Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that “the die shows a number greater than 4” given that “there is at least one tail”?
Answer
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Hint: We start solving the problem by writing all the possible outcomes for this problem. We then find the possibilities for the event “that there is at least one tail” and find the probability of it. We then find the possibilities for the event that “the die shows a number greater than 4” and find the probability of it. We then find the intersection of two events and apply the conditional probability definition to get the required result.
Complete step-by-step answer:
According to the problem, we are tossing a coin and if it shows head, we toss it again but if it shows tail, we throw a die. We need to find the conditional probability of the event that “the die shows a number greater than 4” given that “there is at least one tail”.
We assume that the coin is just tossed twice (the die will not be rolled if we get a tail even in the 2nd toss) and we stop tossing once we get the tail.
Let us write the possibilities of tossing of coin + Rolling the die if we get tails in our first toss: $\left\{ \left( H,H \right),\left( H,T \right),\left( T,1 \right),\left( T,2 \right),\left( T,3 \right),\left( T,4 \right),\left( T,5 \right),\left( T,6 \right) \right\}$. Here H = event of getting head on tossing the coin, T = event of getting tail on tossing the coin, 1 = event of getting 1 on rolling the die, 2 = event of getting 2 on rolling the die, 3 = event of getting 3 on rolling the die, 4 = event of getting 4 on rolling the die, 5 = event of getting 5 on rolling the die and 6 = event of getting 6 on rolling the die.
We know that the tossing coin or rolling die is an independent event. We know that the probability of the intersection of two independent events a, b is $P\left( a\cap b \right)=P\left( a \right).P\left( b \right)$.
We know that the probability of getting head or tail on tossing a coin is $\dfrac{1}{2}$. We also know that the probability of getting any number between 1 to 6 on rolling a die is $\dfrac{1}{6}$.
According to the problem, it is given that there is at least 1 tail. So, the possibilities of this event (A) are $\left\{ \left( H,T \right),\left( T,1 \right),\left( T,2 \right),\left( T,3 \right),\left( T,4 \right),\left( T,5 \right),\left( T,6 \right) \right\}$ ---(1). Let us assume the probability of this event being $P\left( A \right)$. Let us find this probability.
So, we have $P\left( A \right)=P\left( H\cap T \right)+P\left( T\cap 1 \right)+P\left( T\cap 2 \right)+P\left( T\cap 3 \right)+P\left( T\cap 4 \right)+P\left( T\cap 5 \right)+P\left( T\cap 6 \right)$.
Since, the tossing and throwing die are independent events. We get,
$\Rightarrow P\left( A \right)=P\left( H \right).P\left( T \right)+P\left( T \right).P\left( 1 \right)+P\left( T \right).P\left( 2 \right)+P\left( T \right).P\left( 3 \right)+P\left( T \right).P\left( 4 \right)+P\left( T \right).P\left( 5 \right)+P\left( T \right).P\left( 6 \right)$.
\[\Rightarrow P\left( A \right)=\left( \dfrac{1}{2}.\dfrac{1}{2} \right)+\left( \dfrac{1}{2}.\dfrac{1}{6} \right)+\left( \dfrac{1}{2}.\dfrac{1}{6} \right)+\left( \dfrac{1}{2}.\dfrac{1}{6} \right)+\left( \dfrac{1}{2}.\dfrac{1}{6} \right)+\left( \dfrac{1}{2}.\dfrac{1}{6} \right)+\left( \dfrac{1}{2}.\dfrac{1}{6} \right)\].
\[\Rightarrow P\left( A \right)=\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}\].
\[\Rightarrow P\left( A \right)=\dfrac{3+1+1+1+1+1+1}{12}\].
\[\Rightarrow P\left( A \right)=\dfrac{9}{12}\].
\[\Rightarrow P\left( A \right)=\dfrac{3}{4}\] ---(2).
Now, we find the probability of the event “that the die shows a number greater than 4”.
We know that the numbers greater than 4 in a die are 5 and 6 and we need tail to roll a die. Let us assume that this event is B and the possibilities are as follows: $\left\{ \left( T,5 \right),\left( T,6 \right) \right\}$ ---(3).
Let us find the probability of event B i.e., $P\left( B \right)$.
So, we have $P\left( B \right)=P\left( T\cap 5 \right)+P\left( T\cap 6 \right)$.
