An archive the questions from Mark's Fall 2018 Stat 225.

Probabilty and double counting

Mark

Suppose that 55% of UNCA students drinks coffee, 45% drinks beer, and 70% drinks at least one of those two.

  1. What is the probability that a randomly selected UNCA student drinks both coffee and beer?
  2. What is the probability that a randomly selected UNCA student drinks neither coffee nor beer?

You can find more problems like this in section 2.2 of our text.

megan
  1. The probability that a randomly selected UNCA student drinks both coffee and beer is 0.3
  2. The probability that a randomly selected UNCA student drinks neither coffee nor beer is 0.3
joshua

1.)The probability that a randomly selected student drinks both beer and coffee can be inferred from the fact that at least 70% drink one of them so the other 30% is that probability.
ANSWER: 30% - probability of randomly selected student drinking both beer and coffee
2.)The same can be inferred for the percentage that drinks neither since 70% drink at least one that only leaves 30% that could drink neither.
ANSWER: 30% - probability of randomly selected student drinking neither beer or coffee

Mark

I think that @megan and @joshua have the correct answers and joshua has taken it a bit further attempting to write down an explanation. I wonder if we can compute the result using expressions like P(A \cup B) and similar things?

joshua

In terms of P
P(A) = 0.55
P(B) = 0.45
P(A∩B) = 0.70
P(A∪B) = P(A)+P(B)-P(A∩B)
P(A∪B) = 0.55+0.45-0.70
P(A∪B) = 0.30

Mark

So here’s what I’m thinking. We first specify events A and B by

  • A = person chosen drinks coffee, and
  • B = person chosen drinks beer.

Note that this is an important step; we can’t just assume that the reader knows what we mean by A and B. Once we’ve stated that, though, we can re-interpret the first sentence in terms of A and B. The first sentence, again, is:

In terms of A and B, this says:

  • P(A) = 0.55,
  • P(B) = 0.45, and
  • P(A \cup B) = 0.7.

Now, solving our double counting formula for P(A \cap B), we have

\begin{align} P(A \cap B) &= P(A) + P(B) - P(A \cup B) \\ & = 0.55 + 0.45 - 0.7 = 0.3 \end{align}

In words, this P(A \cap B) expresses the probability that a student drinks both coffee and beer so that the answer to number 1 is 0.3.

Finally, the probability that a student drinks neither coffee nor beer can be expressed in symbols as

P((A \cup B)^c) = 1 - P(A \cup B) = 1 - 0.7 = 0.3.
dennis

Define:
Coffee: P(A) = .55
Beer: P(B) = .45
either: P(A uu B)= .7

  1. P(A nn B) = P(A) + P(B) -P(A uuB) =
    .55 +.45 -.7 = .3

  2. P(A^c nn B^c) = 1- P(A uu B)= 1 - .7 = .3