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In a study for a certain type of battery, it was found that the probability of a lifetime X exceeding 4 hours is 0.2. If three such batteries are in use in independently operating systems, find the probability that

(a) Exactly 2 of the batteries lasts 4 hours or more

(b) At least 2 batteries lasts 4 hours or more.

I believe this is a binomial distribution So: $$\mathsf P(X{=}x)~=~{n \choose x} \cdot p^x \cdot (1-p)^{(n-x)}$$

For this one n = 3, and x=2

$$\mathsf P(X{=}2)~=~3\cdot (0.20)^2 \cdot (1-0.20)^1 = 0.096$$

But how would you do an at least type of question?

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    for b) find $P(B=2\vee B=3)=P(B=2)+P(B=3)$ where $B$ denotes the number of batteries that lasts $4$ hours or more. – 2017-02-12

1 Answers 1

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Confirmed.   The count of long-lived batteries among the three is indeed a Binomial Distribution.   You also have the correct probability mass function for that. $$Y\sim\mathcal{Bin}(3, 0.2) ~\iff~ \mathsf P(Y{=}k) = \dbinom 3 k\, (0.2)^k\,(0.8)^{3-k}\;\mathbf 1_{k\in\{0,1,2,3\}}$$

But how would you do an at least type of question?

Since the support if $Y\in\{0,1,2,3\}$ then the event of $Y\geq 2$ is $Y\in\{2,3\}$.   So we just sum the probability masses for the discrete atoms.

$$\mathsf P(Y{\geq}2) ~=~ \mathsf P(Y{=}2)+\mathsf P(Y{=}3)$$