Probability

Question

In a certain group of computer personnel, 65% have insufficient knowledge of hardware, 45% have inadequate idea of software and 70% are in either one (or) both of the two categories. What is the percentage of people who knpw software among those who have a sufficient knowledge of hardware?

• A. 0.35/0.3
• B. 0.3/0.35
• C. 0.3
• D. 0.35

Comments

i think B is ans

Answer 1

IH:-Insufficient Hardware Knowledge

H:-Hardware Knowledge

IH=65 So H=35

IS=45 So S=55

Given  $(IH \cup IS)$ =70

Follow the IMG.

Total - $(IH \cup IS)$ =$(H \cap S)$

100-70=30

They asked $(H \cap S)$ / H

So 30/35 Option B is Ans.

Answer 2

Is this can be an easy method?

P(A)=Insufficient s/w=0.65, P(A')=Sufficient s/w=0.35

P(B)=Insufficient h/w=0.45, P(A')=Sufficient h/w=0.55

P(A U B)=0.7

Now, we want Sufficient s/w and Sufficient h/w= P(A' intersection B')=P(A U B)'=(0.7)'=1-0.7=0.3

Final answer=P(A' intersection B')/P(A')=0.3/0.35

Answer 3

A: Knows hardware

B: Knows software

Given,

P(A') = 65%

P(B') = 45%

P(A' ∪ B') = 70%

Using DeMorgan's law,

P(A ∩ B) = P((A' ∪ B')') = 30%

So, P(B|A) = P(A ∩ B)/P(A) = 30/35