A computer shop sells an electronic game powered by a single battery that has an exponential lifetime with a mean of 5 hours. I purchased 4 batteries for my game. What is the probability that exactly two of my four batteries last longer than $5$ hours.
I know that the distribution of the question is exponential with a given mean $\theta$ of $5$. If we let $T$ be the random variable where $T=t$ is the battery life I get that we have to find $P(T > 5)$ but I am not quite sure how to take in account the fact that you picked two of four.
It's a multiple choice question with the following answers:
$(a) 0.3679$ $(b) 0.4656$ $(c) 0.3245$ $(d) 0.4323$ $(e) 0.2218$
I got
0.3244607. $\mu_x = 5 \Rightarrow \lambda = \frac{1}{5}.$$F(x) = 1-e^{-\lambda x} \Rightarrow F(5) = 1-e^{-1} = 0.6321206.$
Thus the probability of a battery dieing for less than or equal to five hours is 0.6321206. Since you may view the four batteries as four draws from an iid random variable. Your problem becomes: out of four independent observations what is the probability that an event with an individual probability $p=0.6321206$ occurs an exact number of times? This is a binomial distribution with $N = 4, p=0.6321206$. And $Binom(4,2,0.6321206) = 0.3244607$.