Given that first toss is heads, what is probability that all $5$ tosses are heads?

Question

Problem

Toss a Coin $5$ times.
Let $H_1 = $ First Toss is $\mathbb{HEAD}$.
Let $H_A = $ All $5$ Tosses are $\mathbb{HEAD}$.
Find $P(H_A\mid H_1)$

The answer says that $P(H_A\mid H_1) = \frac{1}{16}$, but I don't agree with that.

My Approach

$$P(H_1) = \frac{1}{2}$$ $$P(H_A) = \frac{1}{32}$$ $$P(H_A\mid H_1)=\frac{P(H_A\cap H_1)}{P(H_1)}$$

$$P(H_A∩H_1) = \frac{1}{2} * \frac{1}{32} = \frac{1}{64}$$

$$P(H_A\mid H_1) = \frac{\frac{1}{64}}{\frac{1}{2}} = \frac{1}{32}$$

Why does my answer not match $\frac{1}{16}$? Where did I made mistake?

Answer 1

Let $H_1$ be the event that First Toss is $\mathcal{Heads}$.
Let $H_A$ be the event that All Tosses are $\mathcal{Heads}$.

Now, as per Baye's Theorem
\begin{align} \ P(B |A) &= \frac{P(B).P(A|B)}{P(A)} \\\\ & = \frac{P(B).P(A\cap B)}{P(A).P(B)} \\\\ & = \frac{P(A\cap B)}{P(A)} \\ \end{align}

Thus,

$$P(H_A|H_1) = \frac{P(H_A\cap H_1)}{P(H_1)} \qquad(\text{This can also be derived from definition of Conditional Probability})$$

Since we have $5$ Empty Blanks to fill.

• For Sample Space $(S)$, every Empty Blank can take $\mathcal{Two}$ values $\mathcal{Heads \space or \space Tails}$

  Thus, $n(S)=2\times2\times2\times2\times2 = 32$

• For $H_A \cap H_1$, every Empty Blank can take only $\mathcal{One}$ value $\mathcal{Heads}$

  Thus, $n(H_A \cap H_1)=1\times1\times1\times1\times1 = 1$

• For $H_1$, First Empty Blank can take only $\mathcal{One}$ value $\mathcal{Heads}$, remaining empty Blanks can take $\mathcal{Two}$ values $\mathcal{Heads \space or \space Tails}$

  Thus, $n(H_1)=1\times2\times2\times2\times2 = 16$

$$P(H_A|H_1) = \frac{ \frac {n(H_A \cap H_1)}{n(S)} } {\frac {n(H_1)}{n(S)}}$$

On Putting values

$$P(H_A|H_1) = \frac{1}{16}$$

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