GATE CSE 2018 | Question: 44

Question

Consider Guwahati, $(G)$ and Delhi $(D)$ whose temperatures can be classified as high $(H)$, medium $(M)$ and low $(L)$. Let $P(H_G)$ denote the probability that Guwahati has high temperature. Similarly, $P(M_G)$ and $P(L_G)$ denotes the probability of Guwahati having medium and low temperatures respectively. Similarly, we use $P(H_D),\:  P(M_D)$ and $P(L_D)$ for Delhi. The following table gives the conditional probabilities for Delhi’s temperature given Guwahati’s temperature.

$$
\begin{array}{|c|c|c|c|}
\hline
 & H_D & M_D & L_D \\
\hline
H_G & 0.40 & 0.48 & 0.12 \\
\hline
M_G & 0.10 & 0.65 & 0.25 \\
\hline
L_G & 0.01 & 0.50 & 0.49 \\
\hline
\end{array}
$$

Consider the first row in the table above. The first entry denotes that if Guwahati has high temperature $(H_G)$ then the probability of Delhi also having a high temperature $(H_D)$ is $0.40;$ i.e., $P(H_D \mid H_G) = 0.40$. Similarly, the next two entries are $P(M_D \mid H_G) = 0.48 \text{ and } P(L_D \mid H_G) = 0.12$. Similarly for the other rows.

If it is known that $P(H_G) = 0.2, \: P(M_G) = 0.5$, and $P(L_G)=0.3$, then the probability (correct to two decimal places) that Guwahati has high temperature given that Delhi has high temperature is ________.

Comments

Getting .605

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Can someone check step 2 and 3 and tell  me it is correct or not.

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if you forget bayes theorem at examination hall then you can do like this

use basic rule of probability total no. of favourable outcomes/ total possible outcomes

favourable outcome in this case will be Guwahati has high temperature given that Delhi has high temperature

which is =>0.2*0.40

total possible outcomes in this case are when delhi has high temperature then Guwahati can have high ,medium and low temperature so =>0.2*0.40+0.5*0.10+0.3*0.01

so 0.2*0.40/0.2*0.40+0.5*0.10+0.3*0.01

which is 80/133

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I think best Way to solve this question is 

https://gateoverflow.in/204119/gate-cse-2018-question-44?show=412090#a412090

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@Rishabh1006 then why you are reading it , just ignore or if you are able to delete , delete then , i dont care 

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sol

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ans is 0.6015

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Ans

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easy question

Answer 1

$P\left(H_G\mid H_D\right) = \frac{P(H_G \wedge H_D)}{P(H_D)} = P\left(H_D\mid H_G\right) \times \frac{P(H_G)}{P(H_D)}$

$=0.40 \times 0.2 \times\frac{1}{P(H_D)}$.

$P(H_D) = P\left(H_D \mid H_G\right) .P(H_G)+ P\left(H_D \mid M_G\right).P(M_G)+ P\left(H_D \mid L_G\right).P(L_G) $

$=0.4\times0.2+0.1\times0.5+0.01\times0.3 = 0.133$

$\therefore P\left(H_G\mid H_D\right)$ $\\= 0.40 \times 0.2 \times\frac{1}{P(H_D)} \\=\frac{0.08}{  0.133}\\=0.6015$

Ref: https://en.wikipedia.org/wiki/Bayes%27_theorem

Comments

In question it is mentioned "correct to 2 places", so should it be 0.60?

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I'm getting 0.47 by following the below approach.

Please let me know why we can't solve like this?

I have solved for P(H$_{D}$) as shown below:

P(H$_{D}$) = P(H$_{D}$|H$_{G}$) $\times$ P(H$_{G}$) + P(H$_{D}|\overline{H_{G}}$) $\times$ P($\overline{H_{G}}$)

= (0.40) * (0.2) + (0.10+0.01) * (1-0.2)

= 0.08 + 0.088  = 0.168

Final answer, i;e P(H$_{G}$|H$_{D}$) = 0.08/0.168 = 0.47

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@rohith1001 your $P(H_D)$ is wrong there are three cases and you have to use total probability theorem here-

three cases are there:

1- Delhi temperature high when Guwahati's temperature is high. = $P(H_D∣H_G).P(H_G)$

2- Delhi temperature high when Guwahati's temperature is medium = $P(H_D∣M_G).P(M_G)$

3- Delhi temperature high when Guwahati's temperature is low = $P(H_D∣L_G).P(L_G)$

Answer 2

A Bayes theorem question which can be solved easily using the tree diagram.

Conditional probabilities given in question are represented above in the tree diagram.

The green edges indicate the probability of Guwahati having high, medium or low temperatures.

The blue edges after each yellow edge show the probability of having high, medium or low temperature in Delhi provided, a high, medium or low temperature in Guwahati.

It is asked

probability (correct to two decimal places) that Guwahati has high temperature given that Delhi has high temperature 

which is given by

P(H_G | H_D) = $\frac{P(H_{G} \cap H_{D})}{P(H_{D})}$

P(HG$\cap$H_D) = probability that Delhi has high temperature and Guwahati has high temperature = 0.2*0.4 = 0.08

P(H_D) = Probability that Delhi has high temperature is given by a taken sum of all high-temperature cases in Delhi, regardless of what temperature in Guwahati is(High, Medium or low). i.e. Take the sum of all such conditional probabilities. Marked by yellow bubbles in the diagram.

P(H_D) = (0.2*0.4) + (0.5*0.10) + (0.3*0.01) = 0.133

So, P(H_G | H_D) = $\frac{0.08}{0.133} = 0.601$

Comments

@himgta
See it this way: Temperature readings are taken on 100 random days.
Let us try to establish the total days when Delhi is Hot. There are 3 mutually exclusive and collective exhaustive cases when Delhi is Hot:

1. Guwahati is hot on 20 of of the 100 days. Delhi is Hot on 40% of the days when Guwahati is also hot. So both Guwahati and Delhi are hot on 40% of 20  = 8 days.
2. Guwahati  is Medium for 50 days. Delhi is Hot 10% of these days. That is 5 days.
3. Guwahati is Low for 30 days. Delhi is Hot 1% of these days. That is 0.3 days.

(20+50+30 = 100)
So Delhi is hot for a total of 13.3 days out of the 100 days readings were taken. P(DH) = 0.133 ​​​​​​​

Otherwise you land up overestimating the days Delhi is Hot. Eg. if Delhi was hot on all 20 days when Guwahati was hot (this is highly likely by the way), wouldn't you count these as 100% of 20 days = 20 days. If you don't multiply by 0.2, then you land up counting these days as 100.

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Thank you

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thank you helps a lot

Answer 3

0.601

Answer 4

Here is the answer : 

Comments

wrong answer

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Elaborate, where I am wrong, I would be happy to rectify my mistake.

Answer 5

 Bayes theorem: 

 

Answer 0.6015

Answer 6

Ans