Question

15. 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'


​

Answer 1

Step-by-step explanation:

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})

=

36

1

+

36

1

+

36

1

+

36

1

+

36

1

+

36

1

+

36

1

=

36

7

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)=

P(B)

P(A∩B)

=

36

7

0

=0