Question

Probability and Conditional Independence Suppose there are two types of candidates good candidates G and bad candidates Gº. There are two interviews that a candidate can be selected for: 11 and 12, (I denotes the candidate not getting the first intreview, 15 denotes the candidate not getting the second interview). Here below we list the conditional probabilities for good and bad candidates respectively: Consider the conditional probability table below: Probability Value PIIN 12G) 0.0625 Plin 12G) 0.1875 P(I Ո IS|G) 0.1875 PINIG) 0.5625 PINIG) 0.01 Plin 12G) 0.09 P(In IgG) 0.09 PINIG) 0.81 The probability that a candidate is good is P(G) = 0.2.

Answer 1

Solution-

Since P(G) = 0.2

So, P(G^c) = 1- P(G) = 1- 0.2 = 0.8

Now, considering the above probabilities and probabilities given in the question to answer the following parts-

(a)

It is given that you have a good candidate.

Now,

$P(I_1/G)=P(I_1\cap I_2/G)+P(I_1\cap I_2^c/G)$

On putting the values we get

.....(1)

P(11/G) = 0.0625 +0.1875 = 0.25

Similarly

P(12/G) = P(11n 12/G) + Plin 12/G)

On putting the values we get

...(2)

P(12/G) = 0.0625 +0.1875 = 0.25

Now, using equations (1) and (2), we get

...(3)

P(11/G) * P(12/G) = 0.25 * 0.25 = 0.0625

And we have

$P(I_1\cap I_2/G)= 0.0625$ ...(4)

So, we can say that

P(In 12/G) = P(11/G) * P(12/G) = 0.0625

So, P(I₁/G) and P(I₂/G) are independent.

Hence, if you have good candidate that I₁ and I₂ are independent.

(b)

We know that

$P(I_1\cap I_2)=P(I_1\cap I_2/G)*P(G) +P(I_1\cap I_2/G^c)*P(G^c)$

On putting the values we get

$P(I_1\cap I_2)=0.0625*0.2+ 0.01*0.8$

$P(I_1\cap I_2)=0.0125+0.008=0.0205$ ...(5)

Similarly,

$P(I_1\cap I^c_2)=P(I_1\cap I_2^c/G)*P(G) +P(I_1\cap I_2^c/G^c)*P(G^c)$

On putting the values we get

PI Ո 15) = 0.1875 x 0.2 +0.09 + 0.8

$P(I_1\cap I^c_2)=0.0375+0.072=0.1095$ ...(6)

Similarly,

$P(I_1^c\cap I_2)=P(I_1^c\cap I_2/G)*P(G) +P(I_1^c\cap I_2/G^c)*P(G^c)$

On putting the values we get

$P(I_1^c\cap I_2)=0.1875*0.2+0.09*0.8$

$P(I_1^c\cap I_2)=0.0375+0.072=0.1095$ ...(7)

Similarly,

$P(I_1^c\cap I_2^c)=P(I_1^c\cap I_2^c/G)*P(G) +P(I_1^c\cap I_2^c/G^c)*P(G^c)$

On putting the values we get

$P(I_1^c\cap I_2^c)=0.5625*0.2+0.81*0.8$

$P(I_1^c\cap I_2^c)=0.1125+0.648=0.7605$...(8)

Now, using above equations to find P(I₁) and P(I₂)

Since,

$P(I_1)=P(I_1\cap I_2)+P(I_1\cap I_2^c)$

So,

$P(I_1)=0.0205+0.1095=0.13$.....(9)

Similarly

$P(I_2)=P(I_1\cap I_2)+P(I_1^c\cap I_2)$

So,

$P(I_2)=0.0205+0.1095=0.13$ ...(10)

From equations (9) and (10) , we get

P(I₁)×P(I₂) = 0.13×0.13 = 0.0169

SINCE,

$P(I_1\cap I_2)=0.0205\equiv P(I_1)*P( I_2)=0.0169$

Hence,I₁ and I₂ are not independent events.

The table is completed by using the values from equations (5) to (8) as shown below-

​​

	$I_1$	$I_1^c$	Total
$I_2$	$P(I_1\cap I_2)=0.0205$	$P(I_1^c\cap I_2)=0.1095$	0.13
$I_2^c$	$P(I_1\cap I^c_2)=0.1095$	$P(I_1^c\cap I_2^c)=0.7605$	0.87
Total	0.13	0.87	1