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It is estimat...

Question

It is estimated in a survey that there is a $40 %$ chance that a person will go on a trip to France, and a $60 %$ chance that the person will choose a 5-star hotel for a stay. Using a simulator that generates two-digit numbers using the digits 0 through 9, estimate the probability that a person selected at random will not select France for a trip but stay in a five star hotel.
Note: The ones place of a two-digit number and 1 through 4 represents going to France for a trip.
The tens place of the two-digit number and 1 through 6 represents choosing a 5-star hotel for stay.

Simulator:

26 32 64 08 46 11 77 01 79 67

22 61 87 72 53 49 59 81 68 09

94 78 38 56 42 77 19 48 20 05

02 83 85 82 54 43 69 12 21 91

79 46 98 14 40 45 56 64 96 92

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Solution

The tens place of a two-digit number represents people going to France.

Since there is a $40 %$ chance of a person going to France for a trip, let 1 through 4 represent a person who goes to France for a trip.
Therefore 5 through 9 and 0 represents a person not going on a trip to France.

The ones place of the two-digit number represents people choosing a 5-star hotel for a stay.
Since there is a $60 %$ chance of a person choosing a 5-star hotel for a stay, let 1 through 6 represent the person choosing a 5-star hotel for a stay.

From the simulation, we see that 19 out of 50 people chosen at random will not select France for a trip but stay in a 5-star hotel.

Estimated probability $= \frac{19}{50} \times 100 = 38 %$

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