Since, the tossing and throwing die are independent events. We get,
$\Rightarrow P\left( B \right)=P\left( T \right).P\left( 5 \right)+P\left( T \right).P\left( 6 \right)$.
\[\Rightarrow P\left( B \right)=\left( \dfrac{1}{2}.\dfrac{1}{6} \right)+\left( \dfrac{1}{2}.\dfrac{1}{6} \right)\].
\[\Rightarrow P\left( B \right)=\dfrac{1}{12}+\dfrac{1}{12}\].
\[\Rightarrow P\left( B \right)=\dfrac{1+1}{12}\].
\[\Rightarrow P\left( B \right)=\dfrac{2}{12}\].
\[\Rightarrow P\left( B \right)=\dfrac{1}{6}\] ---(4).
From equations (1) and (3), we can see the intersection of events A and B is B.
So, we have got $A\cap B=B$ ---(5).
We know that the conditional probability of event A given that event B exists is $P\left( {}^{A}/{}_{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$.
Now, we need to find the conditional probability of the event that “the die shows a number greater than 4” given that “there is at least one tail”.
So, we need to find $P\left( {}^{B}/{}_{A} \right)=\dfrac{P\left( B\cap A \right)}{P\left( A \right)}$.
From equation (5), we get
$\Rightarrow P\left( {}^{B}/{}_{A} \right)=\dfrac{P\left( B \right)}{P\left( A \right)}$.
From equations (2) and (4), we get
$\Rightarrow P\left( {}^{B}/{}_{A} \right)=\dfrac{\dfrac{1}{6}}{\dfrac{3}{4}}$.
$\Rightarrow P\left( {}^{B}/{}_{A} \right)=\dfrac{1}{6}\times \dfrac{4}{3}$.
$\Rightarrow P\left( {}^{B}/{}_{A} \right)=\dfrac{2}{9}$.
We have found the conditional probability of the event that “the die shows a number greater than 4” given that “there is at least one tail” as $\dfrac{2}{9}$.
Note: We assumed that the coin is just tossed twice to reduce the calculation time as this will give us an infinite possibility which may not give us the required conditional probability. Whenever we get the probability, we need to start solving by writing the possibilities of the events to reduce confusion and for completing the problem efficiently. Similarly, we can expect problems to find the probability of winning if the condition of heads and tails is given for two sides.
Complete step-by-step answer:
According to the problem, we are tossing a coin and if it shows head, we toss it again but if it shows tail, we throw a die. We need to find the conditional probability of the event that “the die shows a number greater than 4” given that “there is at least one tail”.
We assume that the coin is just tossed twice (the die will not be rolled if we get a tail even in the 2nd toss) and we stop tossing once we get the tail.
Let us write the possibilities of tossing of coin + Rolling the die if we get tails in our first toss: $\left\{ \left( H,H \right),\left( H,T \right),\left( T,1 \right),\left( T,2 \right),\left( T,3 \right),\left( T,4 \right),\left( T,5 \right),\left( T,6 \right) \right\}$. Here H = event of getting head on tossing the coin, T = event of getting tail on tossing the coin, 1 = event of getting 1 on rolling the die, 2 = event of getting 2 on rolling the die, 3 = event of getting 3 on rolling the die, 4 = event of getting 4 on rolling the die, 5 = event of getting 5 on rolling the die and 6 = event of getting 6 on rolling the die.
We know that the tossing coin or rolling die is an independent event. We know that the probability of the intersection of two independent events a, b is $P\left( a\cap b \right)=P\left( a \right).P\left( b \right)$.
We know that the probability of getting head or tail on tossing a coin is $\dfrac{1}{2}$. We also know that the probability of getting any number between 1 to 6 on rolling a die is $\dfrac{1}{6}$.
According to the problem, it is given that there is at least 1 tail. So, the possibilities of this event (A) are $\left\{ \left( H,T \right),\left( T,1 \right),\left( T,2 \right),\left( T,3 \right),\left( T,4 \right),\left( T,5 \right),\left( T,6 \right) \right\}$ ---(1). Let us assume the probability of this event being $P\left( A \right)$. Let us find this probability.
So, we have $P\left( A \right)=P\left( H\cap T \right)+P\left( T\cap 1 \right)+P\left( T\cap 2 \right)+P\left( T\cap 3 \right)+P\left( T\cap 4 \right)+P\left( T\cap 5 \right)+P\left( T\cap 6 \right)$.
