The probability that a customer at a certain bar will purchase a beer

Question

The probability that a customer at a certain bar will purchase a beer is 0.75. If four customers visit the bar on a certain evening, what is the probability that at least one of them will purchase a beer ?

A. 1/256
B. 82/256
C. 1/2
D. 175/256
E. 255/256

A. 1/256
B. 82/256
C. 1/2
D. 175/256
E. 255/256

talwarmeister · Answer

Probability that a customer at a certain bar will purchase a beer = 0.75 or 3/4
Probability that a customer at a certain bar will not purchase a beer then = 1-3/4= 1/4
Probability that none out of 4 men will buy beer= (1/4)^4= 1/256

probability that at least one of them will purchase a beer= 1- Probability that none out of 4 men will buy beer= 1-1/256= 255/256

So E is the answer.

gmatophobia · Answer

P(at least 1) = 1 - P(none)

P(buying a beer) =  \frac{3}{4}

P(not buying a beer) =  \frac{1}{4}

P(at least one of them will purchase a beer) = 1 - (none of the 4 men will purchase a beer)

P(at least one of them will purchase a beer) =  1 - (\frac{1}{4})^4

P(at least one of them will purchase a beer) =  1 - \frac{1}{256}

P(at least one of them will purchase a beer) =  \frac{255}{256}

Option E

achloes · Answer

gmatophobia

Thanks for your explanation. However, I'd like to understand why I'm getting a different answer with my approach.

Probability of buying a beer = 3/4 
Probability of not buying a beer = 1/4

1 out of 4 buys a beer: 3/4 x 1/4 x 1/4 x 1/4 = 3/256

2 out of 4 buy beers: 3/4 x 3/4 x 1/4 x 1/4 = 9/256

3 out of 4 buy beers: 3/4 x 3/4 x 3/4 x 1/4 = 27/256

All 4 buy beers: 3/4 x 3/4 x 3/4 x 3/4 = 81/256

Adding all up as any of these situations are probable = 120/256

Thanks in advance!

gmatophobia · Answer

Hey  achloes

In your approach, you've fixed the probability of a particular customer and have not considered other possibilities that may exist. For example, let's take the case in which only one customer buys beer and the other three do not buy.

Let's assume there are four customers -  P_1 , P_2 .. P_4

In this approach, you've considered that only one of the customers (say  P_1 ) buys the beer and the rest ( P_2  to  P_4 ) do not buy beer.

While this is a correct case to consider, this is NOT the only case. There are three more cases to consider in this arrangement.

For example -  P_2  buys beer and the rest of the customers do not buy, similarly, other cases are -  P_3  buys a beer and the rest don't and   P_4  buys a beer and the rest don't

In a nutshell - the probability of ONLY one customer buying beer and the rest not buying =  P_1  buys and others don't buy +  P_2  buys and others don't buy ..... +  P_4  buys and others don't buy

=  4 * \frac{3}{4} * \frac{1}{4} * \frac{1}{4} * \frac{1}{4}

This expression can be also obtained by computing the below steps -

Step 1) Out of four customers, select one who buys beer = this can be done in   ^4C_1   ways 
Step 2) Calculate the required probability =   \frac{3}{4} * \frac{1}{4} * \frac{1}{4} * \frac{1}{4}

Net probability = Step 1 * Step 2

=  ^4C_1 * \frac{3}{4} * \frac{1}{4} * \frac{1}{4} * \frac{1}{4}

=  4 * \frac{3}{4} * \frac{1}{4} * \frac{1}{4} * \frac{1}{4} = \frac{12}{256}

The above two steps should be followed for each case that you've considered -

Case 2  : 2 out of 4 buys beers

Step 1) Out of four customers, select two who buy beer = this can be done in   ^4C_2   ways 
Step 2) Calculate the required probability =   \frac{3}{4} * \frac{3}{4} * \frac{1}{4} * \frac{1}{4}

Net probability = Step 1 * Step 2

=  ^4C_2 * \frac{3}{4} * \frac{3}{4} * \frac{1}{4} * \frac{1}{4}

=  6 * \frac{3}{4} * \frac{3}{4} * \frac{1}{4} * \frac{1}{4} = \frac{54}{256}

Case 3  : 3 out of 4 buys beers

Step 1) Out of four customers, select three who buy beer = this can be done in   ^4C_3   ways 
Step 2) Calculate the required probability =   \frac{3}{4} * \frac{3}{4} * \frac{3}{4} * \frac{1}{4}

Net probability = Step 1 * Step 2

=  ^4C_3 * \frac{3}{4} * \frac{3}{4} * \frac{3}{4} * \frac{1}{4}

=  4 * \frac{3}{4} * \frac{3}{4} * \frac{3}{4} * \frac{1}{4} = \frac{108}{256}

Case 4  : All 4 buy beers

Step 1) Out of four customers, select all four = this can be done in   ^4C_4   ways 
Step 2) Calculate the required probability =   \frac{3}{4} * \frac{3}{4} * \frac{3}{4} * \frac{3}{4}

Net probability = Step 1 * Step 2

=  ^4C_3 * \frac{3}{4} * \frac{3}{4} * \frac{3}{4} * \frac{3}{4}

=  1 * \frac{3}{4} * \frac{3}{4} * \frac{3}{4} * \frac{3}{4} = \frac{81}{256}

Total Probability =  \frac{81}{256} + \frac{108}{256} + \frac{54}{256} +\frac{12}{256} = \frac{255}{256}

Hope this helps!

achloes · Answer

Perfectly clear - thank you!

Bunuel · Answer

GMATNinja Video Solution:

https://www.youtube.com/watch?v=pLLhBZbS6eo

GMATinsight · Answer

Probability (atleast one of them purchases a beer) = 1- Probability (none of them purchases a beer)  = 1 - (\frac{1}{4})^4 = \frac{255}{256}

Answer: Option E

The probability that a customer at a certain bar will purchase a beer

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The probability that a customer at a certain bar will purchase a beer

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