Probability

Three coins are tossed. Find the probability that they all land heads up when at least one coin shows a head.

Number of elements in sample space $S = n \left(S\right) = 2^{3} = 8$

Let $A$ be the event that all 3 coins land heads up.

Then $A = \left\{H, \; H, \; H\right\}$

Let $B$ be the event that at least one coin shows a head.

Then $\overline{B}$ is the event that all coins land tails up.

Now, $\overline{B} = \left\{T, \; T, \; T\right\}$ $\implies$ $n \left(\overline{B}\right) = 1$

$\therefore \;$ $P \left(\overline{B}\right) = \dfrac{n \left(\overline{B}\right)}{n \left(S\right)} = \dfrac{1}{8}$

$\therefore \;$ $P \left(B\right) = 1 - P \left(\overline{B}\right) = 1 - \dfrac{1}{8} = \dfrac{7}{8}$

$\left(A \cap B\right) = $ event that all coins show heads AND at least one coin shows a head

Now, $\left(A \cap B\right) = \left\{H, \; H, \; H\right\}$ $\implies$ $n \left(A \cap B\right) = 1$

$\therefore \;$ $P \left(A \cap B\right) = \dfrac{n \left(A \cap B\right)}{n \left(S\right)} = \dfrac{1}{8}$

$\therefore \;$ Probability that all coins land heads up when at least one coin shows a head

$= P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)} = \dfrac{1/8}{7/8} = \dfrac{1}{7}$

Probability

One card is drawn from a standard deck of cards. What is the probability that it is a queen if it is known to be a face card?

Number of elements in sample space $S = n \left(S\right) = 52$

Let $A$ be the event of drawing a queen.

Let $B$ be the event of drawing a face card.

Number of elements in $B = n \left(B\right) = 12$

$\therefore \;$ $P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{12}{52}$

$\left(A \cap B\right) = $ event that the face card is a queen

$\because \;$ There are $4$ queen cards, $n \left(A \cap B\right) = 4$

$\therefore \;$ $P \left(A \cap B\right) = \dfrac{n \left(A \cap B\right)}{n \left(S\right)} = \dfrac{4}{52}$

$\therefore \;$ Probability of drawing a queen if the card selected is a face card

$= P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)} = \dfrac{4/52}{12/52} = \dfrac{4}{12} = \dfrac{1}{3}$

Probability

A can hit a target $4$ times in $5$ shots, B $3$ times in $4$ shots and C $2$ times in $3$ shots. They fire a volley. What is the chance that the target is damaged by exactly $2$ hits?

Let $A$ be the event that A hits the target.

Let $B$ be the event that B hits the target.

Let $C$ be the event that C hits the target.

Then, $P \left(A\right) = \dfrac{4}{5}$, $\;$ $P \left(\overline{A}\right) = 1 - P \left(A\right) = 1 - \dfrac{4}{5} = \dfrac{1}{5}$

$P \left(B\right) = \dfrac{3}{4}$, $\;$ $P \left(\overline{B}\right) = 1 - P \left(B\right) = 1 - \dfrac{3}{4} = \dfrac{1}{4}$

$P \left(C\right) = \dfrac{2}{3}$, $\;$ $P \left(\overline{C}\right) = 1 - P \left(C\right) = 1 - \dfrac{2}{3} = \dfrac{1}{3}$

Now, probability that the target is damaged by exactly $2$ hits

$= P \left[\left(A \cap B \cap \overline{C}\right) \cup \left(A \cap \overline{B} \cap C\right) \cup \left(\overline{A} \cap B \cap C\right)\right]$

$= P \left(A\right) \times P \left(B\right) \times P \left(\overline{C}\right) + P \left(A\right) \times P \left(\overline{B}\right) \times P \left(C\right) + P \left(\overline{A}\right) \times P \left(B\right) \times P \left(C\right)$

[$\because \;$ $A$, $B$, $C$ are independent events, $\overline{A}$, $\overline{B}$, $\overline{C}$ are also independent events]

$= \left(\dfrac{4}{5} \times \dfrac{3}{4} \times \dfrac{1}{3}\right) + \left(\dfrac{4}{5} \times \dfrac{1}{4} \times \dfrac{2}{3}\right) + \left(\dfrac{1}{5} \times \dfrac{3}{4} \times \dfrac{2}{3}\right) $

$= \dfrac{26}{60}$

$= \dfrac{13}{30}$

Probability

A problem in Mathematics is given to three students whose chances of solving it are $\dfrac{1}{2}$, $\dfrac{1}{3}$ and $\dfrac{1}{4}$.

