It is given that he walks to school on a particular Tuesday, then we need to find the probability that he will go by bus on Thursday.
So now, we can have two cases.
"Case1:" He will walk on Wednesday too.
Now, it is given that if one day he walks to school, then the probability that he go by bus on next day is 2/5.
So, if one day he walks to school, then the probability that he will go to school by walk on next day too will be "1-(2\/5)=3\/5"
So, the probability that he will walk on Wednesday given that he walked on previous day too will be equal to "3\/5"
And now the probability that he will go by bus on Thursday will be equal to "2\/5" because he walked on previous day.
So, by multiplication theorem, the probability that he will go by bus on Thursday, given that he walks on Tuesday will be equal to "(3\/5)*(2\/5)=6\/25"
"Case2:" He will go by bus on Wednesday
Probability that he will go by bus on Wednesday given that he go by walk on Tuesday will be equal to "2\/5"
Now, the probability that he go by bus on Thursday assuming he go by bus on Wednesday will be equal to "7\/10"
So by multiplication theorem probability that he go by bus on Thursday, given that he walks on Tuesday will
be equal to "2\/5)\u2217(7\/10)=7\/25"
Now combining both the cases by addition theorem, we can find the probability that he go by bus on Thursday given that he go by walk on Tuesday, which will be equal to "(6\/25)+(7\/25)=13\/25"
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There are three boxes, one of which contains a prize. A contestant is given two chances, such that if he chooses the wrong box in the first round, that box is removed from the selection and he then chooses between the two remaining boxes. 1. What is the probability that the contestant wins? 2. Does the contestant’s probability of winning increases on the second round?
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