Since, the tossing and throwing die are independent events. We get,
$\Rightarrow P\left( A \right)=P\left( H \right).P\left( T \right)+P\left( T \right).P\left( 1 \right)+P\left( T \right).P\left( 2 \right)+P\left( T \right).P\left( 3 \right)+P\left( T \right).P\left( 4 \right)+P\left( T \right).P\left( 5 \right)+P\left( T \right).P\left( 6 \right)$.
\[\Rightarrow P\left( A \right)=\left( \dfrac{1}{2}.\dfrac{1}{2} \right)+\left( \dfrac{1}{2}.\dfrac{1}{6} \right)+\left( \dfrac{1}{2}.\dfrac{1}{6} \right)+\left( \dfrac{1}{2}.\dfrac{1}{6} \right)+\left( \dfrac{1}{2}.\dfrac{1}{6} \right)+\left( \dfrac{1}{2}.\dfrac{1}{6} \right)+\left( \dfrac{1}{2}.\dfrac{1}{6} \right)\].
\[\Rightarrow P\left( A \right)=\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}\].
\[\Rightarrow P\left( A \right)=\dfrac{3+1+1+1+1+1+1}{12}\].
\[\Rightarrow P\left( A \right)=\dfrac{9}{12}\].
\[\Rightarrow P\left( A \right)=\dfrac{3}{4}\] ---(2).
Now, we find the probability of the event “that the die shows a number greater than 4”.
We know that the numbers greater than 4 in a die are 5 and 6 and we need tail to roll a die. Let us assume that this event is B and the possibilities are as follows: $\left\{ \left( T,5 \right),\left( T,6 \right) \right\}$ ---(3).
Let us find the probability of event B i.e., $P\left( B \right)$.
So, we have $P\left( B \right)=P\left( T\cap 5 \right)+P\left( T\cap 6 \right)$.
Since, the tossing and throwing die are independent events. We get,
$\Rightarrow P\left( B \right)=P\left( T \right).P\left( 5 \right)+P\left( T \right).P\left( 6 \right)$.
\[\Rightarrow P\left( B \right)=\left( \dfrac{1}{2}.\dfrac{1}{6} \right)+\left( \dfrac{1}{2}.\dfrac{1}{6} \right)\].
\[\Rightarrow P\left( B \right)=\dfrac{1}{12}+\dfrac{1}{12}\].
\[\Rightarrow P\left( B \right)=\dfrac{1+1}{12}\].
\[\Rightarrow P\left( B \right)=\dfrac{2}{12}\].
\[\Rightarrow P\left( B \right)=\dfrac{1}{6}\] ---(4).
From equations (1) and (3), we can see the intersection of events A and B is B.
So, we have got $A\cap B=B$ ---(5).
We know that the conditional probability of event A given that event B exists is $P\left( {}^{A}/{}_{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$.
Now, we need to find the conditional probability of the event that “the die shows a number greater than 4” given that “there is at least one tail”.
So, we need to find $P\left( {}^{B}/{}_{A} \right)=\dfrac{P\left( B\cap A \right)}{P\left( A \right)}$.
From equation (5), we get
$\Rightarrow P\left( {}^{B}/{}_{A} \right)=\dfrac{P\left( B \right)}{P\left( A \right)}$.
From equations (2) and (4), we get
$\Rightarrow P\left( {}^{B}/{}_{A} \right)=\dfrac{\dfrac{1}{6}}{\dfrac{3}{4}}$.
$\Rightarrow P\left( {}^{B}/{}_{A} \right)=\dfrac{1}{6}\times \dfrac{4}{3}$.
$\Rightarrow P\left( {}^{B}/{}_{A} \right)=\dfrac{2}{9}$.
We have found the conditional probability of the event that “the die shows a number greater than 4” given that “there is at least one tail” as $\dfrac{2}{9}$.
Note: We assumed that the coin is just tossed twice to reduce the calculation time as this will give us an infinite possibility which may not give us the required conditional probability. Whenever we get the probability, we need to start solving by writing the possibilities of the events to reduce confusion and for completing the problem efficiently. Similarly, we can expect problems to find the probability of winning if the condition of heads and tails is given for two sides.
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