Official Solution:

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 stores has it. Since the probability that a convenience store has cans of iced tea is \(\frac{1}{2}\) then the probability that a convenience store does not have cans of iced tea is also \(\frac{1}{2}\), (\(1-\frac{1}{2}=\frac{1}{2}\)), so \(P(NNN)=( \frac{1}{2} )^3=\frac{1}{8}\).


Answer: A
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P(BBB) = 1/8 is the probability that he buys in ALL 3 shops, so 1-1/8 is the probability that he will NOT be able to buy in ALL 3 shops, which is not not the same as not buying in any.
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No. We want P(No iced tea, No iced tea, No iced tea). NNN can be arranged only in one way. So, no need to multiply by 3!.
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Out of curiosity -

Is this approach correct ? -

Probability of him not able to buy after 1st store ( Failure in store 1) = 1/2
Probability of him not able to buy after 2nd store (Failure in store 1 x Failure in store 2) = 1/2 x 1/2
Probability of him not able to buy after 3rd store (Failure in store 1 x Failure in store 2 x Failure in store 3) = 1/2 x 1/2 x 1/2

Probability of him not getting in 3 store = Failure of above three probabilities = 1/2 + 1/4 + 1/8 = 7/8

@Bunnel - Can you please confirm whats wrong with above approach?

Oh! after re-reading it I see where I went wrong! Thanks Bunnel!