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Probability

Given $P \left(A\right) = 0.50$, $P \left(B\right) = 0.40$ and $P \left(A \cap B\right) = 0.20$

Verify that

$P \left(A | B\right) = P \left(A\right)$
$P \left(A | \overline{B}\right) = P \left(A\right)$
$P \left(B | A\right) = P \left(B\right)$
$P \left(B | \overline{A}\right) = P \left(B\right)$

By definition, $P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)}$

i.e. $\;$ $P \left(A | B\right) = \dfrac{0.20}{0.40} = \dfrac{1}{2} = 0.50 = P \left(A\right)$
By definition, $P \left(A | \overline{B}\right) = \dfrac{P \left(A \cap \overline{B}\right)}{P \left(\overline{B}\right)}$

Now, $P \left(\overline{B}\right) = 1 - P \left(B\right) = 1 - 0.40 = 0.60$

and $P \left(A \cap \overline{B}\right) = P \left(A\right) - P \left(A \cap B\right) = 0.50 - 0.20 = 0.30$

$\therefore \;$ $P \left(A | \overline{B}\right) = \dfrac{0.30}{0.60} = \dfrac{1}{2} = 0.50 = P \left(A\right)$
By definition, $P \left(B | A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)}$

i.e. $\;$ $P \left(A | B\right) = \dfrac{0.20}{0.50} = \dfrac{2}{5} = 0.40 = P \left(B\right)$
By definition, $P \left(B | \overline{A}\right) = \dfrac{P \left(B \cap \overline{A}\right)}{P \left(\overline{A}\right)}$

Now, $P \left(\overline{A}\right) = 1 - P \left(A\right) = 1 - 0.50 = 0.50$

and $P \left(B \cap \overline{A}\right) = P \left(B\right) - P \left(A \cap B\right) = 0.40 - 0.20 = 0.20$

$\therefore \;$ $P \left(B | \overline{A}\right) = \dfrac{0.20}{0.50} = \dfrac{2}{5} = 0.40 = P \left(B\right)$

Probability

If $P \left(A\right) = 0.4$, $P \left(B\right) = 0.7$ and $P \left(B | A\right) = 0.5$, find $P \left(A | B\right)$ and $P \left(A \cup B\right)$

By definition, $P \left(B | A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)}$

i.e. $\;$ $0.5 = \dfrac{P \left(A \cap B\right)}{0.4}$

$\implies$ $P \left(A \cap B\right) = 0.5 \times 0.4 = 0.2$

Now, $P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)}$

$\implies$ $P \left(A | B\right) = \dfrac{0.2}{0.7} = \dfrac{2}{7}$

By definition, $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$

i.e. $\;$ $P \left(A \cup B\right) = 0.4 + 0.7 - 0.2 = 0.9 = \dfrac{9}{10}$

Probability

A teacher gives her mathematics class $20$ study problems. She selects $10$ to answer on an upcoming test. A student can solve $15$ of the problems.

Find the probability that the student can solve all $10$ problems on the test.
Find the odds that the student will know how to solve $8$ of the problems.

$10$ problems can be selected from $20$ problems in

${^{20}}{C}_{10} = \dfrac{20!}{10! \times 10!} = 184756$ ways.

$\therefore \;$ Number of elements in sample space $S = n \left(S\right) = 184756$

Let $A$ be the event that the student can solve all $10$ problems.

Total number of problems the student can solve $= 15$

$\therefore \;$ $10$ problems can be selected from $15$ problems in

${^{15}}{C}_{10} = \dfrac{15!}{10! \times 5!} = 3003$ ways

$\therefore \;$ Number of elements in $A = n \left(A\right) = 3003$

$\therefore \;$ Probability of event $A = P \left(A\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{3003}{184756} = \dfrac{21}{1292}$
Let $B$ be the event that the student solves $8$ problems.

Total number of problems the student cannot solve $= 5$

$8$ problems can be selected from $15$ problems in

${^{15}}{C}_{8} = \dfrac{15!}{8! \times 7!} = 6435$ ways

$2$ problems which cannot be solved can be selected from the $5$ problems in

${^{5}}{C}_{2} = \dfrac{5!}{3! \times 2!} = 10$ ways

$\therefore \;$ Number of ways in which the student can solve $8$ problems is

$= 6435 \times 10 = 64350$ ways

$\therefore \;$ Number of elements in $B = n \left(B\right) = 64350$

Let $P \left(s\right)$ be the probability of success of event $B$ and $P \left(f\right)$ be the probability of failure of event $B$

$P \left(s\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{64350}{184756} = \dfrac{225}{646}$

$P \left(f\right) = 1- \dfrac{225}{646} = \dfrac{421}{646}$

$\therefore \;$ Odds that the student will know how to solve $8$ of the problems

$= \dfrac{P \left(s\right)}{P \left(f\right)} = \dfrac{225/646}{421/646} = \dfrac{225}{421}$

Probability

Jill uses a combination lock on her locker that has $3$ wheels, each labeled with $10$ digits from $0$ to $9$. The combination is a particular sequence with no digits repeating.

