Word Problem Probability

Question

Jackson can strangely guess the height and weight of people. The probability that he correctly guesses a person's height is 2/3, and the probability that he can correctly guesses a person's weight is 3/5. If Jackson meets three people and his success rate remans constant. What is the probability that Jackson correctly guesses a person's weight and height at least once?

A. 98/125

B. 8/27

C. 49/81

D. 125/144

Answer 1

Assume that guessing height correctly and guessing weight correctly are independent events. Then the probability of guessing height correctly and guessing weight correctly is equal to the product of their probabilities. That is, (2/3)(3/5) = 2/5. This re-defines the event of success to be both height and weight being correctly guessed. This is going to go into a binomial distribution-related calculation.

I am looking for the probability of guessing correctly at least once. Complementary to "at least once" is "exactly 0 times," or "none." It's easier to calculate it this way than figuring it out from the "at least once" perspective.

If you are doing this by the TI-83 or TI-84, you can do this calculation where n = 3 (from the three people), p = 0.4 (this is what we found previously, the probability of guessing height and weight correctly), and x = 0 (this is the number of successes we want to get rid of for the sake of our method with complements:

1 - binompdf(3, 2/5, 0) = 0.784 = 98/125

If you are doing this by hand:

1 - [(nCr)p^rq^n-r]

1 - [(3 C 0)((2/5)⁰(3/5)³]

1 - [(1)(1)(27/125)]

1 - (27/125)

98/125