probability distribution

Question

A arrives at office at 8-10am regularly; B arrives at 9-11 am every day. Probability that one day B arrives before A? [Assume arrival time of both A and B are uniformly distributed]

Comments

there is a 60 min. intersection. what is the answer given?

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answer is 1/8

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@Vaishali.

If B comes at 9, A can come at any time after 9

If B comes at 9.01, A can come at any time after 9.01

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If B comes at 9.59, A can come only at 10.

So, favourable events = 59+ 58+........+2+1

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lets consider sample space.

If A arrives at 8, B can arrive anytime between 9-11 i.e. 120 min

If A arrives at 8.01, B can arrive anytime between 9-11 i.e. 120 min

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If A arrives at 10.00, B can arrive anytime between 9-11 i.e. 120 min

THus, sample space = 120 * 120

So, probablity = 1/8 (approx)

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Sushant not the appropriate way to solve.

Answer 1

B can arrive prior to A in the interval of $9-10$ Am.

In this graph i have changed timings,  that A arrives between $0$ and $2$, and B arrives between $1$ and $3$.

Now we are only interested in area where B arrives prior to A, this will be the area where $(x,y)$ where $x > y$.

Because x which is arrival time of A must be greater than that of B's $(y)$.

Shaded area is the area where $x > y$,

Area of triangle = $\large \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$

Area of sample space (square) = $4$

Probability of B's arrival prior to A's = $\Large \frac{\frac{1}{2}}{4} = \frac{1}{8}$

 

Comments

nice concept :)

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Area of sample space (triangle) = 4

u mean rectangle here ......

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Deepanshu thanks. updated

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its a really nice way!!

thanks @Mk Utkarsh