Question

Suppose 88% 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.

(c) You probabilities in parts a and b should differ slightly. Why do you think this is?

Answer 1

Please note ⁿC_r = n! / [(n-r)!*r!]

Binomial Probability = ⁿC_x * (p)^x * (q)^n-x, where n = number of trials and x is the number of successes.

(a) Here n = 20, p = 0.88, q = 1 – p = 0.12

P(19 or 20) = P(19) + P(20)

P(19) = ²⁰C₁₉ * (0.88)¹⁹ * (0.12)^20-19=1 = 0.2115349

P(20) = ²⁰C₂₀ * (0.88)²⁰ * (0.12)^20-20=0 = 0.0775628

Therefore the required probability = 0.2115349 + 0.0775628 = 0.2891

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(b) Binomial Probability = ⁿC_x * (p)^x * (q)^n-x, where n = number of trials and x is the number of successes. Here p is defined as the probability of the flashligh working which requires 2 batterires of 88% voltage.

Here n = 10, p = 0.88 * 0.88 = 0.7744, q = 1 – p = 0.2256

P(at least 9) = P(9) + P(10)

P(9) = ¹⁰C₉ * (0.7744)⁹ * (0.2256)^10-9=1 = 0.2259577

P(10) = ¹⁰C₁₀ * (0.7744)¹⁰ * (0.22556)^10-10=0 = 0.0775628

Therefore the required probability = 0.2259577 + 0.0775628 = 0.3035

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(c) In part (a) we were finding the probability that out of 20 batterries chosen, the probability of 19 or 20 working. Here then we were using the probability of a single battery fiunctioning.

In part (b) we are using 10 flashlights, which again use 2 batteries each. Therefore we are again using 20 randomly selected batteries, but here we are not using the probability of a single battery as a flashlight requires both batteries to function, and thereore the probability of the flashlight functioning is not the same as the probability of having a battery functioning properly, and thats why there is a slight difference in the values.