Suppose 88% of all batteries from a supplier have acceptable voltages. A certain type of flashlight requires two D batteries, and the flashlight will work only if both its batteries have acceptable voltages. Treat all batteries and flashlights as independent of one another. Suppose 20 randomly selected batteries are placed into 10 flashlights. What is the probability that at least 9 of the flashlights function properly

Answer:

0.3036 = 30.36% probability that at least 9 of the flashlights function properly.

Step-by-step explanation:

For each flashlight, there are only two possible outcomes. Either it works, or it does not. Flashlights are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

Flashlight works correctly if both its batteries have acceptable voltages:

Each battery has an 88% = 0.88 probability of working correctly(having acceptable voltage), so the probability of the flashlight working correctly is given by:

[tex]p = (0.88)^2 = 0.7744[/tex]

10 flashlights.

This means that [tex]n = 10[/tex]

What is the probability that at least 9 of the flashlights function properly?

This is

[tex]P(X \geq 9) = P(X = 9) + P(X = 10)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 9) = C_{10,9}(0.7744)^{9}.(0.2256)^{1} = 0.2260[/tex]

[tex]P(X = 10) = C_{10,10}(0.7744)^{10}.(0.2256)^{0} = 0.0776[/tex]

[tex]P(X \geq 9) = P(X = 9) + P(X = 10) = 0.2260 + 0.0776 = 0.3036[/tex]

0.3036 = 30.36% probability that at least 9 of the flashlights function properly.

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