ShashankDave wrote:

jayshah0621 wrote:

In a recent survey at a local deli, it was observed that 3 out of 5 customers bought a bagel and 5 out of 7 customers bought a coffee. Some customers bought both. If 3 customers are selected, what are the chances that at least 1 customer bought a coffee and a bagel?

A) 27/343

B) 3/7

C) 27/125

D) 279/343

E) 9/125

Source: Self made

I think that the question is flawed. Let me explain two different scenarios here.

Before I do some explaining, let's find out the numbers.

P(Bagel) = 3/5

P(Coffee) = 5/7

Let's say there were 35 customers(LCM of 5 and 7)

P(Bagel) = 21/35

P(Coffee) = 25/35

Therefore, by applying sets, we know that the number of people who would buy both coffee and bagel(provided the situation that every customer buys only bagel or only coffee or both, and there s no customer who buys something else or buys nothing at all), are 11 out of 35.

P(Both) = 11/35

Now let me present two situations here.

1. The total number of customers is 35. In that case the solution provided by someone above would be accurate.

2. The total number of customers are huge and unspecified. In this case, every customer will have the same respective probabilities.

Let me explain this calculation.

P(At least one bought both)

\(1 - \frac{24}{35}*\frac{24}{35}*\frac{24}{35}\)

This is how I did it, and the answer is not listed.

Bunuel,

VeritasPrepKarishma and

IanStewart please chip in. And if my reasoning is wrong then I'm in dire need of help

I think you are right ,we follow this procedure when the total number of customers is explicitly mentioned, but when ratio is given, we take the probability of every customer the same. I am struggling with same kind of question

A string of lights is strung with red, blue, and yellow bulbs in a ratio of two to five to three, respectively. If 6 bulbs are lit in a random sequence, what is the probability that no color is lit more than twice?

A) 9/1000

B) 81/1000

C) 10/81

D) 1/3

E) 80/81

Explanation provided:

The only way no color will be lit more than twice is if we have exactly two of each of the three colors in our set of six. For instance, if we had less than two blues, we would have to have more than two reds or more than two whites. The only way we can have no color with more than two lights is if all three colors have exactly two lights. Thus, the question is really: what is the probability that a set of six has exactly two of each color?

Probability = 2/10*2/10*5/10*5/10*3/10*3/10 * 6!/2!2!2! =81/1000 (Answe B)

But my thought process was 2c2*5c2*3c2/10c6 x6!/2!2!2! (ways of selecting 2 bulbs from each colour multipilied by the number of ways of arranging the selection)

I am unable to understand the flaw in my approach

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