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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. $0.6$
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5 Answers

+50 votes
Best answer

on Wednesday we want cs

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

answer = option C

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Nice, thanks for the visuals
+12 votes
answer - C

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
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+4 votes
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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+1 vote

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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+1 vote
I'll use obvious notations.

$P(M|C)=0.6$

$P(C|M)=0.4$

 

Since she studies either of $M$ or $C$ everyday $P(C|C)=0.4$

 

If she studies $C$ on Monday, we want her to study $C$ on Wednesday, too.

We can do this via:

$C \rightarrow C \rightarrow C$ or $C \rightarrow M \rightarrow C$

$0.4*0.4+0.6*0.4$

=> $0.16+0.24$

=> $0.4$
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