Question

Let $A$denote the event that the midtown temperature in Los Angeles is${70}^{◦}F$, and let$B$denote the event that the midtown temperature in New York is${70}^{◦}F$. Also, let's $C$denote the event that the maximum of the midtown temperatures in New York and in Los Angeles is${70}^{◦}F$. If $P\left(A\right)=.3,P\left(B\right)=.4,$and$P\left(C\right)=.2$, find the probability that the minimum of the two midtown temperatures is${70}^{◦}F$.

Short answer

$P=0.5$

Analise all the events that can be derived from $A,B$and$C$.

Use the Venn diagram.

Step-by-step solution

Given Information.

$A=\left\{t_{LA}=70\right\}$, In an event, the temperature in LA is$70$ degrees Fahrenheit.

$B=\left\{t_{NY}=70\right\}$, an event that the temperature in NY is $70$degrees Fahrenheit.

$C=\left\{\max \left\{t_{LA},t_{NY}\right\}=70\right\}$, an event that the minimum temperatures in LA and NY are $70$degrees Fahrenheit.

Explanation.

What is the probability of an event$D=\left\{\min \left\{t_{LA},t_{NY}\right\}=70\right\}?$

Analise the events:

$$\begin{gathered} AB=\left\{t_{NY}=70,t_{LA}=70\right\} \\ BC=\left\{t_{NY}=70,t_{LA}\le 70\right\} \end{gathered}$$

Because if both$t_{NY}=70$, and$\max \left\{t_{LA},t_{NY}\right\}=70,\Rightarrow t_{LA}\le 70$. The same for$AC$

$AC=\left\{t_{LA}=70,t_{NY}\le 70\right\}$

It $t_{NY}=70,t_{LA}=70,C$also occurs, so:

$ABC=\left\{t_{NY}=70,t_{LA}=70\right\}$

Furthermore:

$$\begin{gathered} A^{c}BC=\left\{t_{NY}=70,t_{LA}<70\right\} \\ A{B}^{c}C=\left\{t_{LA}=70,t_{NY}<70\right\} \\ AB{C}^c=\varnothing \\ A{B}^c{C}^c=\left\{t_{LA}=70,t_{NY}>70\right\} \\ {A}^{c}B{C}^c=\left\{t_{NY}=70,t_{LA}>70\right\} \\ {A}^c{B}^{c}C=\varnothing \end{gathered}$$

Because if neither of $t_{NY}$or$t_{LA}$is$70$, the maximum of the two can't be$70$.

Explanation.

Explanation.

Reading from the Venn diagram (and using$0$):

$P\left(A\right)=P\left(\left\{t_{NY}=70,t_{LA}<70\right\}\right)+P\left(\left\{t_{NY}=70,t_{LA}=70\right\}\right)+P\left(\left\{t_{NY}=70,t_{LA}>70\right\}\right)$

$P\left(B\right)=P\left(\left\{t_{NY}<70,t_{LA}=70\right\}\right)+P\left(\left\{t_{NY}=70,t_{LA}=70\right\}\right)+P\left(\left\{t_{NY}>70,t_{LA}=70\right\}\right)$

$P\left(C\right)=P\left(\left\{t_{NY}=70,t_{LA}<70\right\}\right)+P\left(\left\{t_{NY}=70,t_{LA}=70\right\}\right)+P\left(\left\{t_{NY}<70,t_{LA}=70\right\}\right)$

$P\left(D\right)=P\left(\left\{t_{NY}=70,t_{LA}=70\right\}\right)+P\left(\left\{t_{NY}=70,t_{LA}>70\right\}\right)+P\left(\left\{t_{NY}>70,t_{LA}=70\right\}\right)$

Combining these probabilities:

$$\begin{gathered} P\left(A\right)+P\left(B\right)\&=P\left(C\right)+P\left(D\right) \\ 0.4+0.3\&=0.2+P\left(D\right) \\ P\left(D\right)\&=0.5 \end{gathered}$$