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35 votes

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 $25°C$?

- $0.4$
- $0.6$
- $0.8$
- $0.9$

50 votes

Best answer

**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/ )

61 votes

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)**

6 votes

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 $P(A \mid B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{0.4}{\frac{1}{2}} = 0.8$

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 $P(A \mid B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{0.4}{\frac{1}{2}} = 0.8$

5 votes

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