Consider the experiment of throwing a die,

Question:

Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

Solution:

The outcomes of the given experiment can be represented by the following tree diagram.

The sample space of the experiment is,

$S=\left\{\begin{array}{l}(1, \mathrm{H}),(1, \mathrm{~T}),(2, \mathrm{H}),(2, \mathrm{~T}),(3,1)(3,2),(3,3),(3,4),(3,5),(3,6), \\ (4, \mathrm{H}),(4, \mathrm{~T}),(5, \mathrm{H}),(5, \mathrm{~T}),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}$

Let A be the event that the coin shows a tail and B be the event that at least one die shows 3.

$\therefore \mathrm{A}=\{(1, \mathrm{~T}),(2, \mathrm{~T}),(4, \mathrm{~T}),(5, \mathrm{~T})\}$

$\mathrm{B}=\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)\}$

$\Rightarrow \mathrm{A} \cap \mathrm{B}=\phi$

$\therefore \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0$

Then, $\mathrm{P}(\mathrm{B})=\mathrm{P}(\{3,1\})+\mathrm{P}(\{3,2\})+\mathrm{P}(\{3,3\})+\mathrm{P}(\{3,4\})+\mathrm{P}(\{3,5\})+\mathrm{P}(\{3,6\})+\mathrm{P}(\{6,3\})$

$=\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}$

$=\frac{7}{20}$

Probability of the event that the coin shows a tail, given that at least one die shows 3, is given by P(A|B).

Therefore,

$P(A \mid B)=\frac{P(A \cap B)}{P(B)}$

$=\frac{0}{\frac{7}{20}}=0$

 

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