Question

According to records from 50 years, there is a 2/3 chance of June temperatures being above 80 F. Azul needs to find the probability that it will be above 80 F for 20 or more days this June. There are 30 days in June. If we use a number cube with the numbers 1-6 and assign 1-4 as a temperature above 80 F and 5-6 as a temperature 80 F or below, what is the best way to peform the simulation?

Answer 1

Record of june temperature of last 50 years

Probability of June temperature being above 80° F from last 50 years

`=(2)/(3)`

It means , out of 30 days ,there are only 20 days , in which temperature is above 80°F.

⇒Probability that it will be above 80° F for 20 or more days this June

`=(20+x)/(30)`

where, x is the number of extra days after 20 days in month of june, when temperature is above 80°F.

Value of this Probability will be in between

`(2)/(3)\leq P \leq 1`

Now, what Azul wants

To find the probability that it will be above 80° F for 20 or more days this June.

The Simulation given is

⇒There are 30 days in June. If we use a number cube with the numbers 1-6 and assign 1-4 as a temperature above 80° F and 5-6 as a temperature 80° F or below.

Total number of numbers on number cube ={1,2,3,4,5,6}

Favorable Outcome={1,2,3,4}=Temperature above 80° F

Temperature 80° F or below ={5,6}

⇒Probability of getting number {1,2,3,4} on the number cube when rolled once,that is Temperature above 80° F

`=(4)/(6)\\\\=(2)/(3)`

⇒Probability of getting number {5,6} on the number cube when rolled once, that is Temperature 80° F or below

`=(2)/(6)\\\\=(1)/(3)`

This is Incorrect Simulation, as there are chances that , there can be more than 20 days out of 30 days when Temperature goes above 80° F.But Here we have got Probability of temperature being above 80° F for 20 days only, not more than 20 days, which is
`=(2)/(3)`

Not,a number

`(2)/(3)< \text{Probability} \leq 1`.