Probability question

[prerana.joshi2]

Suppose that the probability that it will rain of Saturday is 0.60. The Probability that it will rain on Sunday is also 0.60. If it rains on one day, there is a0.80 chance that it will rain the next day. What is the chances that it will rain at least one day during the weekend.

My solution

P(A) =
Rain on Saturday* No rain on Sunday + No rain on Sat * Rain on Sun + Rain on Sat * Rain on Sun
P(A) = 0.60*0.20 + 0.40*0.60 + 0.6*0.8
= .12+.24+.48 = 0.84
Or else
P(A) = 1- No rain on Sat* No rain on Sun
P(A) = 1- 0.60*0.60 = 0.84 => My answer which is wrong

I don’t know the right answer, but it should be one of these

0.72
0.40
0.52
0.62

[ericaa.wang]

0.62?

[mithunsam]

Answer should be .84

Total Probability is
P(No Sat, No Sun) + P(Sat, No Sun) + P(No Sat, Sun) + P(Sat, Sun)=
.4*.4 + .6*.2 + .6*.4 + .6*.8 = .16 + .12 + .24 + .48 = 1.00 (That means, we considered all probabilities).

For the particular question, we just have to calculate
P(Sat, No Sun) + P(No Sat, Sun) + P(Sat, Sun) = .6*.2 + .6*.4 + .6*.8 = .84

[shaji]

The fastest approach is indeed the reverse gear i.e. Prob(at least rain in the weekend)=1-Prob(no rain in the weekend).

Prob(No rain on Sunday if no rain on Saturday)=(0.2*0.6)/0.4=0.3
This implies that Prob(no rain in the weekend)=(1-0.3)*0.4=0.28.; and the required Prob=1-0.28=0.72.
The correct answer is indeed 0.72.
An interesting problem with practical importance!!!

[mithunsam]

The fastest approach is indeed the reverse gear i.e. Prob(at least rain in the weekend)=1-Prob(no rain in the weekend).

Prob(No rain on Sunday if no rain on Saturday)=(0.2*0.6)/0.4=0.3
This implies that Prob(no rain in the weekend)=(1-0.3)*0.4=0.28.; and the required Prob=1-0.28=0.72.
The correct answer is indeed 0.72.
An interesting problem with practical importance!!!

Not correct. Probability to rain on Sunday increases when it rains on Saturday. However, the question didn't say that the reverse is true. (You cannot assume this). If that were the case, then the .6 chance to rain on Sunday will become immaterial.

[shaji]

The fastest approach is indeed the reverse gear i.e. Prob(at least rain in the weekend)=1-Prob(no rain in the weekend).

Prob(No rain on Sunday if no rain on Saturday)=(0.2*0.6)/0.4=0.3
This implies that Prob(no rain in the weekend)=(1-0.3)*0.4=0.28.; and the required Prob=1-0.28=0.72.
The correct answer is indeed 0.72.
An interesting problem with practical importance!!!

Not correct. Probability to rain on Sunday increases when it rains on Saturday. However, the question didn't say that the reverse is true. (You cannot assume this). If that were the case, then the .6 chance to rain on Sunday will become immaterial.

It appears that you have missed a vital point that the occurrence of rain on Sunday is conditional on what happens on Saturday.

The reverse is logical inference not an 'assumption', which mathematics does not allow. Notice that the the 0.6 prob to rain on Sunday is critical to evaluate the probabilities, I have done in my post above.

[mithunsam]

The fastest approach is indeed the reverse gear i.e. Prob(at least rain in the weekend)=1-Prob(no rain in the weekend).

Prob(No rain on Sunday if no rain on Saturday)=(0.2*0.6)/0.4=0.3
This implies that Prob(no rain in the weekend)=(1-0.3)*0.4=0.28.; and the required Prob=1-0.28=0.72.
The correct answer is indeed 0.72.
An interesting problem with practical importance!!!

Not correct. Probability to rain on Sunday increases when it rains on Saturday. However, the question didn't say that the reverse is true. (You cannot assume this). If that were the case, then the .6 chance to rain on Sunday will become immaterial.

It appears that you have missed a vital point that the occurrence of rain on Sunday is conditional on what happens on Saturday.

The reverse is logical inference not an 'assumption', which mathematics does not allow. Notice that the the 0.6 prob to rain on Sunday is critical to evaluate the probabilities, I have done in my post above.

I didn't miss that... If it rains on Saturday, Sunday's probability to rain is .8 (P for not to rain is .2). If it doesn't rain on Saturday, Sunday's probability to rain is .6 (P for not to rain is .4).

The question, did state that P(Sunday Rain) will increase, if it rains on Saturday. However, since it didn't state otherwise, P(No rain on Sunday) will remain .4 if there isn't any rain on Saturday. Therefore, P(No rain for both days) will be P(no rain on Sat) * P(no rain on Sun) = .4*.4 =.16.

