Question

question 1 (1*5 points)
a card is drawn randomly from a standard 52 - card deck. find the probability of the given event.
(a) the card drawn is 5
the probability is
(b) the card drawn is a red card
the probability is
(c) the card drawn is not a face card (jack, queen, or king)
the probability is
question 2 (5 points)
suppose a jar contains 12 red marbles and 33 blue marbles. if you reach in the jar and pull out 2
marbles at random at the same time, find the probability that both are red.

question 3 (2*5 points)
giving a test to a group of students, the grades and gender are summarized below

if one student is chosen at random,
find the probability that the student was female or got a \b\.

find the probability that the student got a \c\ given they are male.

question 4 (5 points)
how many different ways can a race with 6 runners be completed? (assume there is no tie.)
your answer is:

question 5 (5 points)
from a group of 10 people, you randomly select 3 of them. what is the probability that they are the
3 oldest people in the group? (assume that they are different in age.)

give your answer as a fraction

Explanation:

Question 1 (a)

Step1: Determine total and favorable cards

Total cards \(n = 52\). Number of \(5\)s: there are \(4\) fives (one for each suit). Favorable \(m=4\).
Probability formula \(P=\frac{m}{n}\).
So \(P=\frac{4}{52}=\frac{1}{13}\).

Question 1 (b)

Step1: Count red cards

Total cards \(n = 52\). Red cards (hearts and diamonds): \(26\). Favorable \(m = 26\).
Using \(P=\frac{m}{n}\), we get \(P=\frac{26}{52}=\frac{1}{2}\).

Question 1 (c)

Step1: Count non - face cards

Total cards \(n = 52\). Face cards (\(J,Q,K\) of each suit): \(3\times4 = 12\). Non - face cards \(m=52 - 12=40\).
By \(P=\frac{m}{n}\), \(P=\frac{40}{52}=\frac{10}{13}\).

Question 2

Step1: Calculate total and red marble combinations

Total marbles \(N=12 + 33=45\). Number of ways to choose \(2\) marbles from \(45\): \(C(N,2)=\frac{45!}{2!(45 - 2)!}=\frac{45\times44}{2\times1}=990\).
Number of red marbles \(n = 12\). Number of ways to choose \(2\) red marbles: \(C(n,2)=\frac{12!}{2!(12 - 2)!}=\frac{12\times11}{2\times1}=66\).
Probability \(P=\frac{C(n,2)}{C(N,2)}=\frac{66}{990}=\frac{1}{15}\).

Question 3 (first part)

Step1: Use the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

Let \(A\) be the event of being female and \(B\) be the event of getting a \(B\).
\(n(A) = 22\), \(n(B)=21\), \(n(A\cap B)=16\), \(n(\text{total}) = 59\).
\(P(A)=\frac{22}{59}\), \(P(B)=\frac{21}{59}\), \(P(A\cap B)=\frac{16}{59}\).
\(P(A\cup B)=\frac{22 + 21-16}{59}=\frac{27}{59}\).

Question 3 (second part)

Step1: Use conditional probability formula \(P(C|A)=\frac{P(A\cap C)}{P(A)}\)

Let \(A\) be the event of being male and \(C\) be the event of getting a \(C\).
\(n(A) = 37\), \(n(A\cap C)=20\).
\(P(C|A)=\frac{20}{37}\).

Question 4

Step1: Use permutation formula \(n!\) for \(n\) distinct objects

For \(n = 6\) runners, the number of ways \(=6! = 720\).

Question 5

Step1: Calculate combination

Number of ways to choose \(3\) people from \(10\): \(C(10,3)=\frac{10!}{3!(10 - 3)!}=\frac{10\times9\times8}{3\times2\times1}=120\).
There is only \(1\) way to choose the \(3\) oldest.
Probability \(P=\frac{1}{120}\).

Answer:

Question 1 (a): \(\frac{1}{13}\)
Question 1 (b): \(\frac{1}{2}\)
Question 1 (c): \(\frac{10}{13}\)
Question 2: \(\frac{1}{15}\)
Question 3 (first part): \(\frac{27}{59}\)
Question 3 (second part): \(\frac{20}{37}\)
Question 4: \(720\)
Question 5: \(\frac{1}{120}\)