A men and a women decided to meet at a certain location. If each of them independently arrives at a time uniformly between 12pm
and 1pm, find the probability that the first to arrive has to wait longer than 10 mins?
1 answer:
Answer:
the probability that the first to arrive has to wait longer than 10 mins is 35/36.
Step-by-step explanation:
First you need to denote by X and Y the time past 12 noon that the man and woman arrive.
Then you have to compute P(X+10<Y)+P(Y+10<X), which by symmetry equals 2P(X+10<Y).
So basically you have to compute P(X+10<Y).
check the attachment for the rest of the step by step.
Answer should be 25/72
But since we want 2× of the above probability 2× 25/72 = 25/36 final answer.
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Answer:
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General Formulas and Concepts:
<u>Algebra I</u>
Point-Slope Form: y - y₁ = m(x - x₁)
- x₁ - x coordinate
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Step-by-step explanation:
<u>Step 1: Define</u>
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Point (4, -7)
<u>Step 2: Write Function</u>
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