Sachin9 wrote:

Bunuel wrote:

Smita04 wrote:

Bunuel, can you please show how to solve this one?

The probability that a convenience store has cans of iced tea in stock is 50%. If James stops by 3 convenience stores on his way to work, what is the probability that he will not be able to buy a can of iced tea?A. \(\frac{1}{8}\)

B. \(\frac{1}{4}\)

C. \(\frac{1}{2}\)

D. \(\frac{3}{4}\)

E. \(\frac{7}{8}\)

James won't be able to buy a can of iced tea if none of the shops has it: P(NNN)=1/2^3=1/8.

Answer: A.

For other versions of this question check this:

the-probability-that-a-convenience-store-has-cans-of-iced-128689.htmlbro bunuel,

how to decide if we have to add or if we have to multiply probabilities?

FUNDAMENTAL PRINCIPAL OF MULTIPLICATION :- if their are two jobs such that 1st job can be completed in 'm' ways and 2nd job can be completed in 'n' ways then two job successively can be done in m X n ways. i.e. doing 1st

AND 2nd Job.

so in our case James didn't get the iced tea in 1st shop AND in 2nd shop AND in 3rd shop. -------- > (1/2) X (1/2) X (1/2) --------> (1/2)^3 ------> 1/8

Take another example :- In a class of 8 boys and 10 girls, teacher wants to select 1 boy

AND 1 girl

number of ways selecting a boy = 8 ---------> number of ways selecting a girl = 10

so number of ways of selecting 1 boy

AND 1 girl = 8 X 10 = 80

FUNDAMENTAL PRINCIPAL OF ADDITION :- if their are two jobs such that 1st job can be completed in 'm' ways and 2nd job can be completed in 'n' ways then either of the job can be performed in m + n ways. i.e. doing 1st

OR 2nd Job.

Take the same example :- In a class of 8 boys and 10 girls, teacher wants to select 1 boy

OR 1 girl (or we can say wants to select a student)

number of ways selecting a boy = 8 ---------> number of ways selecting a girl = 10

so number of ways of selecting 1 boy

OR 1 girl = 8 + 10 = 18

Now in our case if the question had been

what is the probability that james will get the iced tea in exactly one shop he gets tea at 1st shop----------AND---didn't get at 2nd shop---AND---didn't get at 3rd shop------------(1/2) X(1/2) X (1/2)

---------------------------------------------------OR--------------------------------------------------------------------------------+

he didn't get tea at 1st shop---AND---gets at 2nd shop----------AND---didn't get at 3rd shop------------(1/2) X(1/2) X (1/2)

---------------------------------------------------OR--------------------------------------------------------------------------------+

he didn't get tea at 1st shop---AND---didn't get at 2nd shop---AND---gets at 3rd shop-------------------(1/2) X(1/2) X (1/2)

(1/8) + (1+8) + (1/8) = 3/8

After practice you will be able to solve it in a way (1/8)(3!/2!)

Hope it clears!

Regards,

Abhijit

_________________

Live Q&A with B-School Adcoms - 2015 Season

Be the coolest guy in the MBA Forum - Be a threadmaster!

Have a blog? Feature it on GMAT Club!

All MBA Resources All 'Sticky' Topics at one place

Please share your profiles for this application season: 2015 Profiles w/ Admit/Dings Results!

GMAT Club Premium Membership - big benefits and savings

Next Generation GMATClub CATS with Brilliant Analytics.