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Bunuel
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suhasreddy
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Bunuel
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


WHAT IS WRONG WITH THE FOLLOWING APPROACH?

The probability that he will be able to buy a can of iced tea after visiting third store is 1/2*1/2*1/2=1/8.
so the probability that he wont be able to buy = 1-1/8=7/8?

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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Hi Bunuel.

Should we not consider that the person could visit the stores in any order and multiply the probability with 3! ?

Regards
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Hi Bunuel.

Should we not consider that the person could visit the stores in any order and multiply the probability with 3! ?

Regards

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?
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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

Can anyone confirm whats wrong with above approach?

Each successive case after the first one includes the previous one.

1/8 is already the probability that there is no candy in either of the stores.
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Again one of the AND problem.

AND means multiplication.

He goes to first store "AND" he goes to second store "AND" he goes to third store.

1/2 * 1/2 * 1/2 = 1/8
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This is high-quality question
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Bunuel
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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I like the solution - it’s helpful.
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