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Probability

Bag $A$ contains $5$ white, $6$ black balls and bag $B$ contains $4$ white, $5$ black balls. One bag is selected at random and one ball is drawn from it. Find the probability that it is white.

Let $A_1$ be the event of selecting bag $A$.

Let $A_2$ be the event of selecting bag $B$.

Let $B$ be the event of selecting a white ball.

To find: P(of selecting a white ball) $= P \left(B\right)$

Now, $A_1$ and $A_2$ are mutually exclusive and exhaustive events.

$\therefore \;$ $P \left(B\right) = P \left(A_1\right) \times P \left(B | A_1\right) + P \left(A_2\right) \times P \left(B | A_2\right)$ $\;\;\; \cdots \; (1)$

Probability of selecting bag $A = P \left(A_1\right) = \dfrac{1}{2}$

Probability of selecting bag $B = P \left(A_2\right) = \dfrac{1}{2}$

P(selecting a white ball when bag $A$ is selected) $= P \left(B | A_1\right)$

$P \left(B | A_1\right) = \dfrac{{^{5}}{C}_{1}}{{^{11}}{C}_{1}} = \dfrac{5}{11}$

P(selecting a white ball when bag $B$ is selected) $= P \left(B | A_2\right)$

$P \left(B | A_2\right) = \dfrac{{^{4}}{C}_{1}}{{^{9}}{C}_{1}} = \dfrac{4}{9}$

Substituting the values of $P \left(A_1\right)$, $P \left(A_2\right)$, $P \left(B | A_1\right)$ and $P \left(B | A_2\right)$ in equation $(1)$ we have,

$P \left(B\right) = \left(\dfrac{1}{2} \times \dfrac{5}{11}\right) + \left(\dfrac{1}{2} \times \dfrac{4}{9}\right) = \dfrac{89}{198}$

Probability

In a game played with a standard deck of cards, each face card has a value of $10$ points,, each ace has a value of $1$ point, and each number card has a value equal to its number. Two cards are drawn at random. One card is the queen of diamonds. What is the probability that the sum of the cards is greater than $18$?

The first card can be drawn from a pack of cards in $52$ ways.

The second card can be drawn from the pack of cards in $51$ ways ($\because \;$ cards are drawn without replacement).

Let $A$ be the event that the card drawn is a queen of diamonds.

Then $P\left(A\right) = \dfrac{1}{52}$

Let $B$ be the event of drawing a card from the cards with value $9$ ($4$ cards), value $10$ ($4$ cards), or the face cards ($11$ cards :-- $4$ kings, $3$ queens, $4$ jacks)

(This is done so that the sum of the two cards is greater than $18$.)

$\therefore \;$ Number of elements in $B = n \left(B\right) = 19$

$\therefore \;$ $P \left(B\right) = \dfrac{19}{51}$

$\left(A \cap B\right) \;$ is the event that a card is queen of diamonds AND the sum of the two cards is greater than $18$

$\therefore \;$ $P \left(A \cap B\right) = \dfrac{1}{52} \times \dfrac{19}{51}$

$\therefore \;$ P(when one card is queen of diamonds, then the sum of the cards is greater than $18$)

$= P \left(B | A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)} = \dfrac{\dfrac{1}{52} \times \dfrac{19}{51}}{\dfrac{1}{52}} = \dfrac{19}{51}$

Probability

In a classroom, $60\%$ of the students have brown hair, $30\%$ have brown eyes, and $10\%$ have both brown hair and eyes. A student is selected at random.

If the student has brown hair, what is the probability that the student also has brown eyes?
If the student does not have brown hair, what is the probability that the student does not have brown eyes?

Given: $60\%$ of the students have brown hair; $30\%$ have brown eyes; $10\%$ have both brown hair and brown eyes

Let $A$ be the event of selecting a student with brown hair

Then, $P \left(A\right) = \dfrac{60}{100} = \dfrac{6}{10}$

Let $B$ be the event of selecting a student with brown eyes

Then, $P \left(B\right) = \dfrac{30}{100} = \dfrac{3}{10}$

$\left(A \cap B\right)$ is the event of selecting a student with brown hair AND brown eyes

Then, $P \left(A \cap B\right) = \dfrac{10}{100} = \dfrac{1}{10}$

P(if the student has brown hair then the student also has brown eyes)

$= P \left(B | A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)} = \dfrac{1 / 10}{6 / 10} = \dfrac{1}{6}$
P(student does not have brown hair) $= P\left(\overline{A}\right) = 1 - P \left(A\right) = 1 - \dfrac{6}{10} = \dfrac{4}{10}$

