Condition Probability

[mattgad]

A boy always either walks to school or goes by bus. If one day he goes to school by bus, then the probability that he goes by bus the next day is 7/10. If one day he walks to school, then the probability that he goes by bus the next day is 2/5.

(a) Given that he walks to school on a particular Tuesday, draw a tree diagram and hence find the probability that he will go to school by bus on thursday of that week.

(b) Given that the boy walks to school on tuesday and thursday of that week, find the probability that he will also walk to school on Wednesday.

So far, I have made this tree diagram.

For (a), Do I do (W * B * B) + (W * W * B)?

 

[rahian2k]

[deleted]

 

[soroban]

Hello, mattgad!

A boy always either walks to school or goes by bus.
If one day he goes to school by bus, then the probability that he goes by bus the next day is 0.7.
If one day he walks to school, then the probability that he goes by bus the next day is 0.4.

(a) Given that he walks to school on a particular Tuesday, draw a tree diagram
and hence find the probability that he will go to school by bus on thursday of that week.

(b) Given that the boy walks to school on Tuesday and Thursday of that week,
find the probability that he will also walk to school on Wednesday.

In both questions, we are <u>given</u> that he walked on Tuesday.
. . We begin our tree diagram with that event.

Tues                  --walk--
                     /         \
                 0.6/           \0.4
                   /             \
Wed             walk               bus
                 / \              / \
             0.6/   \0.4      0.3/   \0.7
               /     \          /     \
Thurs       walk     bus     walk     bus

Reading down the branches, we have all the possible outcomes and their probabilities:

. . \(\displaystyle Pr(WWW)\,=\,(0.6)(0.6)\,=\,0.36\)
. . \(\displaystyle Pr(WWB)\;=\,(0.6)().4)\,=\,0.24\)
. . \(\displaystyle Pr(WBW)\;=\,(0.4)(0.3)\,=\,0.12\)
. . \(\displaystyle Pr(WBB)\;\;=\,(0.4)(0.7)\,=\,0.28\)


(a) \(\displaystyle Pr(\text{bus Thurs})\:=\r(WWB)\,+\,Pr(WBB)\:=\:0.24\,+\,0.28\:=\:0.52\)


(b) .We will use Bayes' Theorem:
. . . \(\displaystyle Pr(\text{walked Wed }|\text{ walked Thurs})\;=\;\frac{Pr(\text{walked Wed and Thurs})}{Pr(\text{walked Thurs})}\)


. . The numerator is the prob. that he walked all three days: .\(\displaystyle 0.36\)

. . In part (a) we found that \(\displaystyle Pr(\text{bus Thurs})\,=\,0.52\) . . . Hence: \(\displaystyle Pr(\text{walk Thurs})\,=\,0.48\)

The answer is: .\(\displaystyle \frac{0.36}{0.48}\:=\:0.75\)