Intro to Probability - Sample Test

Problem 1

What is the probability that a hand of 5 cards randomly picked from a deck of 52 cards does not contain any spades? A deck of cards consists of 52 cards of 4 different suits (clubs, diamonds, hearts, and spades) and 13 values ($1,2,\dots,10,J,Q,K$) in each suit.

Out of the ${52 \choose 5}$ possible hands, ${39 \choose 5}$ are favorable ($39$ of the cards are of a suit other than spades). Thus the probabilty amounts to $$ \frac{{39\choose 5}}{{52 \choose 5}}\sim 22.15\% $$

Problem 2

A florist has 5 types of flowers in stock. How many floral compositions consisting of 10 flowers can he create?

Each composition is completely described by the number of flowers of each type it contains. Since any given type need not be represented, there are a total of $$ {10+5-1 \choose 5-1}={14 \choose 4}=1001 $$ of compositions since we are choosing 10 objects of 5 types.

Problem 3

Take three events $E,F,G$ with $P(G)\neq 0$ from a probability space and consider the following identity: $$ P(E\cap F|G)=P(E|G)+P(F|G)-P(E\cup F|G). $$ Prove its validity by providing a complete argument or give a counterexample if it does not hold.

By definition of conditional probability it holds that $$ P(A|B)=\frac{P(A\cap B)}{P(B)} $$ for any events $A,B$ of a probability space with $P(B)\neq 0$. Then \begin{align*} P(E\cup F|G)&=\frac{P\bigl((E\cup F)\cap G\bigr)}{P(G)}= \frac{P \bigl((E\cap G)\cup(F\cap G)\bigr) }{P(G)}\\&= \frac{1}{P(G)}\bigl[ P(E\cap G)+P(F\cap G)-P(E\cap F\cap G)\bigr] \\ &=P(E|G)+P(F|G)-P(E\cap F|G) \end{align*} since $(E\cap G)\cap(F\cap G)=E\cap F\cap G$, which shows the validity of the identity.

Problem 4

There are 25 strawberry flavored and 15 raspberry flavored candies in a candy bag. Eating the candies in a random order, what is the probability that all strawberry flavored candies are eaten before the last raspberry flavored one?

The event of interest occurrs if the last candy in the bag is raspberry flavored. Thus the probability amounts to $\frac{15}{25+15}=15/40=3/8$. Alternatively there are $\frac{40!}{25!15!}$ distinguishable orders in which the candies can be eaten. Of these, the favorable ones are those where the last candy is raspberry and add up to $\frac{39!}{25!14!}$. Computing the ratio, we obtain the same result.

Problem 5

Two urns contain 10 red and 5 black balls and a third urn contains 8 red and 12 black balls. If you randomly choose a ball from a randomly chosen urn, what is the probability that it is red?

Let $U$ denote the event of choosing one of the two first urns and $R$ denote the event that a red ball be drawn. Then, by Bayes' formula, \begin{align*} P(R)&=P(R|U)P(U)+P(R|U^\mathsf{c})P(U^\mathsf{c})\\ &=\frac{10}{15}\frac{2}{3}+\frac{8}{20}\frac{1}{3}=\frac{26}{45}. \end{align*}

Problem 6

A basketball coach has to form a co-ed team of 5 players with a captain from a roster of 14 players (8 female and 6 male). How many total teams can she form if a team has to have at least 2 female and 1 male players and the captain needs to be female?

A team will comprise 2, 3, or 4 female players. There are ${8\choose 2}{6\choose 3}$ possibilities for the first case, ${8\choose 3}{6\choose 2}$ for the second, and ${8\choose 4}{6\choose 1}$ for the last. In the first case there are two choices of a captain, 3 and 4, respectively in the remaining two cases. Thus we arrive at a total of $$ 2 {8\choose 2}{6\choose 3}+3 {8\choose 3}{6\choose 2}+4 {8\choose 4}{6\choose 1}=5,320 $$ possibilities.