is two. So the smallest possible value of y is 2. Since there is a chance that a battery could be unacceptable, there is no upper limit to the possible values of y. In other words, y could be any positive integer 2 or greater, i.e. y ∈ {2, 3, 4, 5, ...}.

Calculate P(y=2)

The value P(y=2) will represent the scenario where both tested batteries are acceptable. This would be a simple case of multiplying the independent probabilities together. Since the probability of a battery being acceptable is 0.8, the equation is \(P(y=2) = 0.8 × 0.8 = 0.64\). So, the probability that exactly 2 batteries are required to be tested to find two acceptable batteries is \(0.64\).

Calculate P(y=3)

Calculating P(y=3) is a bit more complex since we have to consider two possibilities: the first two batteries tested could both be unacceptable and the third could be acceptable, or the first battery could be unacceptable and the next two could be acceptable. For the first case, we have probability \(P = 0.2 × 0.2 × 0.8 = 0.032\), for the second case we have probability \(P = 2 × (0.2 × 0.8 × 0.8 = 0.128)\). We multiply the second case by 2 as this event can happen in two ways (either the first or the second battery tested is unacceptable). Adding these two possibilities together, we get \(P(y=3) = 0.032 + 0.128 × 2 = 0.288\).