Veritas Probability Explanation required

to be accepted at least one school =1- not to be accepted at all


not to be accepted at all =(4/5)*(4/5)*(4/5)=64/125

1-64/125=61/125

To be accepted by atleast one school is also equal to OR condition, so why cant it be 1/5 +1/5 + 1/5 = 3/5 ???

Please clarify the confusion???

LalaB wrote:to be accepted at least one school =1- not to be accepted at all


not to be accepted at all =(4/5)*(4/5)*(4/5)=64/125

1-64/125=61/125

To be accepted by atleast one school means (1 -to be accepted to NONE school at all)

if ur get OR condition, then u wont get NONE, since Or means that there is some possibibity for being accepted

karthikpandian19 wrote:

this one is quite easy
step 1: one of 3 schools he will get in :
3!/2!* 20/100*80/100*80/100=48/125
step 2: he will get in 2 in three:
3!/2!*20/100*20/100*80/100=12/125
step 3: he is able to get in all the whole three
3!/3!*20/100*20/100*20/100=1/125
totally =(48+12+1)/125= the answwer

Can any GMAT Expert explain this one?

Still i m confused????

karthikpandian19 wrote:

karthikpandian19 wrote:Can any GMAT Expert explain this one?

Still i m confused????

Ok. Let me try.
Remember: AND implies 'multiplication'. OR implies 'addition'.

Let the names of the three MBA programs be A, B and C.
If for each of A, B, C
P(accepted) = 1/5, then
for each of them
P(not accepted) = 1 - 1/5 = 4/5

Better Method:
P(accepted by atleast one school)
= 1 - P(not accepted by any school)
= 1 - P(not accepted by A) AND P(not accepted by B) AND P(not accepted by C)
= 1 - (4/5)*(4/5)*(4/5)
= 1 - (64/125)
= [spoiler]61/125[/spoiler]

Alternatively(And, this is a long method):

I think you were stuck in the explanation of this method, so I will elaborate it in detail. Read on.

P(accepted by atleast one school)
= P(accepted by one school) OR P(accepted by two schools) OR P(accepted by three schools)

P(accepted by one school)
= [P(accepted by A) AND P(not accepted by B) AND P(not accepted by C)]*[Arrange]
Why and What do we arrange?
Notice that we have considered a Yes-No-No for A,B,C but we can also have a No-Yes-No or a No-No-Yes.
The Yes,No,No can be arranged in (3!/2!) = 3 ways.
So,
P(accepted by one school)
= (1/5)*(4/5)*(4/5)*3
= 48/125

P(accepted by two schools)
= [P(accepted by A) AND P(accepted by B) AND P(not accepted by C)]*[Arrange]
= (1/5)*(1/5)*(4/5)*3
= 12/125

Note that we have arranged a Yes-Yes-No in 3!/2! = 3 ways here.
P(accepted by 3 schools)
= [P(accepted by A) AND P(accepted by B) AND P(accepted by C)]*[Arrange]
= (1/5)*(1/5)*(1/5)*1
= 1/125

Note that there is just one way of arranging Yes-Yes-Yes.
Therefore,
P(accepted by atleast one school)
= (48/125) + (12/125) + (1/125)
= [spoiler]61/125[/spoiler]

Same Answer. Hooraah!