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The probability that a customer at a certain bar will purchase a beer 06 Feb 2023, 07:05
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E
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80% (01:14) correct 20% (01:58) wrong based on 265 sessions

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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
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E
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The probability that a customer at a certain bar will purchase a beer 06 Feb 2023, 10:14
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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.
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The probability that a customer at a certain bar will purchase a beer 11 Feb 2023, 12:55
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Bunuel
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
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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
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The probability that a customer at a certain bar will purchase a beer 09 May 2023, 00:05
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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!
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The probability that a customer at a certain bar will purchase a beer 09 May 2023, 02:39
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achloes
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!
Show more

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 - \(P_1 , P_2 .. P_4\)

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

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!
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The probability that a customer at a certain bar will purchase a beer 13 May 2023, 22:31
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gmatophobia
achloes
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!

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 - \(P_1 , P_2 .. P_4\)

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

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!
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Perfectly clear - thank you!
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