a) To find \( q \), we sum all probabilities to equal 1:
\[ \frac{1}{10} + q + \frac{2}{10} + \frac{3}{10} + \frac{q}{2} + \frac{1}{10} = 1 \]
\[ \frac{7}{10} + q + \frac{q}{2} = 1 \]
\[ q = \frac{1}{5} \]
b) To find \( E(x) \), we calculate the weighted average:
\[ E(X) = 0 * \frac{1}{10} + 1 * \frac{1}{5} + 2 * \frac{2}{10} + 3 * \frac{3}{10} + 4 * \frac{1}{10} + 5 * \frac{1}{10} \]
\[ E(X) = 0 + \frac{1}{5} + \frac{4}{10} + \frac{9}{10} + \frac{4}{10} + \frac{5}{10} = 2.4 \]
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.
Closea) We know that 60 people were surveyed, so it has to be that:
\[ 60 = 13 + 5 + 1 + 2 + 11 + 12 + 13 + a\]
\[ a =3\]
b) \( P(non-fiction) = \frac{18}{60} = \frac{3}{10} \)
c) \( P(only \ one) = \frac{13}{60} + \frac{11}{60} + \frac{12}{60} = \frac{36}{60} = \frac{3}{5} \)
d) \( P(exactly \ two) = \frac{5}{60} + \frac{2}{60} + \frac{3}{60} = \frac{10}{60} = \frac{1}{6} \)
e) \( P(P | N) = \frac{P(P \cap N)}{P(N)} = \frac{\frac{2}{60}}{\frac{18}{60}} = \frac{1}{9} \)
d) \( P(C \cap P) = \frac{5}{60} = \frac{1}{12} \)
CloseSince \( A \) and \( B \) are independent:
\[ P(A \cap B) = P(A) * P(B) \]
By the definition of conditional probability:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Substitute \( P(A \cap B) = P(A) * P(B) \) into the conditional probability formula:
\[ P(A|B) = \frac{P(A) * P(B)}{P(B)} \implies P(A|B) = P(A) \]
Closea) The tree diagram is shown below:
b) \(P(Blue \ and \ Blue) = \frac{3}{8} * \frac{2}{7} = \frac{3}{28}\)
c) \(P(Blue \ and \ Green) = P(BG) + P(GB) =\frac{3}{8} * \frac{5}{7} + \frac{5}{8} * \frac{3}{7} = \frac{15}{28}\)
d) \(P(same) = P(BB) + P(GG) =\frac{3}{28} + \frac{5}{8} * \frac{4}{7} = \frac{13}{28}\)
Closea)
\[ P(\text{Black first}) = \frac{7}{20} \]
\[ P(\text{Yellow second}) = \frac{3}{19} \]
\[ P(\text{Black first and Yellow second}) = \frac{7}{20} * \frac{3}{19} = \frac{21}{380} \]
b)
\[ P(\text{Black and Black}) = \frac{7}{20} * \frac{6}{19} = \frac{21}{190} \]
\[ P(\text{Yellow and Yellow}) = \frac{3}{20} * \frac{2}{19} = \frac{3}{190} \]
\[ P(\text{Red and Red}) = \frac{10}{20} * \frac{9}{19} = \frac{45}{190} \]
\[ P(\text{Same color}) = \frac{21}{190} + \frac{3}{190} + \frac{45}{190} = \frac{69}{190} \]
Closea) We know that 28% of people visited Milan, so:
\[ 0.28 = 0.15 + 0.05 + 0.03 + b \]
\[ b = 0.05 \]
b)
\[ 0.2 = 0.05 + 0.03 + c \]
\[ c = 0.12 \]
c) We know that all probabilities have to add up to 1, so:
\[ 0.3 + 0.05 + 0.15 + 0.03 + 0.05 + 0.12 + 0.11 + a = 1 \]
\[ a = 0.19 \]
d)
\[ 0.15 + 0.3 + 0.05 + 0.03 + 0.11 + 0.12 = 0.76 \]
e)
\[ P(R'|M) = \frac{P(R' \cap M)}{P(M)} = \frac{0.2}{0.28} = 0.71\]
Closea) 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:
\[ P(A \cap B') = P(A) - P(A \cap B) = 0.21 - 0.0903 = 0.120 \]
CloseLet \( x \) 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).
Close| \( x \) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| \( P(X = x) \) | \( \frac{1}{10} \) | \( q \) | \( \frac{2}{10} \) | \( \frac{3}{10} \) | \( \frac{q}{2} \) | \( \frac{1}{10} \) |
