Expected Exponential Times: Extended Memoryless Property?

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I am looking for someone to walk me through the following question:

Q: Mary and Kate compete as a team in a race which consists of a single discipline (running or swimming) and a team discipline (basketball shooting). The race finishes when both single disciplines and the team discipline are finished. Simultaneously, Mary starts running 1000m while Kate starts swimming 200m. Mary takes exponential amount of time with mean 200 seconds to run 1000m, while Kate takes exponential amount of time with mean 250 seconds to swim 200m. The first person to complete their single discipline starts trying to score a basket (as in basketball) from the three-point line, which takes exponential amount of time with mean 60 seconds. When the second person is done with the single disciple in, she can help the first one by also starting to shoot at the basket, and scoring a basket will take exponential amount of time with mean only 30 seconds. Of course, if the first person scores a basket before the second one finishes her single discipline then the race finishes when the second one completes her single discipline. What is the expected team race time of Mary and Kate?

We are given the expected times for Mary and Kate to finish their single disciplines from the mean values. Are the mean values (200 seconds for Kate and 250 seconds for Mary) equivalent to $\lambda_{Kate}$ and $\lambda_{Mary}$? Or do we take 200 seconds and 250 seconds to calculate the $\lambda$ values?

Is our first step figuring out the expected time for the first single discipline to be finished?

Thank you.

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You have three independent random variables. Times for $R$unning, $S$wimming, $B$asketball(one player), and $T$wo player basket ball.

You know $\mathsf E(R), \mathsf E(S), \mathsf E(B), \mathsf E(T)$ and wish to find : $\mathsf E(\max(\min(R,S)+B, \max(R,S)+T))$

When $\vee$ mean max and $\wedge$ means minimum, $\mathsf E( ((R\wedge S)+B)\vee((R\vee S)+T))$

Rearrange as the expectation of a sum of minimum functions of independent random variables. Use the linearity of expectation and known properties of the distribution of the minimum of two independent exponential random variables.


Hint: $~\max(R,S)=R+S-\min(R,S)~$ or $~R\vee S = R+S-(R\wedge S)~$