Question

Consider the experiment of throwing a dice, ‘if a multiple of 3 comes up, throw the dice again and if any other number comes toss a coin’.

Find the conditional probability of the event ‘the coin shows tail’, given that at least one dice shows a 3.

Answer 1

The outcomes of the given experiment can be represented by the following tree diagram. 
The sample space of the experiment is,
S={(1,H),(1,T),(2,H),(2,T),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,H),(4,T),(5,H),(5,T),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}
Let A be the event that the coin shows a tail and B be the event that at least one die shows 3.
∴A={(1,T),(2,T),(4,T),(5,T)}
B={(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)}
⇒A∩B=ϕ
∴P(A∩B)=0
Then, P(B)=P({3,1})+P({3,2})+P({3,3})+P({3,4})+P({3,5})+P({3,6})+P({6,3})

 

\(=\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\)

=7/36

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|B)=\(\frac{P(A∩B}{P(B)}=\frac{0}{\frac{7}{36}}=0\)

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