Section 1.1

  1. Let, \(T\) - Physical Therapist, \(C\) - Chiropractor

28% visit both a physical therapist:

\[P(T \cap C) = 0.28\]

Problem say that the probability of visiting a physical therapist exceeds the probability of visiting a chiropractor by 16%, we write:

\[\begin{equation} P(T) = P(C) + 0.16 \end{equation}\]

Using the fact that 8% visit neither,
\[P((T \cup C)') = 0.08\] therefore \[P(T \cup C) = 0.92\]

Using \(P(T \cup C) = P(T) + P(C) - P(T \cap C)\) we get \(0.92 = P(T) + P(C) - 0.28\)

Now find \(P(C)\) from the second equation and subtitude; we get \(0.92 = P(T) + P(T) - .16 - 0.28\), simplyfy \(P(T) = 0.68\)

    1. \(S = \{HHHH,....,TTTT\}\) use the tric we learn in class to write this sample sapace. 16 outcomes.
    1. 5/16,
  1. 0,
  2. 11/16,
  3. 1/4,
  4. 1/4,
  5. 9/16,
  6. 4/16
  1. Using a Venn diagram would help
  1. \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = .4+.5-.3=.6\)
  2. \(P(A) = P(A\cap B') + P(A\cap B)\), from this we get, \(.4 = P(A\cap B') + .3\), now we get, \(P(A\cap B') = .1\)
  3. \(P(A'\cup B') = P((A\cap B)') = 0.7\)
  1. Using a Venn diagram would help

\(P(A \cup B) =0.76\), \(P(A \cup B') = 0.87\)

\(P(A) + P(B') - P((A\cap B)') = 0.87\)

\(P(A) + P(B) - P((A\cap B)) = 0.76\)

By adding last two equations together we get,

\(2P(A) + 1 - P(A) = 1.63\), hence \(P(A) = 0.63\)

    1. 1,..,36,0,00
  1. \(P(A) = 2/38\)
  2. \(P(B) = 4/38\)
  3. \(P(odd) = (36/2)/38 = 9/19\)

Section 1.2

    1. 4.5.2 = 40, b) 2.2.2 = 8
    1. \(\binom{4}{1}\binom{6}{3} = 80\)
  1. \(\binom{4}{1}2^6 = 256\)
    1. AAAA or NNNN. so 2.
  1. \(\binom{4}{1}\times2 = 8\) (a different team win one of the first 4 games)
  2. \(\binom{5}{2}\times2 = 20\) (a different team(s) win two of the first 5 games)
  3. \(\binom{6}{3}\times2 = 40\) (a different team(s) win three of the first 6 games)
    1. \[\frac{\binom{19}{3}\binom{52-19}{6}}{\binom{52}{9}} = \frac{102486}{351325}=0.2917\]
  1. \[\frac{\binom{19}{3}\binom{10}{2}\binom{7}{1}\binom{3}{0}\binom{5}{1}\binom{2}{0}\binom{6}{2}}{\binom{52}{9}} = \frac{7695}{1236664}=0.00622\]