Cruise Ship Problem

Question

The following data were obtained about 500 passengers on a cruise ship.

1. 272 passengers brought a camera.
2. 218 passengers brought a laptop computer.
3. 196 passengers brought MP3 players.
4. 51 passengers brought a camera and an MP3 player.
5. 77 passengers brought an MP3 player and a laptop computer.
6. 115 passengers brought a camera and a laptop computer.
7. 35 passengers brought a camera, a laptop computer, and an MP3 player.

What percent of these passengers did not bring any of the three items?

Answer 1

In statistics, the percentage of a group is the same as the probability of being in the group. The probability of not in the three groups is the complement of the probability of being in one of the three groups.

We will use this formula: P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P (B and C) - P(A and C) + P(A and B and C)

Let C = buying a camera, L = buying a laptop, and M = buying an MP3. So: P(C or L or M) = P(C) + P(L) + P(M) - P(C and L) - P (L and M) - P(C and M) + P(C and L and M). (Note the denominator for each of these probabilities is going to be 500.)

P(C or L or M) = (272 + 218 + 196 - 115 - 77 - 51 + 35)/500 = 478/500

Therefore, P(none) = 1 - 478/500 = 22/500 = 0.044. or 4.4%