Question

Mike wants to go fishing this weekend to nearby lake. His neighbour alice (is the one mike was hoping to ask out since long time) is also planing to go to the same spot for fishing this weekend. The probability that it will rain this weekend is p1. There are two possible ways to reach the fishing spot (bus or train). The probability that mike will take the bus is pmb and that alice will take the bus is pab. Travel plans of both are independent of each other and rain. What is the probability prs that mike and alice meet each other only (should not meet in bus or train) in a romantic setup (on a lake in rain)? Input constraints (0p11), (0pab1), (0pmb1) input format first line: pmb second line: pab third line: p1 output format prs, rounded up to six decimal places.

Answer 1

To find out: probability of Mike and Alice meeting only in a romantic setup(on a lake in rain).

ANSWER:

◆ Probability of rain = p1

◆ For them to not meet during commute, there are two possibilities:

1. Mike takes bus and Alice takes train Probability = Pmb × (1 - Pab)

2. Mike takes train and Alice takes bus

probability = Pab × (1 - Pmb)

So total probability of them not meeting during commute and meeting only on the lake = Pmb × (1 - Pab) + Pab × (1 - Pmb) = Pab + Pmb - 2•Pab•Pmb

● Thus probability of Mike and Alice meeting in the lake while raining

= p1 × (Pab + Pmb - 2•Pab•Pmb)


I hope they meet!♡♡♡

Answer 2

Answer:

The probability that both Mike and Alice will take the bus, and so meet on the bus, is (0.2)(0.5)= 0.1. The probability both Mike and Alice will take the train, and so meet on the train, is (1- 0.2)((1- 0.5)= (0.8)(0.5)= 0.4. The probability they will meet on either the train or the bus is 0.1+ 0.4= 0.5 so the probability they will not meet on bus or train is 1- 0.5= 0.5. Assuming that they do meet, the probability it is not on the train or bus is 1- 0.5= 0.5. The probability it is also raining is 0.5(0.2)= 0.1.

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