Probability Sure Questions: Bayes' Theorem and Independent Events

This lesson works through three Class 12 probability questions: a draw with replacement, a Bayes' theorem problem, and a test of independence for events from three coin tosses.

Two balls drawn with replacement

A box contains $6$ red and $4$ blue balls, so the total is $10$. Two balls are drawn at random with replacement, which means each draw is independent and the probabilities stay the same.

The single-draw probabilities are:

$P(\text{red}) = \tfrac{6}{10}, \qquad P(\text{blue}) = \tfrac{4}{10}$

Two red balls

$P(\text{2 red}) = \tfrac{6}{10} \times \tfrac{6}{10} = \tfrac{36}{100} = \tfrac{9}{25}$

Two blue balls

$P(\text{2 blue}) = \tfrac{4}{10} \times \tfrac{4}{10} = \tfrac{16}{100} = \tfrac{4}{25}$

One red and one blue ball

The red and blue can come in either order, so add both cases:

$P(\text{1 red, 1 blue}) = \tfrac{6}{10}\cdot\tfrac{4}{10} + \tfrac{4}{10}\cdot\tfrac{6}{10} = \tfrac{24}{100} + \tfrac{24}{100} = \tfrac{48}{100} = \tfrac{12}{25}$

Which bag did the red ball come from? (Bayes' theorem)

Bag 1 has $2$ white and $4$ red balls (total $6$). Bag 2 has $5$ white and $3$ red balls (total $8$). A bag is chosen at random and one ball is drawn; it turns out to be red. We want the probability it came from bag 2.

Let $E_1$ and $E_2$ be the events of choosing bag 1 and bag 2, and let $E$ be the event that the drawn ball is red.

$P(E_1) = \tfrac{1}{2}, \qquad P(E_2) = \tfrac{1}{2}$

$P(E \mid E_1) = \tfrac{4}{6}, \qquad P(E \mid E_2) = \tfrac{3}{8}$

By Bayes' theorem:

$P(E_2 \mid E) = \frac{P(E_2)\,P(E \mid E_2)}{P(E_1)\,P(E \mid E_1) + P(E_2)\,P(E \mid E_2)}$

$P(E_2 \mid E) = \frac{\tfrac{1}{2}\cdot\tfrac{3}{8}}{\tfrac{1}{2}\cdot\tfrac{4}{6} + \tfrac{1}{2}\cdot\tfrac{3}{8}} = \frac{\tfrac{3}{16}}{\tfrac{1}{3} + \tfrac{3}{16}}$

The denominator is $\tfrac{1}{3} + \tfrac{3}{16} = \tfrac{16 + 9}{48} = \tfrac{25}{48}$, so:

$P(E_2 \mid E) = \frac{\tfrac{3}{16}}{\tfrac{25}{48}} = \tfrac{3}{16} \times \tfrac{48}{25} = \tfrac{9}{25}$

Independence of events for three coin tosses

Three coins are tossed. The sample space has $8$ outcomes:

$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

Define the events:

• $E$: three heads or three tails $= \{HHH, TTT\}$, so $P(E) = \tfrac{2}{8}$.
• $F$: at least two heads $= \{HHH, HHT, HTH, THH\}$, so $P(F) = \tfrac{4}{8}$.
• $G$: at most two heads (everything except $HHH$), so $P(G) = \tfrac{7}{8}$.

Pair E and F

$E \cap F = \{HHH\}, \quad P(E \cap F) = \tfrac{1}{8}$

$P(E)\,P(F) = \tfrac{2}{8} \times \tfrac{4}{8} = \tfrac{8}{64} = \tfrac{1}{8}$

Since $P(E \cap F) = P(E)\,P(F)$, the events $E$ and $F$ are independent.

Pair F and G

$F \cap G = \{HHT, HTH, THH\}, \quad P(F \cap G) = \tfrac{3}{8}$

$P(F)\,P(G) = \tfrac{4}{8} \times \tfrac{7}{8} = \tfrac{7}{16}$

Since $\tfrac{3}{8} \ne \tfrac{7}{16}$, the events $F$ and $G$ are dependent.

Pair E and G

$E \cap G = \{TTT\}, \quad P(E \cap G) = \tfrac{1}{8}$

$P(E)\,P(G) = \tfrac{2}{8} \times \tfrac{7}{8} = \tfrac{7}{32}$

Since $\tfrac{1}{8} \ne \tfrac{7}{32}$, the events $E$ and $G$ are dependent.

Key takeaways

• With replacement, draws are independent, so multiply the single-draw probabilities and add over the possible orders.
• Bayes' theorem flips a conditional probability: it finds the chance of a cause given an observed effect.
• Two events are independent only when $P(A \cap B) = P(A)\,P(B)$; otherwise they are dependent.