Lunch with Muhammad

For the fourth customer, ordering a previously unordered item is equivalent to ordering a unique item – an item nobody else chooses – within the first four customers, since he is last. Note that the order in which menu items are chosen do not affect whether a customer orders a unique item, so the answer will be the same if we let the fourth customer order first.

There are five items, so the chance that the fourth customer's item is not chosen is $\frac {4}{5}$ . With three other customers, the desired probability is then simply $(\frac {4}{5})^3 = \frac {64}{125}$ , so the answer is $189$ .

We need to find the probability that a menu item will not be chosen by the first three customers.

The probability that an item will not be chosen is just the complement of the probability that it will be chosen.

Since each item has 1/5 chance of being chosen, there is 4/5 chance that it will not be chosen.

So there is (4/5)^3 chance it will not be chosen by the first, second, and third customer.

We now want our fourth customer to choose the menu that has not been chosen yet by the first three. The probability of this happening is 1/5.

We will now have:

(1/5)(4/5)^3

But we have to take in consideration that the above probability can happen to any one of the 5 menu items. So we need to multiply it by 5.

(5)(1/5)(4/5)^3

= 64/125 = a/b

so a +b = 64 + 125 = 189.

O total de formas dos quatro clientes pedirem é: N=5×5×5×5 pois cada um tem cinco possíveis escolhas de prato.

O total de formas dos quatro pedirem e quarto comprar um prato diferente dos demais é: n=4×4×4×5 podemos pensar que o quarto escolheu seu prato primeiro, então podia escolher entre as cinco possibilidades depois os outros três escolheram seus pedidos entre as outras quadro opções. Dessa forma a probabilidade é: $\frac {n}{N}$ = $\frac {4×4×4×5}{5×5×5×5}$ = $\frac {4³}{5³}$ , então a=4³=64; b=5³=125; a+b=189

Given that the 4th customer orders one item uniformly at random which is independent of the others, the probability that each other customer randomly picks a different item is $\frac{4}{5}$ . Therefore, the probability that all three other customers pick a different item than the 4th customer is:

$\frac{4}{5}\cdot\frac{4}{5}\cdot\frac{4}{5} = \frac{64}{125} = \frac{a}{b}$ .

$a+b = 189$ .

First we count all cases:
The total is the four customers choose anything, so each one has five options So, we have 625

Second we count the favorable cases:
1) the four customers choose diferent options So, 120
2) two customers choose the samr option and one chose a diferent option So, 60x3 = 180
3) three customers choose the same option So 20

And now we divide the favorable cases and all cases to have the probability. So, 320÷625 = 64÷125, a+b is 189.

First we find the following probabilities.

i) Probability that first 3 customers all choose distinct items .................................................................................................. For this to happen, no matter what the first customer has chosen, the second customer has to choose a different item. So the second customer has $4$ possible choices out of a total of $5$ items. Also no matter what the second customer has chosen, the third customer again has to choose a different item. So the third customer has $3$ possible choices out of a total of $5$ items. So probability of the first three customers choosing three distinct items = $4/5 * 3/5$ .

ii) Probability that the first three customers choose the same item ................................................................................................................ No matter what the first customer has chosen, the second customer has to choose that item out of a total of $5$ items and the third customer also has to choose that item out of a total of $5$ items. So probability = $1/5 * 1/5$

iii) Probability of first three customers choosing exactly 2 distinct items ........................................................................................................................ We won't evaluate this probability directly. We know that the first three customers can either choose 1 distinct item, 3 distinct items, or 2 distinct items. Hence the probabilities of choosing 1 distinct item, 2 distinct items, and 3 distinct items must add up to $1$ . Hence probability of choosing $2$ distinct items = $1 - 1/5*1/5 - 4/5*3/5 = 12/25$

Now we know that the fourth customer chooses a previously un-chosen item. This can be divided into three cases.

Case 1:- the first three customers choose 3 different items .................................................................................................... Then the fourth customer can choose from the $2$ remaining items. Probability of this is $2/5$ . Hence total probability = $2/5 * 3/5 * 4/5$

Case 2:- the first three customers choose 2 different items ..................................................................................................... Then the fourth customer can choose from the $3$ remaining items. Probability of this is $3/5$ . Hence total probability = $12/25 * 3/5$

Case 3:- the first three customers choose 1 distinct item ................................................................................................... Then the fourth customer can choose from the $4$ remaining items. Probability of this = $4/5$ . Hence total probability = $1/5 * 1/5 * 4/5$ .

Combining all the cases, we find out that the final probability will be equal to $1/5 * 1/5 * 4/5 + 12/25 * 3/5 + 2/5 * 3/5 * 4/5 = 64/125$ . Hnce $a= 64$ , $b= 125$ , and $a+b= 64+125= 189$ .

After 3 orders, probability of 3, 2, and 1 different choices of food having been ordered is 12/25, 12/25 and 1/25 respectively.

Probability of a previously unordered item being ordered is therefore

(12/25 \times 2/5) + (12/25 \times 3/5) + (1/25 \times 4/5) = 64/125

64 + 125 = 189

Suppose the first 3 customers choose 3 different items, then first one can choose in 5 ways, second in 4 ways and third in 3 ways. Hence, no. of favorable cases are 5x4x3=60. Needless to say, total cases are 5x5x5=125. Hence, the probability is $\frac {60}{125}$ . Now, the 4th person can choose with probability $\frac {2}{5}$ .

Hence, the probability of first case is $\frac {120}{625}$ .

Considering the second case, where only 2 items are selected, the items have to be distinct... Hence, ways of selecting the repeated and non-repeated items is 5x4=20 ways. Now, we can choose 2 people from 3 who choose the same item in 3C2=3 ways. Hence no. of favorable cases is 20x3=60. Hence, the probability is $\frac {60}{125}$ . Now, the fourth person can choose with probability $\frac {3}{5}$ .

Hence , probability of second case is $\frac {180}{625}$ .

Considering the third case where all the previous 3 customers choose the same item. That item can be chosen in 5 ways , hence, the probability is $\frac {5}{125}$ . Hence, the probability is $\frac {5}{125}$ . Now, the fourth person can choose with probability $\frac {4}{5}$ ways.

Hence, probability of the third case is $\frac {20}{625}$ .

Adding up all the cases and reducing to lowest form, the probability comes out to be $\frac {64}{125}$ . Hence the answer is 64+125=189.

The number N of all possibilities is given by 5^4 (each one of the four persons can select five different items and the order of this process is relevant).

Now, the number n of the possibilities considering the problem restrictions is given by 5 * (4)^3. First, we fix the item that will be selected by the fourth customer (5 possibilities). Then, the first three customers have four possibilities each one ( they can not select the item that will be selected by the 4th).

Therefore, the probability of the problem restrictions is given by n/N, so n/N=(4^3/5^3).

a=4^3; b=5^3; a+b=189