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Exercises 10.5 Exercises

1.

Alice has courtside tickets to see the Charlotte Hornets. She has a Muggsy Bogues bobblehead on her desk and has Kemba Walker’s jersey. (Those are, or were, Hornets players.) Which of these two events is more likely: that Alice works in real estate, or that Alice works in real estate and is a Hornets fan?

2.

Of a particular basketball player’s shots, \(60\%\) are worth 3 points. In general they make \(40\%\) of their shots. Of the player’s 3-point shots, \(30\%\) make it in the basket.

  1. What is the probability that a shot is worth 2 points? (Assume all shots are worth 2 or 3 points.)

  2. What is the probability that a shot is worth 3 points and makes it?

  3. What is the probability that a shot is worth 3 points or makes it?

  4. Of the shots that make it, how many are 3-point shots?

3.

A twenty-sided die is rolled. (The sides are labeled \(1-20\text{.}\))

  1. What is the probability that the result is at least \(6\text{?}\)

  2. What is the probaility that the result is prime given that it is at least \(6\text{?}\)

  3. What is the probability that the result is at least \(6\) and prime?

  4. What is the probability that the result is at least \(6\) or prime?

4.

What is the probability that a randomly-chosen binary string of length \(10\) has exactly \(3\) ones?

5.

A six-character alphanumeric password is made.

  1. What is the probability that the password contains all letters?

  2. What is the probability that the password has no repeated characters?

  3. What is the probability that the password has all letters given that it has no repeated characters?

  4. What is the probability that the password has repeated characters?

6.

A particular illness has an incidence rate of \(0.3\%\text{,}\) meaning that the probability anyone in the population has it is only \(3\) out of \(1000\text{.}\) Suppose that in specimens containing the illness, a test gives a positive \(80\%\) of the time. Furthermore the same test has a \(10\%\) false positive rate (positive if the specimen does not contain the illness).

  1. Let \(A\) be the event that an individual is sick with the illness and let \(B\) be the event they test positive. Explain why, in general (regardless of what \(A\) and \(B\) represent)

    \begin{equation*} P(B) = P(B \cap A) + P(B \cap \overline{A}). \end{equation*}

  2. Calculate \(P(A|B)\text{,}\) the probability that an individual is sick given they test positive.

  3. Calculate \(P(A|B)/P(A)\text{.}\) This ratio represents the factor by which a belief the individual is sick, quantified by a probability, should increase given the evidence of a positive test. Bayes’ theorem (which you used unknowingly in part (b)) quantifies the way belief can change with evidence. For more, check out this article.

Solution.

  1. This is a consequence of the additivity axiom of probability, since \((B \cap A) \cup (B \cap \overline{A}) = B\) and \((B \cap A)\) and \((B \cap \overline{A})\) are disjoint. To be less mathy about it, any outcome in \(B\) is either shared with \(A\) or not shared with \(A\text{.}\)

  2. The formula for conditional probability is

    \begin{equation*} P(A|B) = \dfrac{P(A \cap B)}{P(B)}. \end{equation*}

    The above part hints that we should calculate \(P(B)\text{,}\) the probability of a positive result, by decomposing it into true positives and false positives.

    \begin{align*} P(B) \amp = P(B \cap A) + P(B \cap \overline{A})\\ \amp = P(B|A)P(A) + P(B|\overline{A})P(\overline{A})\\ \amp = 0.8\times 0.003 + 0.1\times 0.997\\ \amp = 0.0024 + 0.0997\\ \amp = 0.1021. \end{align*}

    As part of the above calculation we calculated the proportion of overall true positives, \(P(A \cap B) = 0.8\times 0.003 = 0.0024\text{.}\) Therefore,

    \begin{equation*} P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \frac{0.0024}{0.1021} = 0.0235. \end{equation*}

    This result should be interpreted as follows: If you test positive, there is a \(2.35\%\) chance you have the illness. That's not much, right? Well...

  3. The odds ratio is

    \begin{equation*} \frac{P(A|B)}{P(A)} = \frac{0.0235}{0.003} = 7.83. \end{equation*}

    The correct interpretation here is that if you test positive for the illness, you are nearly eight times more likely to have it. If a test has a low incidence rate, a positive test doesn't automatically mean you have it; it means you are much more likely to have it and should take the appropriate precautions.

    Bayes' theorem says that we can always compute such an odds ratio when thinking about conditional probabilities. Given some data, our belief in whether or not a statement is true either goes up or goes down by some ratio.

7.

Consider the random variable \(X\) whose probability mass function is given below.

Table 10.5.1.
\(x\) \(1\) \(2\) \(3\) \(4\) \(5\) \(6\) \(7\)
\(P[X=x]\) \(7/28\) \(6/28\) \(5/28\) \(4/28\) \(3/28\) \(2/28\) \(1/28\)

  1. What is the probability that \(X\) takes on a value that's at least 5?

  2. What is the expected value of \(X\text{?}\)

8.

Gallant has a subscription to “free coffee” at Soakin' Bagels, but he still often tips the employees. (Goofus pays full price for coffee and does not tip.) If he doesn't tip it's usually because he doesn't have any \(\$1\) bills on hand. The following table gives the pmf of \(X\text{,}\) Gallant's tip each time he goes to Soakin' Bagels.

Table 10.5.2.
\(x\) \(\$0\) \(\$1\) \(\$2\) \(\$3\)
\(P[X=x]\) \(0.1\) \(0.5\) \(0.35\) \(0.05\)

  1. Verify that the above pmf is indeed a pmf.

  2. What is the probability that Gallant tips?

  3. What is the average value of Gallant's tip each morning?

9.

Suppose \(100\) electronic parts are randomly sampled. It is known that these electronic parts have a \(3\%\) failure rate. Let \(Y\) represent the number of defective parts in the sample of \(100\text{.}\)

  1. What is the probability that exactly \(5\) parts in the sample are defective? Make sure that the correct formula for the answer is included in your response.

  2. What is the expected number of defective parts in the sample? Give an exact answer.

10.

Roughly \(38.5\%\) of a standard deck of 52 cards has a rank of \(10\) (the 10, Jack, Queen, King, and Ace). Suppose that you draw seven cards independently (meaning that each card is returned to the deck and that the deck is shuffled between draws). What is the probability that exactly four of the seven cards drawn will have a rank of \(10\text{?}\) Make sure that the correct formula for the answer is included in your response.