1. What is the probability that the problem is solved?
2. What is the probability that exactly one of them will solve it?

Let $A$ be the event that the problem is solved by the first student.

Let $B$ be the event that the problem is solved by the second student.

Let $C$ be the event that the problem is solved by the third student.

Given: $\;$ $P \left(A\right) = \dfrac{1}{2}$, $\;$ $P \left(B\right) = \dfrac{1}{3}$, $\;$ $P \left(C\right) = \dfrac{1}{4}$

Then, $P \left(\overline{A}\right) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$, $\;$ $P \left(\overline{B}\right) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$, $\;$ $P \left(\overline{C}\right) = 1 - \dfrac{1}{4} = \dfrac{3}{4}$

1. Probability that the problem is solved:
   
   $\begin{aligned} P \left(\text{problem is solved}\right) & = P \left(\text{problem is solved by at least one of them}\right) \\\\ & = P \left(A \cup B \cup C\right) \\\\ & = 1 - P \left(\overline{A \cup B \cup C}\right) \\\\ & = 1 - P \left(\overline{A} \cap \overline{B} \cap \overline{C}\right) \;\;\; \left[\text{By DeMorgan's law}\right] \\\\ & = 1 - P \left(\overline{A}\right) \times P \left(\overline{B}\right) \times P \left(\overline{C}\right) \\\\ & \left[\because \; \text{A, B, C are independent events, } \right. \\ & \hspace{2cm} \left. \overline{A}, \overline{B}, \overline{C} \text{ are also independent events}\right] \\\\ & = 1 - \dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4} \\\\ & = \dfrac{3}{4} \end{aligned}$
   
   
2. Exactly one of them solves the problem:
   
   $P \left(\text{exctly one solves the problem}\right)$
   
   $= P \left[\left(A \cap \overline{B} \cap \overline{C}\right) \cup \left(\overline{A} \cap B \cap \overline{C} \right) \cup\left(\overline{A} \cap \overline{B} \cap C\right)\right] $
   
   $ = P \left(A \cap \overline{B} \cap \overline{C}\right) + P \left(\overline{A} \cap B \cap \overline{C}\right) + P \left(\overline{A} \cap \overline{B} \cap C\right) $
   
   $ \left[\because \; \left(A \cap \overline{B} \cap \overline{C}\right), \; \left(\overline{A} \cap B \cap \overline{C}\right), \; \left(\overline{A} \cap \overline{B} \cap C\right) \right.$
   $\hspace{3cm} \left. \text{ are mutually exclusive events} \right]$
   
   $= P \left(A\right) \times P \left(\overline{B}\right) \times P \left(\overline{C}\right) + P \left(\overline{A}\right) \times P \left(B\right) \times P \left(\overline{C}\right) $
   $\hspace{5cm} + P \left(\overline{A}\right) \times P \left(\overline{B}\right) \times P \left(C\right) $
   
   $ \left[\because \; \text{A, B, C are independent events, } \overline{A}, \overline{B}, \overline{C} \text{ are also independent events}\right] $
   
   $ = \left(\dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4}\right) + \left(\dfrac{1}{2} \times \dfrac{1}{3} \times \dfrac{3}{4}\right) + \left(\dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{1}{4}\right) $
   
   $ = \dfrac{11}{24} $

Probability

One bag contains $5$ white and $3$ black marbles. Another bag contains $4$ white and $6$ black marbles. If one marble is drawn from each bag, find the probability that

1. both are white;
2. both are black;
3. one is white and one is black.

Let bag $B_1$ contain $5$ white and $3$ black marbles.

