Can someone please give me a hand; Bayes Theorem?

[asanz89]

According to statistics Canada, 31.5% of all Canadians are in the 0-24 age bracket. 13.7% are in the 25-34 age bracket, 16.2% are in the 34-44 age bracket and 38.6% are in the 45+ age bracket. suppose a market research study shows that Canadians use their leisure time in different ways according to age. to be more specific, of those who are in the 45+ age group, 39% read more than 10 hours a week. Those in the 0-24 age group only 11% read more then 10+ hrs a week. The corresponding percentages for those people who read more than 10 hrs per week are 24% for those 25-34 and 27% for the 35-44 age group.

Suppose that a Canadian is randomly selected.

So I made a chart/table of the age groups and % of Canadians + the % of Canadians that read 10+ hours per week hoping it would help me out, but it didnt.

a)what is the probability that he/she reads more than 10 hours per week?

b)what ist he probability that he/she belongs to a certain age group if he/she reads more than 10 hours per week?
(calculate the probabilities that he/she belongs to 0-24 age group, 25-34 and so on)

...This is the last question of a first assignment and Im having some trouble. Can someone please get me on the right track? Thank you.

 

[soroban]

Hello, asanz89!

According to the Canadian census:

. . \(\displaystyle \begin{array}{c|| c|c|} \text{Age} & \text{Pop'n} & \text{read}> 10\text{hrs} \\ \hline 0-24 & 31.5\% & 10\% \\ 25-34 & 13.7\% & 24\% \\ 35-44 & 16.2\% & 27\% \\ 45+ & 38.6\% & 39\% \end{array}\)

Suppose that a Canadian is randomly selected.

a) What is the probability that he/she reads more than 10 hours per week?

\(\displaystyle \begin{array}{ccccc} P\bigg[\text{(age 0-24)} \wedge (>10\text{ hrs})\bigg] &=& (0.315)(0.10) &=& 0.03150 \\ \\[-3mm] P\bigg[\text{(age 25-34)}\wedge (>10\text{ hrs})\bigg] &=& (0.137)(0.24) &=& 0.03288 \\ \\[-3mm] P\bigg[\text{(age 35-44)} \wedge (>10\text{ hrs})\bigg] &=& (0.162)(0.27) &=& 0.04374 \\ \\[-3mm] P\bigg[\text{(age 45+)} \wedge (> 10\text{ hrs)}\bigg] &=& (0.386)(0.39) &=& 0.15054 \\ \\[-3mm] \hline \\[-3mm] & & \text{Total:} && 0.25866 \end{array}\)


\(\displaystyle \text{Therefore: }\;P(> 10\text{ hrs}) \;=\;0.25866\)

b) What is the probability that he/she belongs to a certain age group if he/she reads more than 10 hours per week?
. .(Calculate the probabilities that he/she belongs to 0-24 age group, 25-34, and so on.)

\(\displaystyle \begin{array}{ccccccc}P\bigg[\text{(0-24)}\,\bigg|\,(> 10\text{ hrs})\bigg] &=& \dfrac{P\bigg[\text{(0-24)} \wedge (> 10\text{ hrs})\bigg]}{P(> 10\text{ hrs})} &=& \dfrac{0.03150}{0.25866} & \approx & 0.122 \\ \\ P\bigg[\text{(25-34)}\,\bigg |\,(> 10\text{ hrs})\bigg] &=& \dfrac{P\bigg[\text{(25-34)} \wedge (> 10\text{ hrs})\bigg]}{P(> 10\text{ hrs})} &=& \dfrac{0.03288}{0.25866} & \approx & 0.127 \\ \\ P\bigg[\text{(35-44)}\,\bigg|\,(> 10\text{ hrs})\bigg] &=& \dfrac{P\bigg[\text{(35-44)} \wedge (> 10\text{ hrs})\bigg]}{P(> 10\text{ hrs})} &=& \dfrac{0.04374}{0.25866} & \approx & 0.169 \\ \\ P\bigg[\text{(45+)}\,\bigg|\,(> 10\text{ hrs})\bigg] &=& \dfrac{P\bigg[\text{(45+)} \wedge (> 10\text{ hrs})\bigg]}{P(> 10\text{ hrs})} &=& \dfrac{0.15054}{0.25866} & \approx & 0.582 \end{array}\)

 

[asanz89]

Thank you very much for the clear response and how prompt it was. You guys have set up a great site and forums here. Thanks again!

-asanz89