The probability that a convenience store has cans of iced tea in stock

Question

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}

Bunuel · Accepted Answer

Official Solution:

The probability that a convenience store has cans of iced tea in stock is 50%. If James stops by three 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 have it in stock. Given that the probability a convenience store has cans of iced tea is \(\frac{1}{2}\), the probability a store does not have it is also \(\frac{1}{2}\), (\(1-\frac{1}{2}=\frac{1}{2}\)). Therefore, the probability that all three stores don't have the iced tea is \(P(NNN)=( \frac{1}{2} )^3=\frac{1}{8}\).

Answer: A

For other versions of this question check this: https://gmatclub.com/forum/the-probabil ... 28689.html

AbhayPrasanna · Answer

Let the stores be A, B and C. The probability that each store does NOT have iced tea is 1/2

Therefore the probability that all 3 stores do not have ice tea in stock = 1/2 * 1/2 * 1/2

Ans: 1/8

144144 · Answer

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}

Guys, can someone explain me how is it possible that the chances for him to buy and not to buy is the same probability?

if he have 1/8 chance not to find it, how much chance he will have to find it and why. thanks.

fluke · Answer

Since the probability of the event occurring is equally likely as to the probability of the event not occurring, then both the chances of finding the can or not finding the can after visiting 3 stores will be 1/8.

I see it like this;

If he goes to the first store and finds the can. He won't go to the next store, right!!

But, if the question says; he visited 3 stores for the can, we can infer that he didn't find it in the first two stores.

P(Finding) = 50% = 1/2

P(Not finding) = 1/2

He visited 3 stores; he didn't find it;

He didn't find in 1st store AND he didn't find it in 2nd store AND he didn't find it in 3rd store
1/2*1/2*1/2(this last 1/2 is the probability for Not finding the can even in the 3rd store) = 1/8

He visited 3 stores; he found it(in the 3rd store);
He didn't find in 1st store AND he didn't find it in 2nd store AND he FOUND it in 3rd store
1/2*1/2*1/2(this last 1/2 is the probability for successfully finding the can) = 1/8

Bunuel · Answer

No, above is not correct. James will be able to buy a can of iced tea if  at least one  of the shops has it and the probability that  at least one  of the shops has an iced tea is 7/8 (so it is indeed 1-1/8=7/8).

Each shop has 2 options either to have an iced tea or not, so there are 2*2*2=8 outcomes possible. James will not be able to buy a can of iced tea if neither of the shops has it, P=1/2*1/2*1/2 in all other 7 options at least one has it thus he'll be able to buy it.

144144 · Answer

WOW amazing explanation. thanks as usual.

144144 · Answer

so the exact way to calculate will be (only for my understanding)

find on the 1st store - 1/2
2nd store - 1/2*1/2 = 1/4
3rd store - 1/2*1/2*1/2=1/8

so 1/2+1/4+1/8 = 7/8 for him to find it.

am i right?
 
thanks.

Bunuel · Answer

Direct probability approach:
One store has an iced tea: P(YNN)=3*1/2*1/2^2=3/8 (multiplying by 3 as scenario YNN can occur in 3!/2!=3 ways);
Two shops have an iced tea: P(YYN)=3*1/2^2*1/2=3/8;
All three shops have an iced tea: P(YYY)=1/2^3=1/8;

P=3/8+3/8+1/8=7/8.

gmat1220 · Answer

This one from Gmatprep so I remember it. It is analogous to the above question. I think :wink:

On his drive to work, Leo listens to one of 3 radio stations, A, B, or C. He first turns to A. If A is playing a song he likes, he listens to it; if not, he turns to B. If B is playing a song that he likes, he listens to it; if not, he turns to C. If C is playing a song he likes, he listens; if not, he turns off the radio. For each station, the probability is 0.3 that at any given moment the station is playing a song Leo likes. On his drive to work, what is the probability that Leo ill hear a song he likes? a. 0.027 b. 0.09 c. 0.417 d. 0.657 e. 0.9 D

Bunuel · Answer

Discussed here: radio-station-104659.html and can be solved as 144144 suggested in this post: m03-110379.html#p884977 The desired probability is the sum of the following events: A is playing the song he likes - 0.3; A is not, but B is - 0.7*0.3=0.21; A is not, B is not, but C is - 0.7*0.7*0.3=0.147; P=0.3+0.21+0.147=0.657. Or: 1-the probability that neither of the stations is playing the song he likes: P=1-0.7*0.7*0.7=0.657. Answer: D.

Sachin9 · Answer

bro bunuel,

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

Narenn · Answer

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

adityamurlidharan · Answer

Bruenell, I believe that the solution should be 1/2. The person is going to three stores, that means probability of not getting the can in the first two stores is 1 and the last one is 1/2. hence it should be 1*1*1/2

Bunuel · Answer

Why is the probability of not getting ice tea in the first two shops 1? The question clearly says that "The probability that a convenience store has cans of iced tea in stock is 1/2", so the probability of not having is also 1/2. P(No, No, No) = 1/2^3 = 1/8.

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The probability that a convenience store has cans of iced tea in stock

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The probability that a convenience store has cans of iced tea in stock

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