Let bag $B_2$ contain $4$ white and $6$ black marbles.

1. Let $A$ be the event of drawing a white marble from bag $B_1$
   
   Let $B$ be the event of drawing a white marble from bag $B_2$
   
   Then, probability of drawing white marbles from both the bags $= P \left(A \cap B\right)$
   
   $\because \;$ $A$ and $B$ are independent events, $P \left(A \cap B\right) = P \left(A\right) \times P \left(B\right)$
   
   Now, $P \left(A\right) = \dfrac{5}{8}$, $\;$ $P \left(B\right) = \dfrac{4}{10}$
   
   $\therefore \;$ $P \left(A \cap B\right) = \dfrac{5}{8} \times \dfrac{4}{10} = \dfrac{1}{4}$
   
   
2. Let $C$ be the event of drawing a black marble from bag $B_1$
   
   Let $D$ be the event of drawing a black marble from bag $B_2$
   
   Then, probability of drawing black marbles from both the bags $= P \left(C \cap D\right)$
   
   $\because \;$ $C$ and $D$ are independent events, $P \left(C \cap D\right) = P \left(C\right) \times P \left(D\right)$
   
   Now, $P \left(C\right) = \dfrac{3}{8}$, $\;$ $P \left(D\right) = \dfrac{6}{10}$
   
   $\therefore \;$ $P \left(C \cap D\right) = \dfrac{3}{8} \times \dfrac{6}{10} = \dfrac{9}{40}$
   
   
3. Let $E$ be the event of drawing a white marble from bag $B_1$
   
   Let $F$ be the event of drawing a black marble from bag $B_2$
   
   Let $G$ be the event of drawing a black marble from bag $B_1$
   
   Let $H$ be the event of drawing a white marble from bag $B_2$
   
   Then, probability of drawing a white marble from one bag and a black marble from the other bag $= P \left(E \cap F\right) + P \left(G \cap H\right)$
   
   $\because \;$ $E$, $F$ are independent events and $G$, $H$ are independent events, we have
   
   $= P \left(E \cap F\right) + P \left(G \cap H\right) = P \left(E\right) \times P \left(F\right) + P \left(G\right) \times P \left(H\right)$
   
   Now, $P \left(E\right) = \dfrac{5}{8}$, $\;$ $P \left(F\right) = \dfrac{6}{10}$, $\;$ $P \left(G\right) = \dfrac{3}{8}$, $\;$ $P \left(F\right) = \dfrac{4}{10}$
   
   $\begin{aligned} \therefore \; P \left(E \cap F\right) + P \left(G \cap H\right) & = \dfrac{5}{8} \times \dfrac{6}{10} + \dfrac{3}{8} \times \dfrac{4}{10} \\\\ & = \dfrac{42}{80} \\\\ & = \dfrac{21}{40} \end{aligned}$

Probability

Two cards are drawn one by one at random from a deck of $52$ playing cards. What is the probability of getting two jacks if

1. the first card is replaced before the second is drawn;
2. the first card is not replaced before the second card is drawn.

Let $A$ be the event of drawing a jack in the first draw.

Let $B$ be the event of drawing a jack in the second draw.

1. When the first card is replaced before the second card is drawn:
   
   Number of elements in sample space $S = n \left(S\right) = 52$
   
   $\because \;$ there are $4$ jacks
   
   $n \left(A\right) = 4$, $n \left(B\right) = 4$
   
   Event $A$ will not affect the probability of occurrence of event $B$
   
   i.e. $\;$ $A$ and $B$ are independent events.
   
   $\begin{aligned} \therefore \; \text{Required probability} = P \left(A \cap B\right) & = P \left(A\right) \times P \left(B\right) \\\\ & = \dfrac{n \left(A\right)}{n \left(S\right)} \times \dfrac{n \left(B\right)}{n \left(S\right)} \\\\ & = \dfrac{4}{52} \times \dfrac{4}{52} \\\\ & = \dfrac{1}{169} \end{aligned}$
   
   
2. When the first card is not replaced before the second card is drawn:
   
   In the first draw, there are $4$ jacks and $52$ cards in total.
   
