Probility_questions

[amanvij.1989]

I am facing some doubts in probability ;It would be really great if anyone could help ........

Q1)Probability of raining on tuesday,wednesday and thursday is 1/6 for each day and probability of raining on monday is 2/5 and that for friday is 4/5 .What is the probability that the rain will occur in any of the given 5 days ?

a)5/72
b)7/72
c)13/72
d)67/72
e)1

The Answer is(d) 67/72

I did'nt understood y ?


Q2)Two distinctive numbers are taken from 1.2.3---28 ,what is the probability that their sum is less than 13?

a)1/21
b)4/63
c)5/63
d)1/9
e)8/63

Answer is (c) 5/63

What i am getting is 55/189

[rdm750]

I am facing some doubts in probability ;It would be really great if anyone could help ........
Q2)Two distinctive numbers are taken from 1.2.3---28 ,what is the probability that their sum is less than 13?

a)1/21
b)4/63
c)5/63
d)1/9
e)8/63

Answer is (c) 5/63

What i am getting is 55/189

denominator is 28C2 = (28*27) / 2
numerator well not an elegant way to do it but u can line up and count all the ways each of the respective sums can be computed <13 quickly..
(note numbers 1- 11 add up to <13 and sums of 1 and 2 are 0 since integers have to be non negative and distinct)

Sum ways to get the sum
12 5
11 5
10 4
9 4
8 3
7 3
6 2
5 2
4 1
3 1
2 0 integers have to be non negative and distinct
1 0 integers have to be non negative and distinct

Total ways sum<13 = 5+5+4+4+3+3+2+2+1+1 = 30

P (sum<13) = (Total ways sum<13) / (total ways to select 2 integers from 28)
= 30 / ( 28 C 2)
= ( 30 *2 ) / (28 *27)
= 5 / 63

[rdm750]

I am facing some doubts in probability ;It would be really great if anyone could help ........

Q1)Probability of raining on tuesday,wednesday and thursday is 1/6 for each day and probability of raining on monday is 2/5 and that for friday is 4/5 .What is the probability that the rain will occur in any of the given 5 days ?

a)5/72
b)7/72
c)13/72
d)67/72
e)1

The Answer is(d) 67/72

I did'nt understood y ?

The first one seems easier than the second one.......

P(rain will occur in any of the 5 days) = 1 - P(Rain not occuring at all)
rain not occurring: 5/6 on tuesday,wednesday and thursday
rain not occurring: friday is 1/5
rain not occurring: Monday is 3/5
P(Rain not occuring at all) = (5/6)*(5/6)*(5/6)*(1/5)*(3/5)
= 5 / 72
There is only way to get M,T,W,T,F no rain (5C5) =1 as well
so P(rain will occur in any of the 5 days) = 1 - 5/72 = 67/72

The Longer route would sum up rain occuring on each possible combinations of days multiplied by their respective probabilities and summing them up to get 67/72 (tedious)

Great problems....are they from MGRE?? I haven't seen them on my mgre tests......

Regards,
Rohit M

[amanvij.1989]

Hey thanks a ton Rohit ,i was solving them wrongly...BTW these are from Jamboree course material...
Will be posting more of these .. :)
Thanks once again ...TC

[jen]

Hi, this is Jennifer (instructor). I agree with Rohit's response.

1 - (probability of NO RAIN) = answer

The probability of no rain = 3/5 x 5/6 x 5/6 x 5/6 x 1/5.

How do you know to do this?

We call this the "1-x" strategy, which you will find covered in Manhattan GRE Strategy Guide #5, "Word Problems."

You do this whenever calculating the probability of failure is easier than calculating the probability of success.

[amanvij.1989]

Hey Jen ,
Thanks for your generous help.....

[tommywallach]

We're very glad to help!

-t

P.S. Thankfully, you won't get more than one tough probability question on the GRE (if that!), so definitely don't stress about them too much!