# CONDITIONAL PROBABILITY

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65% of the employees in a firm are graduates and 35% did not receive any graduate trainig.
Among the graduates, 40% have an experience of less than 10 years. Similarly, 18% of the non graduate emplyees have an experience of more than 10 years.

The first question is construct the corresponding contignency tables.
answer: i did it like that

EMPLOYESS 65% 35% (total=100)
less thn 10 yrs 40% 25%(total 65)
mre than 10 years 17% 18% (total 35)

The toal of grduates column is 122, for non graduats is 78% for the total column is 200

The second question :what is the probability that an employee chosen randomly is a graduate or has an experience of more than 10 years?

for that second question:should i take 65/200 and 17/200?if yes,should i calculate it in terms of its percentage?? i mean 65/100*200 ??

• CONDITIONAL PROBABILITY -

I interpret the problem different from you. I take
"Among the graduates, 40% have an experience of less than 10 years. "
to mean 40% among the 65% (graduates) have <10 years experience, so (65-26)=39% among the graduates have >10 years experience.

Similarly,
"18% of the non graduate emplyees have an experience of more than 10 years"
would mean 18% of 35%(non-grad) have >10 years experience, or 6.3% of the total have >10 years, leaving (35-6.3)=28.7% have <10 years experience.

The table then becomes:
====G====NG
<10y 26% 28.7% total = 54.7%
>10y 39% 6.3% total = 45.3%
--------65% 35%

We look for Probability of an employee being a graduate given he has more than 10 years experience.
We have learned that conditional probability P(A|B) prob. of A given B
P(A|B)=P(A∩B)/P(B)
So
P(G|>10y)=P(G ∩ >10y)/P(>10y)
=0.39/(0.453)
=0.861 approx.

P(G

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