Julia is studying potential relationships between one having a pick-up truck and attached ear lobes. Assume a survey of 201 randomly selected persons revealed the following data:
The probability of selecting a person with no pick-up truck is 0.784.
The probability of selecting a person with detached ear lobes is 0.684.
The probability of selecting a person with no pick-up truck and detached ear lobes is 0.536.
If you were to choose a person at random from the sample, determine the probability they have a pick-up truck given that they have attached ear lobes.
(After doing another attempt on my own I ended with the answer of 0.069. Can someone check if this is correct?)
We would have the best luck of using our information of the total number of people and getting a venn diagram. So let's do that.
Favorable/possible = 0.784, so f/201 = 0.784. Thus, there are 201-157.584 = 43.416 people who have pickup trucks.
Favorable / possible = 0.684, so f/201 = 0.684. Thus, there are 201-137.484=63.516 people who have attached earlobes (ik right?)
favorable / possible = f/#of attached earlobes = f/63.516, and 43.416 + 63.516 = 106.932, so there are 0 people that have both attached earlobes and a pick up truck? I'm probably wrong, just this question is an incorrect question because it doesn't have integer number of people for each case.
See the Venn below:
Total people with attached ealobes = 68 + 248 = 316
68 of these have a p/u 68 / 316 = 21.5%
That is surely incorrect, as you have stated that the number of people surveyed with attached earlobes is greater than the number of people actually surveyed, which is not possible.
This Probability is based on percentages
example people with no pu truck is given as .784
the diagram shows
(248+536) / (148 + 68 + 248 + 536) = .784
you just have to scale the numbers
if it makes you feel better, you can just divide all of the numbers in the Venn diagram by 1000
.148 .068 .248 .536