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 \[3\] .

Answer 1

Hint: We have to find the conditional probability of the event of the coin showing a tail given that at least one die shows a \[3\] . We solve this question using the concept of conditional probability. First we will create two sample spaces for the events i.e. one for the possibilities of tail and other with the possibilities of at least one \[3\] . Then we will find the value of the probability of both the events of sample spaces . Then , using the sample space and the formula for conditional probability of the events we will find the required probability .

Complete step-by-step solution:
Given :
Let \[A\] be the event that the coin shows a tail and \[B\] be the event that at least one die shows \[3\] .
Now , we will make the samples space for the two events .
As given in the question we will toss a coin only if the number shown when the die is thrown is not a multiple of \[3\] .
So , the possible values on the die to toss a coin are : \[1,2,4,5\] .
Now , the sample space for event \[A\] can be represented as :
\[A = \left\{ {\left( {1,T} \right),\left( {2,T} \right),\left( {4,T} \right),\left( {5,T} \right)} \right\}\]
Now for the sample space of event \[B\] , we are given that at least one \[3\] , so it is only possible when we get a \[3\] on the first throw and any number on the second throw and if we get a \[6\] on the first throw then a \[3\] on the second throw .
The sample space for event B can be represented as :
\[B = \{ \left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,4} \right),(3,5),\left( {3,6} \right),\left( {6,3} \right)\} \]
Now we will find the common terms of the two events .
The common terms of the two events is given as :
\[A \cap B = \Phi \]
\[P\left( {A \cap B} \right) = 0\]
Now , we know that the formula of conditional probability for the event that the coin shows a tail , given that at least one die shows a \[3\] is given as :
\[P\left( {A|B} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}\]
Now , putting the value of \[P\left( {A \cap B} \right)\] in the formula , we get
\[P\left( {A|B} \right) = 0\]
[We don’t need to find the value of the probability of event \[B\] as we got the value of \[P\left( {A \cap B} \right)\] as zero , so any value of the probability won’t change the value of \[P\left( {A|B} \right)\] . So , we don’t need to find the value of \[P\left( B \right)\]]
Hence , the conditional probability of the event the coin shows a tail , given that at least one die shows a \[3\] is \[0\] .

Note: We find the relation by using the conditional probability formula . If \[A\] and \[B\] are mutually exclusive events , then \[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right)\] . The basic property of probability is that the probability of an event can never be greater than \[1\] .
If two events \[A\] and \[B\] are independent , then
\[P\left( {E \cap F} \right) = P\left( E \right) \times P\left( F \right)\]
\[P\left( {E|F} \right) = P\left( E \right)\] , \[P\left( F \right) \ne 0\]
\[P\left( {F|E} \right) = P\left( F \right)\] , \[P\left( E \right) \ne 0\]