Project 1 Solutions

Problem 1

(a)

Treat A and B as one block. The block can be AB or BA.

\[ 2\cdot5!=240 \]

(b)

Count all permutations and subtract those with E last.

\[ 6!-5!=720-120=600 \]

(a) and (b)

If (a) and (b) were combined, count all cases where E is not last and AB / BA occupy two adjacent slots.

\[ 2\cdot(5!-4!)=192 \]

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Problem 2

Order does not matter, so use a combination (“choose”).

\[ \binom{7+8}{6}=5,005 \]

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Problem 3

The possible committee compositions are 3 women/3 men or 4 women/2 men.

\[ \binom{8}{3}\binom{7}{3} + \binom{8}{4}\binom{7}{2}=3,430 \] —

Problem 4

Two events are independent if one of the following is shown:

• \(P(A\cap B)=P(A)P(B)\)
• \(P(A|B)=P(A)\)
• \(P(B|A)=P(B)\)

Here, \(P(A\cap B)=P(A)P(B)\) is shown.

Find P(A)

Use the complement: neither card is a spade.

\[ P(A) = 1-\frac{\binom{39}{2}}{\binom{52}{2}} = \frac{15}{34} \]

Find P(B)

Use the complement: neither card is a queen.

\[ P(B) = 1-\frac{\binom{48}{2}}{\binom{52}{2}} = \frac{2}{13} \]

Find P(AB)

Using De Morgan’s Laws:

\[ P(A\cap B) = 1-P\left((A\cap B)'\right)=1-P(A'\cup B') \]

\(P(A'\cup B')\) is the probability of no spades and no queens. Take the sum of both probabilities and subtract the probability of the intersection.

• No spades: 52-13=39 options
• No queens: 48 options
• No spades and no queens: 52 - 13 - 4 + 1 = 36 options

\[ P(A\cap B) = 1- \left(\frac{\binom{39}{2}}{\binom{52}{2}} + \frac{\binom{48}{2}}{\binom{52}{2}} - \frac{\binom{36}{2}}{\binom{52}{2}}\right)=\frac{29}{442} \]

Now,

\[ P(A)P(B)=\frac{15}{34}\times\frac{2}{13}=\frac{30}{442}\neq\frac{29}{442} \]

Therefore A and B are not independent. Intuitively, this makes sense because both events share the Queen of Spades.