The Gateway to Computer Science Excellence

+16 votes

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?

- $0.24$
- $0.36$
- $0.4$
- $0.6$

+50 votes

Best answer

+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

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

+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

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

+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=$0.24+.16=0.4$

+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 $C \rightarrow M \rightarrow C$

$0.4*0.4+0.6*0.4$

=> $0.16+0.24$

=> $0.4$

$P(M|C)=0.6$

$P(C|M)=0.4$

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 $C \rightarrow M \rightarrow C$

$0.4*0.4+0.6*0.4$

=> $0.16+0.24$

=> $0.4$

52,345 questions

60,469 answers

201,795 comments

95,272 users