The probability that a customer at a certain restaurant will purchase

Question

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

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

Bunuel · Accepted Answer

GMATNinja Video Solution:

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gmatophobia · Answer

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

Eeekta · Answer

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

Bunuel · Answer

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!.

ronnnie · Answer

formula : nCk* p(a)^k * p(b)^n-k

n = 4 , k = 3, p(a) = 0.75 = 3/4 , p(b) = 1-0.75 = 0.25 = 1/4

hence , placing them in the formula : 4c3* (3/4)^3 * (1/4)^4-3 ==> 27/64 (answer)

sarthak1701 · Answer

Random probability is: (3/4)^3 *1/4 but we have to multiply this with 4!/3! to adjust for the position of the last customer (he could be the 1st, 2nd, 3rd or the 4th). Ans. 27/64

purse1234 · Answer

Hi everyone

I do not understand why we have to multiply with the probability of a person will not purchase the pie (1/4)

Can anyone please clarify?

Bunuel · Answer

The question asks for the probability that exactly 3 out of 4 customers will purchase a pie. This means 3 customers will buy the pie (each with a probability of 3/4), and 1 customer will not buy a pie (with a probability of 1/4). We multiply by the probability of the one customer not purchasing to account for that specific outcome, where 3 purchase and 1 doesn't.

We further multiply by 4!/3! to account for all the cases where 3 buy and one does not: 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!.

So, we get 3/4 * 3/4 * 3/4 * 1/4 * 4!/3! = 27/64.

Answer: C.

The probability that a customer at a certain restaurant will purchase

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The probability that a customer at a certain restaurant will purchase

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