Question

An unbiased coin is tossed twice. The four possible outcomes are equiprobable. If A is the event “both head or tail have occurred” and B is the event: “at most one tail is observed”, then find P(A), P(B), P(A/B) and P(B/A)?

Answer 1

Answer on Question #42088, Math, Statistics and Probability

An unbiased coin is tossed twice. The four possible outcomes are equiprobable. If A is the event "both head or tail have occurred" and B is the event: "at most one tail is observed", then find P(A), P(B), P(A/B) and P(B/A)?

Solution:

Experiment - a set of conditions:

1) Coin is unbiased

2) Coin is tossed twice

Denote events: head = 1

$$\text{tail} = 0$$

Elementary events: $\omega = (a, b)$, where $a, b \in \{0, 1\}$ — permutation with repetitions

Then space of elementary events: $\Omega = \{\omega = (a,b)\} \Rightarrow |\Omega| = 2^2 = 4 < \infty$

Take the maximal $\sigma$ — algebra $\mathfrak{A} = \mathfrak{A}_{\text{max}}$

Then probability $\mathbb{P}(\cdot)$ — classical type.

$$A = \{(0, 1), (1, 0)\} \Rightarrow |A| = 2 \Rightarrow \mathbb{P}(A) = \frac{|A|}{|\Omega|} = \frac{2}{4} = \frac{1}{2}$$ $$B = \{(0, 1), (1, 0), (0, 0)\} \Rightarrow |B| = 3 \Rightarrow \mathbb{P}(B) = \frac{|B|}{|\Omega|} = \frac{3}{4}$$ $$A \cap B = \{(0, 1), (1, 0)\} \Rightarrow |A \cap B| = 2 \Rightarrow \mathbb{P}(A \cap B) = \frac{|A \cap B|}{|\Omega|} = \frac{2}{4} = \frac{1}{2}$$ $$\mathbb{P}(A/B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{4}{6} = \frac{2}{3}$$ $$\mathbb{P}(B/A) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(A)} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$$

Answer:

1) $\mathbb{P}(A) = \frac{1}{2}$

2) $\mathbb{P}(B) = \frac{3}{4}$

3) $\mathbb{P}(A/B) = \frac{2}{3}$

4) $\mathbb{P}(B/A) = 1$

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