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Question

If a real number x is chosen at random in the interval [0, 3], and a real number y is chosen at random in the interval [0, 4],what is the probability that x < y ?

(A) 1/2

(B) 7/12

(C) 5/8

(D) 2/3

Comments

Similar QS:

1. https://gateoverflow.in/67750/probability-distribution

2. https://gateoverflow.in/67746/probability-distribution

Answer 1

It is a question based on areal probability actually..Let us see how to do it..Consider a graph with x ranging from 0 to 3 along x axis and y ranging from 0 to 4 along y axis..

Now we need to find the favorable area in this case..

Total area  =  3 * 4  =  12 sq units..

Now for favorable area we need to look into the graph..The graph looks like :

The area upon which we are interested i.e. comprising the coordinates (x,y) where x < y and x and y are real numbers hence the concept of area comes into picture..

Hence the shaded area  =  1/2 * (sum of parallel edges) * perpendicular distance between them

                                   =   1/2 * (1 + 4) * 3

                                   =    15 / 2

Hence P(x < y)             =    Shaded area / Total area

                                   =    (15/2) / 12

                                   =     15 / 24

                                   =     5 / 8

Hence C) is the correct answer..

Comments

habib its given "real number" but not "integer"
there can be infinite such pairs which satisfy x<y isnt it?

---

U r right @Anusha...Thanks for rectification..Same mistake I did in gatebook test..:)

Answer 2

when y is between 0 and 3 we have two symmetric cases  x<y or y<x (P(y=x )negligible since continuous random variable ).. so P(x<y)=1/2 and when y is between 3 and 4 P(x<y)=1.

P(y is between 0 and 3)=3/4 and P(y is between 3 and 4)=1/4

So (1/2)*(3/4)+(1/4)*1=5/8

Comments

https://gmatclub.com/forum/if-a-real-number-x-is-chosen-at-random-in-the-interval-and-a-re-210013.html

GMAT question

solution is also there, in that forum.