What is the probability of someone guessing the correct combination?
If the digits can be repeated, what are the odds against someone guessing the combination?

Let $A$ be the event of guessing the correct combination when digits are not repeated

Since no digits are repeated,

the first digit can be selected (from the digits $0$ to $9$) in $10$ ways;

the second digit can be selected in $9$ ways;

and the third digit can be selected in $8$ ways.

$\therefore \;$ Number of ways of selecting the $3$ digits $= 10 \times 9 \times 8 = 720$ ways.

$\therefore \;$ Number of elements in sample space $S = n\left(S\right) = 720$

The correct combination can be selected in $1$ way.

$\therefore \;$ Number of elements in event $A = n \left(A\right) = 1$

$\therefore \;$ Probability of guessing the correct combination $= P \left(A\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{1}{720}$
Let $B$ be the event of guessing the correct combination when the digits are repeated

Then, the first digit can be selected (from the digits $0$ to $9$) in $10$ ways;

the second digit can also be selected in $10$ ways;

and the third digit can also be selected in $10$ ways.

$\therefore \;$ Number of ways of selecting the $3$ digits $= 10 \times 10 \times 10 = 1000$ ways

$\therefore \;$ Number of elements in sample space $S = n \left(S\right) = 1000$

The correct combination can only be selected in $1$ way.

$\therefore \;$ Number of elements in event $B = n \left(B\right) = 1$

Let $P \left(s\right)$ be the probability of success of event $B$ and $P \left(f\right)$ be the probability of failure of event $B$

$P \left(s\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{1}{1000}$

$P \left(f\right) = 1- \dfrac{1}{1000} = \dfrac{999}{1000}$

$\therefore \;$ Odds against guessing the correct combination

$= \dfrac{P \left(f\right)}{P \left(s\right)} = \dfrac{999 / 1000}{1 / 1000} = \dfrac{999}{1}$

Probability

Of $27$ students in a class, $11$ have blue eyes, $13$ have brown eyes, and $3$ have green eyes. If $3$ students are chosen at random, what are the odds of each event occurring?

$2$ have brown eyes and $1$ has blue eyes
only $1$ has green eyes

$3$ students can be selected from $27$ students in

${^{27}}{P}_{3} = \dfrac{27!}{24!} = 27 \times 26 \times 25 = 17,550$ ways

$\therefore \;$ Number of elements in sample space $S = n \left(S\right) = 17550$

Let $A$ be the event of selecting $2$ students who have brown eyes AND $1$ student who has blue eyes

$2$ students with brown eyes can be selected from $13$ students in

${^{13}}{P}_{2} = \dfrac{13!}{11!} = 13 \times 12 = 156$ ways

$1$ student with blue eyes can be selected from $11$ students in $11$ ways

Amongst themselves, the $3$ students can be selected in $3$ ways

$\therefore \;$ Number of ways of selecting $2$ students with brown eyes and $1$ student with blue eyes

$= 156 \times 11 \times 3 = 5148$ ways

$\therefore \;$ Number of elements in $A = n\left(A\right) = 5148$

Let $P \left(s\right)$ be the probability of success of event $A$ and $P \left(f\right)$ be the probability of failure of event $A$

$P \left(s\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{5148}{17550} = \dfrac{286}{975}$

$P \left(f\right) = 1- \dfrac{286}{975} = \dfrac{689}{975}$

$\therefore \;$ Odds of event $A = \dfrac{P \left(s\right)}{P \left(f\right)} = \dfrac{286/975}{689/975} = \dfrac{22}{53}$
Let $B$ be the event of selecting $3$ students such that only $1$ student has green eyes

$1$ student with green eyes can be selected from $3$ students in ${^{3}}{P}_{1} = 3$ ways

Remaining $2$ students can be selected from the remaining $24$ students in

${^{24}}{P}_{2} = \dfrac{24!}{22!} = 24 \times 23$ ways

Amongst themselves, these $3$ students can be selected in $3$ ways

$\therefore \;$ Number of ways of selecting $3$ students such that only $1$ student has green eyes

$= 3 \times 24 \times \times 23 \times 3 = 4968$ ways

$\therefore \;$ Number of elements in $B = n\left(B\right) = 4968$

Let $P \left(s\right)$ be the probability of success of event $B$ and $P \left(f\right)$ be the probability of failure of event $B$

$P \left(s\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{4968}{17550} = \dfrac{92}{325}$

$P \left(f\right) = 1- \dfrac{92}{325} = \dfrac{233}{325}$

$\therefore \;$ Odds of event $B = \dfrac{P \left(s\right)}{P \left(f\right)} = \dfrac{92/325}{233/325} = \dfrac{92}{233}$

Probability

A box contains $1$ green, $2$ yellow and $3$ red marbles. Two marbles are drawn at random without replacement. What are the odds of each event occurring?

not drawing yellow marbles;
drawing marbles of two different colors.

