GATE IT 2006 | Question: 1

Question

In a certain town, the probability that it will rain in the afternoon is known to be $0.6$. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to $25°C$, the probability that it will rain in the afternoon is $0.4$. The temperature at noon is equally likely to be above $25°C$, or at/below $25°C$. What is the probability that it will rain in the afternoon on a day when the temperature at noon is above $25°C$?

• A. $0.4$
• B. $0.6$
• C. $0.8$
• D. $0.9$

Comments

Given,

P(A)=0.6

P(A')=0.4

P(A|<=25)=0.4

P(<=25)=0.5

P(>25)=0.5

P(A|>25)=?

P(A|>25)=P(A,>25)/P(>25)

P(A,>25)=?

 

P(A,<=25)=P(A|<=25)*P(<=25)

P(A,<=25)=0.4*0.5

P(A,<=25)=0.2

Law of total probability

P(A)=P(A,>25)+P(A,<=25)

0.6=P(A,>25)+0.2

P(A,>25)=0.4

P(A|>25)=P(A,>25)/P(>25)

P(A|>25)=0.4/0.5

P(A|>25)=0.8

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Ans

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In this Question I missed the Pr(rain) =0.6 which is given ..And I assume that how it is possible to solve this Question then I came across solution ..I got to know that I missed that part .

Answer 1

Answer is C) $0.8$

$P$(rain in afternoon) $= 0.5\times  P($rain when temp $\leq 25) + 0.5 \times P($ rain when temp $> 25 )$
$0.6 = 0.5\times 0.4 + 0.5\times P($ rain when temp $> 25 )$
so,
$P$( rain when temp $> 25$ ) $= 0.8$

( Answer courtesy- Pradeep Pandey sir - https://gateetude.wordpress.com/category/gate-computer-science/information-technology-solutions/ )

Answer 2

This is a question of Total Probability where after happening on one event E1, the probability of another event E2 happening or not happening is added together to get the probability of happening of Event E2.

Given P(Rain in noon) =0.6 (This is total probability given).

"The temperature at noon is equally likely to be above 25°C, or at/below 25°C."

means P(Temp less than or 25) = P(Temp >25) =0.5

P(Rain in noon) = P(Temp $\leq$ 25) * P(Rain | Temp $\leq$ 25) + P(Temp $>$ 25) * P(Rain| Temp $>$ 25)

0.6= (0.5*0.4) + (0.5*X)

X=0.8 Ans (C)

Comments

This should be best answer

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Nice analysis. Got to learn a lot from your answer. Especially the tree method in solving probability questions.

Answer 3

Let $\color{blue}{P(A) = \text{ Prob. that it rains at noon}}$ and $\color{blue}{P(B) = \text{Prob. that temp. at noon is greater than 25}}$.
Given, $P(\bar B) = P(B) = \dfrac{1}{2}$ and $P(A\mid \bar B) = 0.4 = \dfrac{P(A\cap \bar B)}{P(\bar B)}$. So, $\color{blue}{P(A\cap \bar B) = 0.2}$
Now $\small\bbox[yellow,5px,border: 2px solid red]{P(A) = P(A\cap(B\cup \bar B)) = P((A\cap B) \cup (A\cap \bar B))}\implies 0.6 = P(A\cap B) + \color{blue}{0.2}\implies \color{red}{P(A\cap B) = 0.4}$.

$\small\bbox[5px,border: 2px solid red]{\text{Note: } P((A\cap B) \cap (A\cap \bar B)) = 0}$
The final answer would then be $P(A \mid B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{0.4}{\frac{1}{2}} = 0.8$

Answer 4

Given that,  P(rain in the afternoon ) = 0.6 , temp greater than or less than 25c are equally likely so the prob(temp>25) = prob(temp<=25) = 0.5 , P(rain in the afternoon ∣ temp<=25) = 0.4 .

We need to find out the value of P(rain in the afternoon ∣ temp> 25) .

Apply conditional property

P(rain in the afternoon ) =P(rain in the afternoon ⋂ temp<= 25) +  

P(rain in the afternoon ⋂ temp>25)

0.6  =  P(temp<=25).P(rain in the afternoon ∣ temp<= 25) + 

P(temp>25).P(rain in the afternoon ∣ temp> 25)

0.6  =  0.5⨉0.4 + 0.5 ⨉ P(rain in the afternoon ∣ temp> 25)

P(rain in the afternoon ∣ temp> 25) = 0.8

The correct answer is (C) 0.8

Answer 5

Ans

Answer 6

• R = it rains in the afternoon
• H = temperature at noon is above 25°C
• L = temperature at noon is at or below 25°C

Given information:

• P(R) = 0.6 (overall probability of afternoon rain)
• P(R|L) = 0.4 (probability of rain given temp ≤ 25°C)
• P(H) = P(L) = 0.5 (temperature is equally likely to be above or at/below 25°C)

Find: P(R|H)

I'll use the law of total probability. Since the temperature must be either above or at/below 25°C:

P(R) = P(R|H) × P(H) + P(R|L) × P(L)

Substituting the known values:

0.6 = P(R|H) × 0.5 + 0.4 × 0.5

0.6 = 0.5 × P(R|H) + 0.2

0.4 = 0.5 × P(R|H)

P(R|H) = 0.4 / 0.5 = 0.8