Question

QUESTION 6 Match the description below with its probability. Out of the 39 costumes for the school play, 19 need to be altered to fit an actor. 15 costumes need to be shortened and 18 costumes need to be taken in. What is the probability that a randomly selected 2 costume needs both shortened and taken in? A. 7 A marble is chosen from bag containing 3 red, 1 blue, 5 green, and 5 yellow marbles. What is the 14 probability of choosing a red or blue marble? B. 39 In a group of 20 students, 20 write for the school newspaper, 20 are on the yearbook committee, and 27 24 are part of both groups. What is the probability of C. 50 randomly choosing a writer from the newspaper or a member of the yearbook committee? 4 Out of 100 car owners surveyed, 58 said they own D. 5 an SUV. 33 of the SUV owners bought their car used. 13 of the non-SUV owners bought their car used. What is the probability that a randomly selected person from the survey bought a new car? -

Answer 1

1. There are a total of n(T) = 39 costumes. Of these n(A) = 19 require alteration. The alteration can be of two types: shortening and taken in. n(S) = 15 costumes require shortening, and n(I) = 18 costumers require being taken in.

We are to determine the probability that a randomly selected costume requires both shortening and taken in. This will be the probability

PS I)

We can use the formula for the union of sets, which is:

$\\P(S\cup I) = P(S) + P(I) - P(S\cap I)\\\\ \therefore S\cup I = A\\\\ \therefore P(A) = P(S) + P(I) - P(S\cap I)\\\\ \therefore \frac{n(A)}{n(T)} = \frac{n(S)}{n(T)} + \frac{n(I)}{n(T)} - P(S\cap I)\\\\ \therefore P(S\cap I) =\frac{n(S)}{n(T)} + \frac{n(I)}{n(T)} - \frac{n(A)}{n(T)}\\\\$

Substituting the values:

$\\\therefore P(S\cap I) =\frac{15}{39} + \frac{18}{39} - \frac{19}{39}\\\\ \therefore P(S\cap I) =\frac{14}{39}$

Therefore, the correct option is B.

2. The bag has four types of marbles. These are red, blue, green, and yellow marbles. The number of marbles of each color are n(R) = 3, n(B) = 1, n(G) = 5, n(Y) = 5.

We are to determine the probability of choosing a red or a blue marble. Therefore, one marble is drawn randomly from the bad. The number of ways that this marble would be red or blue is $\\\binom{3}{1}+ \binom{1}{1}$

There are a total of n(T) = n(R) + n(B) + n(G) + n(Y) = 3 + 1 + 5 + 5 = 14.

Therefore, the total number of ways of choosing one marble from a total of 14 marbles is $\\\binom{14}{1}$

Hence, the probability of choosing a red or a blue marble is:

$\\P(R\cup B) = \frac{\binom{3}{1}+ \binom{1}{1}}{\binom{14}{1}}\\\\ \therefore P(R\cup B) = \frac{3+ 1}{14}\\\\ \therefore P(R\cup B) = \frac{4}{14}\\\\ \therefore P(R\cup B) = \frac{2}{7}\\\\$

Hence, the correct answer is A.

3. There are a total of n(T) = 20 students. The number of students who write for school newspaper is n(N) = 20. The number of students who write for yearbook committee is n(C) = 20. The number of students that belong to both the groups is n(N and C) = 24.

We are to determine the probability of randomly choosing a writer from newspaper or a member of yearbook committee. This is the probability $P(N\cup Y)$

This can be determined using the formula for the union of sets.

$\\\therefore P(N\cup Y) = P(N) + P(Y) - P(N\cap Y)\\\\ \therefore P(N\cup Y) = \frac{n(N)}{n(T)} + \frac{n(Y)}{n(T)} - \frac{n(N\cap Y)}{n(T)}\\\\$

Substituting the values:

$\\\therefore P(N\cup Y) = \frac{20}{20} + \frac{20}{20} - \frac{24}{20}\\\\ \therefore P(N\cup Y) = \frac{16}{20}\\\\ \therefore P(N\cup Y) = \frac{4}{5}\\\\$

Hence, the correct answer is D.

4. There are a total of n(T) = 100 car owners. Of these, n(S) = 58 own SUV. Therefore, the number of owners who do not own SUV is n(NS) = n(T) - n(S) = 100 - 58 = 42.

Of the SUV owners, n(U|S) = 33 have bought used cars. Therefore, the number of SUV owners who bought new cars is n(N|S) = n(S) - n(U|S) = 58 - 33 = 25.

Of the non-SUV owners, n(U|NS) = 13 have bought used cars. Therefore, the number of non-SUV owners who bought new cars is n(N|NS) = n(NS) - n(U|NS) = 42 - 13 = 29.

Therefore, the number of users that bought a new car is n(N) = n(N|S) + n(N|NS) = 25 + 29 = 54.

The probability of a randomly selected person having bought a new car would be:

$P(N) = \frac{n(N)}{n(T)} = \frac{54}{100} = \frac{27}{50}$

Hence, the correct answer is C.

asda

Answer 2

One letter tile is randomly drawn from a set of 26 alphabet tiles representing letters A through Z.  What is the probability that the letter drawn is a vowel or is in the word "now" (y does not count as a vowel here)?  Give your answer as a fraction in reduced form.