P(if the student does not have brown hair then the student also does not have brown eyes)

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

Now,

$\begin{aligned} P \left(\overline{A} \cap \overline{B}\right) & = P \left(\overline{A \cup B}\right) \\\\ & = 1 - P \left(A \cup B\right) \\\\ & = 1 - \left[P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)\right] \\\\ & = 1 - \left[\dfrac{6}{10} + \dfrac{3}{10} - \dfrac{1}{10}\right] \\\\ & = 1 - \dfrac{8}{10} \\\\ & = \dfrac{2}{10} \end{aligned}$

$\therefore \;$ $P \left(\overline{B} | \overline{A}\right) = \dfrac{2 / 10}{4 / 10} = \dfrac{2}{4} = \dfrac{1}{2}$

Probability

A container holds $3$ green marbles and $5$ yellow marbles. One marble is randomly drawn and discarded. Then a second marble is drawn. Find the probability that the second marble is yellow, given that the first marble was green.

Total number of marbles $= 8$

The first marble can be selected in $8$ ways; the second marble can be selected in $7$ ways.

Let $A$ be the event of drawing the first marble (green).

Let $B$ be the event of drawing the second marble (yellow).

$\therefore \;$ P(selecting a green marble) $= P \left(A\right) = \dfrac{3}{8}$

$1$ yellow marble can be selected from $5$ yellow marbles in $5$ ways

$\therefore \;$ P(selecting a yellow marble) $= \dfrac{5}{7}$

$\left(A \cap B\right) = \;$ event of selecting a green \textbf{AND} a yellow marble

$\therefore \;$ $P \left(A \cap B\right) = \dfrac{3}{8} \times \dfrac{5}{7}$

Now, P(selecting a yellow marble given that the first marble is green)

$= P \left(B|A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)} = \dfrac{\dfrac{3}{8} \times \dfrac{5}{7}}{\dfrac{3}{8}} = \dfrac{5}{7}$

Probability

Two game tiles, numbered $1$ through $9$, are selected at random from a box without replacement. If their sum is even, what is the probability that both the numbers are odd?

$2$ game tiles can be selected from $9$ tiles in ${^{9}}{P}_{2} = \dfrac{9!}{7!} = 9 \times 8 = 72 \;$ ways

$\therefore \;$ Number of elements in sample space $S = n \left(S\right) = 72$

Let $A$ be the event that both the tiles are odd.

Let $B$ be the event that the sum of numbers on the two tiles is even.

This happens when both the selected tiles are odd \textbf{OR} both are even.

There are $5$ odd tiles (1, 3, 5, 7, 9) and $4$ even tiles (2,4,6,8).

$2$ Odd tiles can be selected from $5$ tiles in ${^{5}}{P}_{2} = \dfrac{5!}{3!} = 5 \times 4 = 20 \;$ ways

$2$ even tiles can be selected from $4$ tiles in ${^{4}}{P}_{2} = \dfrac{4!}{2!} = 4 \times 3 = 12 \;$ ways

$\therefore \;$ Number of elements in $B = n \left(B\right) = 20 + 12 = 32$

$\therefore \;$ $P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{32}{72}$

$\left(A \cap B\right)$ is the event that both the tiles are odd numbered \textbf{AND} their sum is even.

$\therefore \;$ $n \left(A \cap B\right) = 20$

$\therefore \;$ $P \left(A \cap B\right) = \dfrac{n \left(A \cap B\right)}{n \left(S\right)} = \dfrac{20}{72}$

Now, P(that both the numbers are odd given that their sum is even)

$= P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)} = \dfrac{20/72}{32/72} = \dfrac{20}{32} = \dfrac{5}{8}$

Probability

A city council consists of six people of party $P_1$, two of whom are women, and six of party $P_2$, four of whom are men. A member is chosen at random. If the member chosen is a man, what is the probability that he is from party $P_1$?

Total number of members in city council $= 12$

$\therefore \;$ Number of elements in sample space $S = n \left(S\right) = 12$

Total number of men in the council $= 8$; $\;$ total number of women in the council $= 4$

Let $A$ be the event of selecting a member from party $P_1$

Let $B$ be the event that the member chosen is a man

Then, $n \left(B\right) = 8$

$\therefore \;$ $P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{8}{12} = \dfrac{2}{3}$

$\left(A \cap B\right) = \;$ event of selecting party $P_1$ AND a man

Party $P_1$ can be selected in $1$ way;

A man from $P_1$ can be selected in $4$ ways

$\therefore \;$ Number of ways of selecting party $P_1$ $\;$ AND $\;$ a man $= 1 \times 4 = 4$ ways

$\therefore \;$ Number of elements in $\left(A \cap B\right) = 4$

$\therefore \;$ $P \left(A \cap B\right) = \dfrac{n \left(A \cap B\right)}{n \left(S\right)} = \dfrac{4}{12} = \dfrac{1}{3}$

$\therefore \;$ P(selecting party $P_1$ when the member is a man)

$= P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)} = \dfrac{1/3}{2/3} = \dfrac{1}{2}$

Probability

A pair of number cubes is thrown. Find each probability given that their sum is greater than or equal to $9$.