   $\because \;$ The card drawn is not replaced, for the second draw there are only $3$ jacks and $51$ cards in total.
   
   $\therefore \;$ We have $P \left(A\right) = \dfrac{4}{52}$, $P \left(B | A\right) = \dfrac{3}{51}$
   
   By definition, $P \left(B | A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)}$
   
   $\begin{aligned} \therefore \; P \left(A \cap B\right) & = P \left(A\right) \times P \left(B | A\right) \\\\ & = \dfrac{4}{52} \times \dfrac{3}{51} \\\\ & = \dfrac{1}{221} \end{aligned}$

Probability

Given $P \left(A\right) = 0.45$ and $P \left(A \cup B\right) = 0.75$

Find $P \left(B\right)$ if

1. $A$ and $B$ are mutually exclusive;
2. $A$ and $B$ are independent events;
3. $P \left(A | B\right) = 0.50$;
4. $P \left(B | A\right) = 0.50$

1. If $A$ and $B$ are mutually exclusive events, then $P \left(A \cap B\right) = 0$
   
   By definition, $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$
   
   $\because \;$ $A$ and $B$ are mutually exclusive events,
   
   $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right)$
   
   $\therefore \;$ $P \left(B\right) = P \left(A \cup B\right) - P \left(A\right) = 0.75 - 0.45 = 0.30 = \dfrac{3}{10}$
   
   
2. If $A$ and $B$ are independent events, then $P \left(A \cap B\right) = P \left(A\right) \times P \left(B\right)$
   
   By definition, $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$
   
   $\because \;$ $A$ and $B$ are independent events,
   
   $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A\right) \times P \left(B\right)$
   
   $\implies$ $P \left(B\right) = \dfrac{P \left(A \cup B\right) - P \left(A\right)}{1 - P \left(A\right)}$
   
   i.e. $\;$ $P \left(B\right) = \dfrac{0.75 - 0.45}{1 - 0.45} = \dfrac{0.30}{0.55} = \dfrac{30}{55} = \dfrac{6}{11}$
   
   
3. By definition, $P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)}$
   
   i.e. $\;$ $P \left(A | B\right) \times P \left(B\right) = P \left(A \cap B\right)$ $\;\;\; \cdots \; (1)$
   
   Now, $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$
   
   i.e. $\;$ $P \left(A \cap B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cup B\right)$ $\;\;\; \cdots \; (2)$
   
   $\therefore \;$ We have from equations $(1)$ and $(2)$,
   
   $P \left(A\right) + P \left(B\right) - P \left(A \cup B\right) = P \left(A | B\right) \times P \left(B\right)$
   
   i.e. $\;$ $P \left(B\right) = \dfrac{P \left(A\right) - P \left(A \cup B\right)}{P \left(A | B\right) - 1}$
   
   i.e. $\;$ $P \left(B\right) = \dfrac{P \left(A \cup B\right) - P \left(A\right)}{1 - P \left(A | B\right)}$
   
   i.e. $\;$ $P \left(B\right) = \dfrac{0.75 - 0.45}{1 - 0.50} = \dfrac{0.30}{0.50} = \dfrac{3}{5} = 0.60$
   
   
4. By definition, $P \left(B | A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)}$
   
   i.e. $\;$ $P \left(B | A\right) \times P \left(A\right) = P \left(A \cap B\right)$ $\;\;\; \cdots \; (3)$
   
   Now, $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$
   
   i.e. $\;$ $P \left(A \cap B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cup B\right)$ $\;\;\; \cdots \; (4)$
   
   $\therefore \;$ We have from equations $(3)$ and $(4)$,
   
   $P \left(A\right) + P \left(B\right) - P \left(A \cup B\right) = P \left(B | A\right) \times P \left(A\right)$
   
   i.e. $\;$ $P \left(B\right) = P \left(B | A\right) \times P \left(A\right) - P \left(A\right) + P \left(A \cup B\right)$
   
   i.e. $\;$ $P \left(B\right) = 0.50 \times 0.45 - 0.45 + 0.75 = 0.525$