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Aishwarya studies either computer science or mathematics everyday. If she studies computer science on a day, then the probability that she studies mathematics the next day is $0.6$. If she studies mathematics on a day, then the probability that she studies computer science the next day is $0.4$. Given that Aishwarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday?

1. $0.24$
2. $0.36$
3. $0.4$
4. $0.6$

Thanks for the clarification. I can understand that solution is provided considering probability of studying CS on monday is 1

My concern is how do we decide that ? This phrase is “either computer science or mathematics everyday” is bit ambiguous. Does it mean on a particular day it can be either CS (or) maths but not both as its given as “either-or” ?

If thats the case, then in what scenario we can assume that P(CS) = 0.5 ? Usually in toss of coin when question says outcome can be either head (or) tail, we do a similar assumption that P(H) = P(T) = 0.5. So, wanted to know why a similar approach cant be applied here

@sk91 Question is quite clear, in the question it is mentioned that “Aishwarya studies either computer science or mathematics everyday” but probability is not mentioned so we can’t just assume 0.5, maybe there is probability of studying CS is 0.6 or 0.5 or 0.4 we don’t know, since it is not mentioned that there is equal probability of studying CS and Math on a given day.

But what we know is, if she studies CS on one day then on next day she will study Math will have probability 0.6. And if  she study Math next day she will study CS is 0.4. So probability of studying any sub is dependent on what she has studies last day. It is just co-incidence that the probability of studying CS is 0.4 independent of what she have studied last day.
Question could have been framed in other way that if she studies CS today then probability of studying Math next day is 0.6 and if she studies Math today then probability of studying CS next day is 0.3 or anything then the below mentioned approach should be used.

@sk91 In these type of questions, we can’t assume equi-probability unless mentioned. And one more tip is to use the tree method in all conditional probability qs. They work like charm.

on Wednesday we want cs

required probability = $0.6 \times 0.4 + 0.4 \times 0.4 = 0.4$

Nice, thanks for the visuals

prob of studying CS on Tue = 0.4

prob of studying CS on Wed given that it was studied on Tue = 0.4 x 0.4 = 0.16

prob of studying Math on Tue = 0.6

prob of studying CS on Wed given that Math was studied on Tue = 0.6 x 0.4 = 0.24

prob = 0.16 + 0.24 = 0.4
I think both the events are independent events bcoz the prob she studies cs any particular day is .4 and she studies Maths any particular day is  .6 .

Given that she study cs on Mon then,∣

prob(Cs on Wed) = P(Cs on Tue , Cs on Wed) +P(Maths on Tue,Cs on Wed)

=P(Cs on Tue).P(Cs on Wed)+P(Maths on Tue).P(Cs on Wed)

=.4*.4 +.6*.4 =.4

In short, prob(Cs on Wed ∣ Cs on Mon ) = prob(Cs on Wed) = .4      // Independent
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Given that Aishwarya studied CS on Monday,so possible scenarios are-:

$\text{MON TUE WED}$

$\text{1.CS MATHS MATHS}$

$\text{2.CS MATHS CS}$

$\text{3.CS CS MATHS}$

$\text{4.CS CS CS}$

case $1$ and $3$ will be discarded as the question is asking the probability that she studies computer science on Wednesday

Given that

• $\text{CS} \rightarrow \text{MATHS}=0.6$
• $\text{MATHS} \rightarrow CS=0.4$
• $\text{CS} \rightarrow \text{CS} =0.4$
• $\text{Maths} \rightarrow \text{Maths} =0.6$

Considering Case $2$ and $4$ we need to find their respective probability.

case $2 -: 0.6 \times 0.4$

case $4-: 0.4 \times 0.4$

Required proabaility=$0.24+.16=0.4$

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