$P \left(\text{numbers match}\right)$
$P \left(\text{numbers match or sum is even}\right)$

Number of elements in sample space $S = n \left(S\right) = 36$

Let $C$ be the event that the sum of the numbers on the two cubes is greater than or equal to $9$.

i.e. $\;$ $C = \left\{\left(3, \; 6\right), \; \left(4, \; 5\right), \; \left(4, \; 6\right), \; \left(5, \; 4\right), \; \left(5, \; 5\right), \right.$

$\hspace{3cm}$ $\left. \left(5, \; 6\right), \; \left(6, \; 3\right), \; \left(6, \; 4\right), \; \left(6, \; 5\right), \; \left(6, \; 6\right) \right\}$

$\therefore \;$ Number of elements in $C = n \left(C\right) = 10$

$\therefore \;$ $P \left(C\right) = \dfrac{n \left(C\right)}{n \left(S\right)} = \dfrac{10}{36}$

Let $A$ be the event that the numbers on the two cubes match.

i.e. $\;$ $A = \left\{\left(1, \; 1\right), \; \left(2, \; 2\right), \; \left(3, \; 3\right), \; \left(4, \; 4\right), \; \left(5, \; 5\right), \; \left(6, \; 6\right) \right\}$

$\left(A \cap C\right) = \;$ event that the numbers on both the cubes match AND the sum of the numbers is greater than or equal to $9$

$\therefore \;$ $\left(A \cap C\right) = \left\{\left(5, \; 5\right), \; \left(6, \; 6\right) \right\}$

$\therefore \;$ Number of elements in $\left(A \cap C\right) = n \left(A \cap C\right) = 2$

$\therefore \;$ $P \left(A \cap C\right) = \dfrac{n \left(A \cap C\right)}{n \left(S\right)} = \dfrac{2}{36}$

$\therefore \;$ P(numbers on the two cubes match given that their sum is greater than or equal to 9)

$= P \left(A | C\right) = \dfrac{P \left(A \cap C\right)}{P \left(C\right)} = \dfrac{2/36}{10/36} = \dfrac{2}{10} = \dfrac{1}{5}$
Let $B$ be the event that the numbers on the two cubes match or the sum is even

i.e. $\;$ $B = \left\{\left(1, \; 1\right), \; \left(2, \; 2\right), \; \left(3, \; 3\right), \; \left(4, \; 4\right), \; \left(5, \; 5\right), \; \left(6, \; 6\right), \right. $

$\hspace{2cm}$ $\left. \left(1, \; 3\right), \; \left(1, \; 5\right), \; \left(2, \; 4\right), \; \left(2, \; 6\right), \; \left(3, \; 1\right), \; \left(3, \; 5\right) \right. $

$\hspace{3cm}$ $\left. \left(4, \; 2\right), \; \left(4, \; 6\right), \; \left(5, \; 1\right), \; \left(5, \; 3\right), \; \left(6, \; 2\right), \; \left(6, \; 4\right) \right\}$

$\left(B \cap C\right) = \;$ event that the numbers on the two cubes match or the sum is even AND the sum of the numbers is greater than or equal to $9$

$\therefore \;$ $\left(B \cap C\right) = \left\{\left(4, \; 6\right), \; \left(5, \; 5\right), \; \left(6, \; 4\right), \; \left(6, \; 6\right) \right\}$

$\therefore \;$ Number of elements in $\left(B \cap C\right) = n \left(B \cap C\right) = 4$

$\therefore \;$ $P \left(B \cap C\right) = \dfrac{n \left(B \cap C\right)}{n \left(S\right)} = \dfrac{4}{36}$

$\therefore \;$ P(numbers on the two cubes match or sum is even given that their sum is greater than or equal to 9)

$= P \left(B | C\right) = \dfrac{P \left(B \cap C\right)}{P \left(C\right)} = \dfrac{4/36}{10/36} = \dfrac{4}{10} = \dfrac{2}{5}$