IBM
Company
Numerical Ability
Probability
In a software company 4/10 of people know C++, 2/4 of them know java, 2/8 of them know neither,find the total possibility to know C++?
Read Solution (Total 7)
-
- ans is 1/4....
possibility of knowing c++= 4/10
possibility of knowing java=2/4
there are 4 possibilities,
a person knows java but not c++=2/4*(1-(4/10))=3/10
a person knows c++ but not java= 4/10*(1-(2/4))=1/5
possibility of knowing neither =2/8
therefore, probability of knowing any one language only= 3/10+1/5=1/2
therefore ,possibility of knowing both= total probability- probability of knowing neither- probability of knowing anyone language only
total probability is always 1
so, required probability= 1 -(2/8) -(1/2) =1/4
- 8 years agoHelpfull: Yes(29) No(5)
- c++ known= 4/10
java known=2/4
neither=2/8
both known=1-2/8=6/8
Only c++= 6/8(both)-2/4=2/8=1/4 - 7 years agoHelpfull: Yes(21) No(0)
- Assume 100 persons in a company
person knows c++=4/10=100*4/100=20 persons
person knows java=2/4=100*2/4=50 persons
neither=2/8=100*2/8=25 persons
according to our assumption 100 person=20+50+25=95 person covered(5 remaining)
so,total possibility=20 person(c++)+5(remaining)=25 persons
which is 25/100=1/4
- 8 years agoHelpfull: Yes(12) No(19)
- c++ knows 2/5
java knows 1/2
knows nothing(jon snow :p) 1/4
if total people = 20(LCM of 5,2,4)
c++ knows = 8
java knows =10
knows nothing = 5
therefore, (Java)U(C++) =(20-5)=15
knows both the language(intersection of java and c++) = (10+8)-15=3
therefore possibility to know c++ = 8+3=11
ANS is 11/20 - 8 years agoHelpfull: Yes(5) No(5)
- it tells you in the question.... the opening statement - therefore 4/10
- 8 years agoHelpfull: Yes(2) No(5)
- 7/30
n(c++unionjava)=n(c++)+n(java)-n(c++intersectionjava)
n(c++unionjava)=4/10+2/4-6/8 - 7 years agoHelpfull: Yes(1) No(0)
- 1/2...........
- 6 years agoHelpfull: Yes(0) No(0)
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