Latest Infosys Aptitude Question SOLUTION: Persons A and B. Person A picks a random no. from 1 to 1000.Then person B picks a random no. from 1 to 1000. What is the probability of B getting no. greater then what A has picked

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Persons A and B. Person A picks a random no. from 1 to 1000.Then person B picks a random no. from 1 to 1000. What is the probability of B getting no. greater then what A has picked.

Asked In Infosys (14 years ago)

Solved Tijo Jose Read Solution (5)

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Tijo Jose  (14 years ago)

Probability of A choosing 1 and B greater = (1/1000)*(999/1000);
Probability of A choosing 2 and B greater = (1/1000)*(998/1000);
Probability of A choosing 3 and B greater = (1/1000)*(997/1000); ...;
Probability of A choosing 998 and B greater = (1/1000)*(2/1000);
Probability of A choosing 999 and B greater = (1/1000)*(1/1000);
Total = (1/1000)*(1/1000)*(999+998+997+...+2+1) = (1/1000)*(1/1000)*(999*1000/2) = 0.4995

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karthik ks  (8 years ago)

EASIEST WAY S SOLVING USING FORMULA ;
prob of B getting greaterr no than A= P(B)/ ( p(A)+p(B))
P(A)=1/1000 P(B)=1/1000
Therefore ans = (1/1000) / (2/1000) = 1/2

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Tijo Jose  (14 years ago)

Pbbl of A choosing 1 = (1/1000). Pbbl of B choosing >1 = (999/1000). Total = (1/1000)*(999/1000) and so on. Total pbbl = (1/1000)*(999/1000)+(1/1000)*(998/1000)+...+(1/1000)*(1/1000) = (1/1000)*(1/1000)*(999+998+...+2+1) = 0.4995

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Wafa  (14 years ago)

999/2000

Is this solution Helpfull? Yes (3) | No (1)  

Tijo Jose  (14 years ago)

(1/1000)*(999/999) + (1/1000)*(998/999) + ... + (1/1000)*(1/999) = (1/1000)*(1/999)*[999+998+...+2+1] = (1/1000)*(1/999)*[(999*1000)/2] = (1/2)

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