Question

Suppose a girl throws a die. If she gets 5 or 6, she tosses a coin 3 times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or a tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Answer 1

Hint : Let us assume that throwing a die and getting 1, 2, 3 or 4 is an event Q, throwing a die and getting 5 or 6 is an event R and tossing a coin and getting exactly one head as an event S. Now, write the probability of getting exactly one head given that a die threw could be 1, 2, 3 or 4. This probability we are going to find by Bayes’ theorem with the formula $P\left( S| Q \right)=\dfrac{P\left( S \right)P\left( Q| S \right)}{P(Q)}$ and write the probability of getting exactly one head given that a die threw could be 5 or 6. This probability we are going to find by Bayes’ theorem with the formula $P\left( S| R \right)=\dfrac{P\left( S \right)P\left( R| S \right)}{P\left( R \right)}$ .Now, divide the probability of getting 5 or 6 on the die given that exactly one head has obtained by the sum of the probability of getting 5 or 6 on the die given that exactly one head has obtained and the probability of getting 1, 2, 3 or 4 on the die given that exactly one head has obtained. The formula we are going to use is:
$P\left( Q| S \right)=\dfrac{P\left( Q \right)P\left( S| Q \right)}{P\left( Q \right)P\left( S| Q \right)+P\left( R \right)P\left( S| R \right)}$.

Complete step by step solution :
Let us assume throwing a die and getting 1, 2, 3 or 4 is an event Q.
Let us assume throwing a die and getting 5 or 6 is an event R.
Let us assume that getting exactly one head is an event S.
We are going to calculate the probability of throwing a die and getting 1, 2, 3 or 4.
$P\left( Q \right)=\dfrac{4}{6}=\dfrac{2}{3}$
The probability of throwing a die and getting 5 or 6 is:
$P\left( R \right)=\dfrac{2}{6}=\dfrac{1}{3}$
The probability of getting exactly one head is:
$P\left( S \right)=\dfrac{1}{2}$
The probability of getting exactly one head given that a die threw could be 1, 2, 3 or 4 is:
$P\left( S| Q \right)=\dfrac{P\left( S \right)P\left( Q| S \right)}{P(Q)}$
$P\left( S| Q \right)=\dfrac{1}{2}$
The probability of getting exactly one head given that a die threw could be 5 or 6 is:
$P\left( S| R \right)=\dfrac{P\left( S \right)P\left( R| S \right)}{P\left( R \right)}$
$P\left( S| R \right)=\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times 3=\dfrac{3}{8}$
Now, we have to find the probability of getting 1, 2, 3 or 4 on a die given that exactly one head has been obtained.
The above probability can be written as$P\left( Q| S \right)$.
$P\left( Q| S \right)=\dfrac{P\left( Q \right)P\left( S| Q \right)}{P\left( Q \right)P\left( S| Q \right)+P\left( R \right)P\left( S| R \right)}$
$P\left( Q| S \right)=\dfrac{\dfrac{2}{3}\times \dfrac{1}{2}}{\dfrac{2}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{3}{8}}$
$P\left( Q| S \right)=\dfrac{\dfrac{1}{3}}{\dfrac{1}{3}+\dfrac{1}{8}}=\dfrac{8}{11}$
Hence, the probability that she threw 1, 2, 3 or 4 given that exactly one head has been obtained is $\dfrac{8}{11}$.

Note : You might think how we got the probability of getting exactly one head given that a die threw could be 5 or 6 which is shown above as:
$P\left( S| R \right)=\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times 3=\dfrac{3}{8}$
In the question, it is given that in a throw of a die if you got 5 or 6 then the girl tossed the coin three times. Let H be the probability of getting head and T is the probability of getting tail.
The number of cases which shows tossing a coin three times and getting exactly one head is:
(HTT, THT, TTH)
The probability of getting a head or a tail is equal to$\dfrac{1}{2}$. So, the total probability will be:
$\begin{aligned} & \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}+\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}+\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2} \\\ & =\dfrac{1}{8}\times 3=\dfrac{3}{8} \\\ \end{aligned}$