Therefore, P(Rain on at least 1 day) = 1-.16 =.84

[shaji]

"I didn't miss that... If it rains on Saturday, Sunday's probability to rain is .8 (P for not to rain is .2)." Good U did and the understanding is correct.
"f it doesn't rain on Saturday, Sunday's probability to rain is .6 (P for not to rain is .4).". This is erroneous!. See my computation above. The correct probability is 0.3 and not 0.6.

It should alert you of the fallacy since your solution 0.84 doesn't feature in the answer choices. Interesting! the question setter didn't think of this trap.

[mithunsam]

"I didn't miss that... If it rains on Saturday, Sunday's probability to rain is .8 (P for not to rain is .2)." Good U did and the understanding is correct.
"f it doesn't rain on Saturday, Sunday's probability to rain is .6 (P for not to rain is .4).". This is erroneous!. See my computation above. The correct probability is 0.3 and not 0.6.

It should alert you of the fallacy since your solution 0.84 doesn't feature in the answer choices. Interesting! the question setter didn't think of this trap.

You are saying P(No rain on Sun, if no rain on Sat) = .3. That means, P(Rain on Sun, if no rain on Sat) = 1-.3 =.7. This is not correct. Because, the problem itself says that the P(Rain on Sun) is .6 and it will increase to .8 if it rains on Saturday.

If there is no rain on Saturday, then the events are independent. Therefore, there will not be any change in probabilities, if it didn't rain on Saturday.

[kthom83]

"I didn't miss that... If it rains on Saturday, Sunday's probability to rain is .8 (P for not to rain is .2)." Good U did and the understanding is correct.
"f it doesn't rain on Saturday, Sunday's probability to rain is .6 (P for not to rain is .4).". This is erroneous!. See my computation above. The correct probability is 0.3 and not 0.6.

It should alert you of the fallacy since your solution 0.84 doesn't feature in the answer choices. Interesting! the question setter didn't think of this trap.

.3 doesn't seem correct to me. Since it is not a GMAT problem, I would read too much into the choices the author provided. What is the source? By the by, how you will arrive at 1 when you add all the probabilities together?

[shaji]

"You are saying P(No rain on Sun, if no rain on Sat) = .3". This is not what I meant.I agree there was a typo in my initial post. which should read-"Prob(Rain on Sunday if no rain on Saturday)=(0.2*0.6)/0.4=0.3. P(No rain on Sun, if no rain on Sat) = 0.7.. Regret the confusion in my first response. Notice the correction in my later post.

I repeat, for the last time,"the problem itself says that the P(Rain on Sun) is .6 and it will increase to .8 if it rains on Saturday. " is the wrong understanding. Prob 0.8 is the probability to rain on Sunday if it rains on Saturday. So 0.8 is the conditional probability to rain on Sunday not the absolute probability..

Notice the fallacy: 1-(0.4*0.4)=0.84.This is true if and only if chances for rain on Saturday & Sunday are independent.

[poussin0072000]

Shaji > I think your approach was right... but...

0.2*0.6 is Prob(No rain on Sunday if rain on Saturday) * Prob(rain on Saturday), which is also = Prob(No rain on Sunday AND rain on Saturday)

but what you were looking for was Prob(No rain on Sunday if no rain on Saturday) right?

well,
Prob(No rain on Sunday if no rain on Saturday)= Prob(No rain on Sunday AND no rain on Saturday) / Prob(No rain on Saturday)

so to sum up, you calculated
Prob(No rain on Sunday AND rain on Saturday) / Prob(No rain on Saturday)
instead of
Prob(No rain on Sunday AND no rain on Saturday) / Prob(No rain on Saturday)

Your approach was right, but at the end we get 0.84

-------

By the way 0.72 is the prob it rains one day (only one) in the w-e, which is Prob(Rain on Saturday)+Prob(Rain on Sunday) - Prob(Rain on both days)
As Prob(Rain on both days) = Prob(Rains on Sunday if rains on Saturday)*Prob(Rains on Saturday)=0.8*0.6
Then
Prob(Rains on one day only) = 0.6+0.6-0.8*0.6 = 0.72

[shaji]

Poussin- You got the argument correct! A perfect case of stimulating mathematical logic. Perfect indeed, but with a slight error.

"(Rains on one day only+ Rain on both days) = 0.72 is the required probability!!-"What is the chances that it will rain at least one day during the weekend??.
A quick audit of all possibilities To establish the point
Probability to rain on both days=0.48(0.8*0.6)
Probability to rain on Sunday and not on Saturday=0.12(0.2*0.6)
Probability to rain on Saturday and not on Sunday=0.12(0.3*0.4)
Probability no rain on both days=0.28(0.4*0.7)

[kthom83]

Probability to rain on Sunday and not on Saturday=0.12(0.2*0.6)

If it doesn't rain on Saturday, then the events are independent. P(Not Rain on Saturday) will remain as .4 and P(Rain on Sunday) will remain as .6. Combined, P(No Rain on Sat, Rain on Sun) will be .4*.6=.24

[shaji]

[Kthom83-"If it doesn't rain on Saturday, then the events are independent."-Incorrect.Whatever happens on Sunday is conditional to what happens on Saturday, otherwise, refer to my earlier post, the problem becomes trivial!