$2$ marbles can be drawn from $1 + 2 + 3 = 6$ marbles (without replacement) in

${^{6}}{P}_{2} = \dfrac{6!}{4!} = 6 \times 5 = 30$ ways

$\therefore \;$ Number of elements in sample space $= n \left(S\right) = 30$

Let $A$ be the event of not drawing yellow marbles

Then $2$ marbles can be drawn from the remaining $4$ marbles in ${^{4}}{P}_{2} = \dfrac{4!}{2!} = 12$ ways

$\therefore \;$ Number of elements in $A = n \left(A\right) = 12$

Let $P \left(s\right)$ be the probability of success of event $A$ and $P \left(f\right)$ be the probability of failure of event $A$

$P \left(s\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{12}{30} = \dfrac{2}{5}$

$P \left(f\right) = 1- \dfrac{2}{5} = \dfrac{3}{5}$

$\therefore \;$ Odds of event $A = \dfrac{P \left(s\right)}{P \left(f\right)} = \dfrac{2 / 5}{3 / 5} = \dfrac{2}{3}$
Let $B$ be the event of drawing marbles of two different colors

Then, the possibilities for event $B$ are

($1$ green AND $1$ yellow) OR ($1$ green AND $1$ red) OR ($1$ yellow AND $1$ red)

$1$ green AND $1$ yellow marble:

$1$ green marble can be selected in $1$ way

$1$ yellow marble can be selected from $2$ yellow marbles in $2$ ways

Amongst themselves, $1$ green AND $1$ yellow marble can be selected in in $2$ ways

$\therefore \;$ $1$ green AND $1$ yellow marble can be selected in $1 \times 2 \times 2 = 4$ ways

$1$ green AND $1$ red marble:

$1$ green marble can be selected in $1$ way

$1$ red marble can be selected from $3$ red marbles in $3$ ways

Amongst themselves, $1$ green AND $1$ red marble can be selected in in $2$ ways

$\therefore \;$ $1$ green AND $1$ red marble can be selected in $1 \times 3 \times 2 = 6$ ways

$1$ yellow AND $1$ red marble:

$1$ yellow marble can be selected from $2$ yellow marbles in $2$ ways

$1$ red marble can be selected from $3$ red marbles in $3$ ways

Amongst themselves, $1$ yellow AND $1$ red marble can be selected in in $2$ ways

$\therefore \;$ $1$ green AND $1$ red marble can be selected in $2 \times 3 \times 2 = 12$ ways

$\therefore \;$ Two marbles of two different colors can be selected in $4 + 6 + 12 = 22$ ways

$\therefore \;$ Number of elements in $B = n \left(B\right) = 22$

Let $P \left(s\right)$ be the probability of success of event $B$ and $P \left(f\right)$ be the probability of failure of event $B$

$P \left(s\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{22}{30} = \dfrac{11}{15}$

$P \left(f\right) = 1- \dfrac{11}{15} = \dfrac{4}{15}$

$\therefore \;$ Odds of event $B = \dfrac{P \left(s\right)}{P \left(f\right)} = \dfrac{11 / 15}{4 / 15} = \dfrac{11}{4}$

Probability

The probability that a new ship will get an award for its design is $0.25$, the probability that it will get an award for the efficient use of materials is $0.35$, and that it will get both awards is $0.15$. What is the probability, that

it will get at least one of the two awards;
it will get only one of the awards.

Let $D$ be the event that a ship gets an award for its design.

Let $M$ be the event that a ship gets an award for the efficient use of materials.

Then, $\;$ $P \left(D\right) = 0.25$, $\;$ $P \left(M\right) = 0.35$ $\;$ and $\;$ $P \left(D \cap M\right) = 0.15$

Let $A$ be the event that the ship gets at least one of the two awards

$\begin{aligned} P \left(A\right) & = P \left(D \;\; OR \;\; M\right) \\\\ & = P \left(D \cup M\right) \\\\ & = P \left(D\right) + P \left(M\right) - P \left(D \cap M\right) \\\\ & = 0.25 + 0.35 - 0.15 \\\\ & = 0.45 \end{aligned}$
Let $B$ be the event that the ship gets only one of the awards

$\begin{aligned} P \left(B\right) & = P \left(\text{only D OR only M}\right) \\\\ & = P \left[\left(D \cap \overline{M}\right) \cup \left(\overline{D} \cap M\right)\right] \\\\ & = P \left(D \cap \overline{M}\right) + P \left(\overline{D} \cap M\right) \\\\ & = \left\{P \left(D\right) - P \left(D \cap M\right)\right\} + \left\{P \left(M\right) - P \left(D \cap M\right)\right\} \\\\ & = \left\{0.25 - 0.15\right\} + \left\{0.35 - 0.15\right\} \\\\ & = 0.10 + 0.20 \\\\ & = 0.30 \end{aligned}$