On a given day in the rainy season, it may rain $$70 %$$ of the time. If the rains, the chance that a village fair make a loss on that day is $$80 %$$ . However, if it does not rain, the chance that the fair will make a loss on that day is only $$10 %$$ . If the fair has not made a loss on a given day in the rainy season, what is the probability that it has not rained on that day?

In a given day in the rainy season, it may rain of the time. If the rains, chance that a village fair mak

In a given day in the rainy season, it may rain $70 \%$ of the time. If the rains, chance that a village fair make a loss on that day is $80 \%$ .However, if it does not rain, chance that the fair will make a loss on that day is only $10 \%$ . If the fair has not made a loss on a given day in the rainy season, what is the probability that it has not rained on that day?
Option: 1
$\frac{3}{10}$

Option: 2
$\frac{9}{11}$

Option: 3
$\frac{14}{17}$

Option: 4
$\frac{27}{41}$

Answers (1)

Let $\mathrm{A=\{}$ Fair has not made a loss $\mathrm{\}}$

$\mathrm{E_1 =\{\text { it rains }\} \text {, then } P\left(E_1\right)=0.7 }$
$\mathrm{E_2 =\{\text { it does not rains }\} \text { then } P\left(E_2\right) }$

$\mathrm{ =0.3 }$
$\mathrm{P\left(A / E_1\right)=P[ }$ Fair has not made a loss on a rainy $\mathrm{\text { day }]=20 \%=0.2}$
$\mathrm{P\left(A / E_2\right)=P}$ [Fair has not made loss on non rainy day ] $\mathrm{=90 \%=0.9}$

So by Baye's theorem;
$\mathrm{P\left(E_2 / A\right)= \frac{P\left(E_2\right) \cdot P\left(A / E_2\right)}{P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)}}$
$\mathrm{=\frac{03 \times 0.9}{07 . \times 0.2 \times 0.3 \times 0.9}=\frac{27}{41}}$