Detailed Answer

a) It will have to be all combinations of all possible choices, therfore with given menu choices we must have that the first option is Steak-Fries (SF), Chicken-Fries (CF), and so on. Thus, the final sample space is: SF, CF, TF, SP, CP, TP, SL, CL, TL

b) Since we are only looking at Pasta, the possible choices is everything from point (a) which contains Pasta, so: SP, CP, TP.

c) The Venn diagram is shown below:

d) Now again, we only look at Pasta, but we have a new dimension, which are drinks. Therefore, our answer from part (b) needs to be expanded to contain the drinks for each scenario, so it will be: SP1, CP1, TP1, SP2, CP2, TP2, SP3, CP3, TP3.

Detailed Answer

a) The elements of \(A \cap B\) are cards in the spade suit that are also Kings: {K-spade}.

b) There's 52 cards in the deck of cards and only 1 matches our criteria, so: \(\ P(A \cap B) = \frac{1}{52}\).

c) We can make use of the formula: \(\ P(A \cup B) = P(A) + P(B) - P(A \cap B)\).

So then, \(\ P(A \cup B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52}\).

d) The elements of \(A \cup B\) are all the cards in the spade suit and all the cards that are Kings: {Ace-spade, 2-spade, 3-spade, ..., K-spade, K-heart, K-club, K-diamond}.

e) This is the event when a card is both a spade and a heart, which cannot happen. Therefore, the probability of this event is 0.

Detailed Answer

a) \(P(F) = \frac{n(F)}{72} = \frac{28}{72} = \frac{7}{18}\)

b) \(P(F ∩ read) = \frac{20}{72} = \frac{5}{18}\)

c) \(P(read) = P(F ∩ read) + P(R ∩ read) + P(H ∩ read) =\)

\( = \frac{20}{72} + \frac{16}{72} + \frac{12}{72} = \frac{48}{72} = \frac{2}{3}\)

d) \(P(not \ read) = 1 - P(read) = 1 - \frac{2}{3} = \frac{1}{3}\)

e) \(P(H|read) = \frac{P(H \ ∩ \ read)}{P(read)} = \frac{12}{72} : \frac{2}{3} = \frac{1}{4}\)

Detailed Answer

a) The tree diagram is shown below:

b) \(P(BB) = \frac{3}{8} * \frac{2}{7} = \frac{3}{28}\)

c) \(P(B \ and \ G) = P(BG) + P(GB) =\frac{3}{8} * \frac{5}{7} + \frac{5}{8} * \frac{3}{7} = \frac{15}{28}\)

d) \(P(same) = P(BB) + P(GG) =\frac{3}{28} + \frac{5}{8} * \frac{4}{7} = \frac{13}{28}\)

Detailed Answer

a) \( P(5) = 1 - P(1) - P(2) - P(3) - P(4) - P(6) \)

\( = 1 - \frac{1}{8} - \frac{1}{4} - \frac{1}{8} - \frac{1}{5} - \frac{1}{10} = \frac{1}{5} \)

b) \( P(\text{even number}) = P(2) + P(4) + P(6) = \frac{1}{4} + \frac{1}{5} + \frac{1}{10} = 0.55 \)

c) Since 6 is an even number, \( P(6 \cap \text{even}) \) is simply \( P(6) \). Then:

\( P(6 \mid \text{even}) = \frac{P(6 \cap \text{even})}{P(\text{even})} = \frac{\frac{1}{10}}{0.55} = 0.18 \)

Detailed Answer

a) A can be easily found by subtracting: \(1 - 0.52 = 0.48\)

b) The probability of a person being left-handed is \( 0.09 \), and the probability of a person having dark hair is \( b \), so: \(P(L ∩ D) = 0.09b\)

c) (i) We know that there are two possiblities of people having dark hair, mainly being right-handed and having dark hair, and being left-handed and having dark hair. Their respective probabilities are:

\[P(R ∩ D) = 0.91 * 0.52 = 0.4732\]

\[P(L ∩ D) = 0.09b\]

So now, let's add them up and equalise to 51%:

\[0.4732 + 0.09b = 0.51\]

\[0.4732 + 0.09b = 0.51\]

\[b \approx 0.409\]

c) (ii) As it was done in part (a): \(1 - 0.409 = 0.591\)

Detailed Answer

a) The probability that a randomly selected student passed their exam is given by the ratio of students who passed to the total number of students: \( \)

\[ P(\text{passed}) = \frac{61 + 90}{200} = 0.755 \]

b) Given that a student passed, the probability that the student was sitting Mathematics is found using conditional probability:

\[ P(\text{Mathematics} | \text{passed}) = \frac{P(\text{Mathematics} ∩ \text{passed})}{P(\text{passed})}\]

\[ P(\text{Mathematics} | \text{passed}) = \frac{61}{61 + 90} = 0.404 \]

c) (i) The probability that two randomly selected students were sitting Physics is:

\[ P(\text{Two-Physics}) = \frac{90}{200} * \frac{89}{199}= 0.201 \]

It has to be remembered here, that after the first student is selected, he can no longer be a part of the selection poll, so the total number of students has to decrease to 199.

c) (ii) The probability that four randomly selected students were sitting Physics is:

\[ P(\text{Four-Physics}) = \frac{90}{200} * \frac{89}{199}* \frac{88}{198}* \frac{87}{197}= 0.0395 \]

Same logic applies here, after the first person is selected, the total decreases to 199, then, after the second person is selected, the total goes down to 198, and so on.

Detailed Answer

a) Since \( A \) and \( B \) are independent

\[ P(A \cap B) = P(A) * P(B) = 0.21 * 0.43 = 0.0903 \]

b) Using the formula for the union of three events:

\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) \]

\[ 0.9 = 0.21 + 0.43 + P(C) - 0.0903 - 0.12 \]

\[ P(C) = 0.560 \]

c) \( P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{0.0903}{0.43} = 0.21 \)

d) Knowing that:

\[ P(A) = P(A \cap B) + P(A \cap B') \]

We can rearrange this formula:

\[ P(A \cap B') = P(A) - P(A \cap B) = 0.21 - 0.0903 = 0.120 \]

Detailed Answer

Let \( x \) be the number of black socks. Then the probability of picking two black socks without replacement can be calculated as:

\[ P(\text{two black socks}) = \frac{x}{10} * \frac{x-1}{9} \]

We know this probability is \( \frac{1}{3} \), so:

\[ \frac{x}{10} * \frac{x-1}{9} = \frac{1}{3} \]

\[ \frac{x(x-1)}{90} = \frac{1}{3} \]

\[ x(x-1) = 30 \]

\[ x^2 - x - 30 = 0 \]

\[ (x-6)(x+5) = 0 \]

Thus, \( x = 6 \) (since \( x = -5 \) is not a valid solution for the number of socks).