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The probability that a customer at a certain restaurant will purchase

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Bunuel
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The probability that a customer at a certain restaurant will purchase [#permalink] New post   06 Feb 2023, 07:38
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The probability that a customer at a certain restaurant will purchase a pie is 0.75. If four customers visit the restaurant on a certain evening, what is the probability that exactly three of them will purchase a pie ?

A. 27/256
B. 3/64
C. 27/64
D. 3/4
E. 32/81
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C
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gmatophobia
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The probability that a customer at a certain restaurant will purchase [#permalink] New post   11 Feb 2023, 12:48
Bunuel wrote:
The probability that a customer at a certain restaurant will purchase a pie is 0.75. If four customers visit the restaurant on a certain evening, what is the probability that exactly three of them will purchase a pie ?

A. 27/256
B. 3/64
C. 27/64
D. 3/4
E. 32/81


The probability that a person will purchase the pie = \(\frac{3}{4}\)

The probability that a person will not purchase the pie = \(\frac{1}{4}\)

Exact three of them purchase a pie = \((\frac{3}{4})^3 * (\frac{1}{4}) * \frac{4!}{3!}\)

= \(\frac{3^3 }{ 4^3}\)

= \(\frac{27}{64}\)

Option C
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The probability that a customer at a certain restaurant will purchase [#permalink] New post   23 Mar 2023, 06:42
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gmatophobia wrote:
Bunuel wrote:
The probability that a customer at a certain restaurant will purchase a pie is 0.75. If four customers visit the restaurant on a certain evening, what is the probability that exactly three of them will purchase a pie ?

A. 27/256
B. 3/64
C. 27/64
D. 3/4
E. 32/81


The probability that a person will purchase the pie = \(\frac{3}{4}\)

The probability that a person will not purchase the pie = \(\frac{1}{4}\)

Exact three of them purchase a pie = \((\frac{3}{4})^3 * (\frac{1}{4}) * \frac{4!}{3!}\)

= \(\frac{3^3 }{ 4^3}\)

= \(\frac{27}{64}\)

Option C



How did you get 4!/3!?
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Eeekta
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The probability that a customer at a certain restaurant will purchase [#permalink] New post   03 Jun 2023, 06:40
gmatophobia wrote:
Bunuel wrote:
The probability that a customer at a certain restaurant will purchase a pie is 0.75. If four customers visit the restaurant on a certain evening, what is the probability that exactly three of them will purchase a pie ?

A. 27/256
B. 3/64
C. 27/64
D. 3/4
E. 32/81



Hi I am unable to understand why 4!/3!?

We usually do this when we have similar items/letters/people etc. right, but here none of it is the case.

Can you please clarify this?

Bunuel could you please help? Thank you
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Bunuel
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Re: The probability that a customer at a certain restaurant will purchase [#permalink] New post   03 Jun 2023, 08:38
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Eeekta wrote:
gmatophobia wrote:
Bunuel wrote:
The probability that a customer at a certain restaurant will purchase a pie is 0.75. If four customers visit the restaurant on a certain evening, what is the probability that exactly three of them will purchase a pie ?

A. 27/256
B. 3/64
C. 27/64
D. 3/4
E. 32/81



Hi I am unable to understand why 4!/3!?

We usually do this when we have similar items/letters/people etc. right, but here none of it is the case.

Can you please clarify this?

Bunuel could you please help? Thank you


The scenario where exactly three out of the four customers purchase a pie can occur in four different ways: YYYN (this would mean that the first customer purchases the pie, as do the second and third customers, but the fourth customer does not), YYNY, YNYY, and NYYY. As you can see, this situation can occur in four different ways. Essentially, these represent the number of ways we can arrange the four letters Y, Y, Y, and N, which is 4!/3!.
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Re: The probability that a customer at a certain restaurant will purchase [#permalink]
03 Jun 2023, 08:38
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