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Suppose the probability that a U.S. resident has
traveled to Canada is 0.18, to Mexico is 0.09, and to both
countries is 0.04. What's the probability that an American
chosen at random has
a) traveled to Canada but not Mexico?
b) traveled to either Canada or Mexico?
c) not traveled to either country?

Respuesta :

francocanacari

A) The probability that an American has traveled to Canada but not Mexico is 16.38%.

B) The probability that an American has traveled to either Canada or Mexico is 27%.

C) The probability that an American has not traveled to either Canada or Mexico is 73%.

Supposing that the probability that a U.S. resident has traveled to Canada is 0.18, to Mexico is 0.09, and to both countries is 0.04, to determine what's the probability that an American chosen at random has A) traveled to Canada but not Mexico, B) traveled to either Canada or Mexico, and C) not traveled to either country, the following calculations must be made:

A)

  • Probability that he has gone to Canada x probability that he has not gone to Mexico = X
  • 0.18 x (1 - 0.09) = X
  • 0.18 x 0.91 = X
  • 0.1638
  • Therefore, the probability that an American has traveled to Canada but not Mexico is 16.38%.

B)

  • Probability that he went to Canada + probability that he went to Mexico = X
  • 0.18 + 0.09 = X
  • 0.27 = X
  • Therefore, the probability that an American has traveled to either Canada or Mexico is 27%.

C)

  • Probability that he did not go to Canada + probability that he did not go to Mexico = X
  • 1 - 0.18 + 1 - 0.09 = X
  • 0.73 = X
  • Therefore, the probability that an American has not traveled to either Canada or Mexico is 73%.

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