Question

The probability of rain today is 60 percent and the independent probability of rain tomorrow is 75 percent. Calculate the probability that it will rain neither today nor tomorrow.
(a) 10 %
(b) 15 %
(c) 45 %
(d) 55 %
(e) 65 %

Answer 1

Hint: We are asked to find the probability of rain neither today nor tomorrow. We will first use that event as an independent. So, using \[P\left( X \right).P\left( Y \right)=P\left( X\cap Y \right),\] we will find the probability of \[P\left( X\cap Y \right).\] Here we take X as rain today and Y as rain tomorrow. Once we have \[P\left( X\cap Y \right),\] we will use \[P\left( X\cup Y \right)=P\left( X \right)+P\left( Y \right)-P\left( X\cap Y \right)\] followed by \[P\left( {{X}^{'}}\cup {{Y}^{'}} \right)=1-P\left( X\cup Y \right).\] Using these we will find our required probability.

Complete answer:
We are given the probability of rain today is 60 % and the probability of rain tomorrow is 75 %. We are asked to calculate the probability of rain neither today nor tomorrow. Let us denote X as rain today and Y as rain tomorrow. Now, as the probability of rain today is 60 percent, so we have,
P(X) = 60 %
On simplifying, we get,
\[\Rightarrow P\left( X \right)=\dfrac{60}{100}=0.60\]
Similarly, the probability of rain tomorrow is 75 percent. So, we have,
P(Y) = 75 %
On simplifying, we get,
\[\Rightarrow P\left( Y \right)=\dfrac{75}{100}=0.75\]
Now, we are asked to find the probability of rain neither today nor tomorrow. We know that X’ means ‘Not X’. So,
\[P\left( \text{Rain neither today nor tomorrow} \right)=P\left( {{X}^{'}}\cap {{Y}^{'}} \right)\]
By De Moivre's theorem, we have,
\[P\left( {{X}^{'}}\cup {{Y}^{'}} \right)=P{{\left( X\cup Y \right)}^{'}}......\left( i \right)\]
So, we have to first find \[P\left( X\cup Y \right).\]
We know that,
\[P\left( X\cup Y \right)=P\left( X \right)+P\left( Y \right)-P\left( X\cap Y \right)\]
We have P(X) = 0.6 and P(Y) = 0.75 and as X and Y are independent events, so we know,
\[P\left( X\cap Y \right)=P\left( X \right).P\left( Y \right)\]
So, we get,
\[\Rightarrow P\left( X\cap Y \right)=0.6\times 0.75\]
\[\Rightarrow P\left( X\cap Y \right)=0.45\]
Now, using these we get,
\[P\left( X\cup Y \right)=P\left( X \right)+P\left( Y \right)-P\left( X\cap Y \right)\]
\[\Rightarrow P\left( X\cup Y \right)=0.6+0.75-0.45\]
\[\Rightarrow P\left( X\cup Y \right)=1.35-0.45\]
So, we get,
\[\Rightarrow P\left( X\cup Y \right)=0.90\]
Now, as we know,
\[P{{\left( X\cup Y \right)}^{'}}=1-P\left( X\cup Y \right)\]
Using the above value, we get,
\[\Rightarrow P{{\left( X\cup Y \right)}^{'}}=1-0.90\]
\[\Rightarrow P{{\left( X\cup Y \right)}^{'}}=0.10\]
So,
\[P{{\left( X\cup Y \right)}^{'}}=0.10.......\left( ii \right)\]
So, from (i) and (ii), we get,
\[P\left( {{X}^{'}}\cup {{Y}^{'}} \right)=0.10\]
This implies it is 10 %

Hence, option (a) is the right answer.

Note:
Students must remember changing the decimal to a percentage, we will multiply the number by 100, so \[0.1=0.1\times 100=10\text{ percent}\text{.}\] Also, only when the events are independent then only we use \[P\left( X \right).P\left( Y \right)=P\left( X\cap Y \right).\] This is not true if X and Y are not independent.