Question

Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three 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 1 head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Answer 1

Here, we need to find the probability that she threw 1, 2, 3 or 4 with the die, given that she obtained exactly 1 head. Let $A$ be the event of getting exactly one head, ${E_1}$ be the event of getting a 5 or a 6 on the die, and ${E_2}$ be the event of getting a 1, 2, 3, or 4 on the die. You need to apply Bayes theorem to get the probability $P\left( {{E_2}|A} \right)$.
Formula Used: Bayes theorem states that $P\left( {{E_i}|A} \right) = \dfrac{{P\left( {{E_i}} \right)P\left( {A|{E_i}} \right)}}{{\sum\limits_{j = 1}^n {P\left( {{E_j}} \right)P\left( {A|{E_j}} \right)} }}$ where $i = 1,2,3, \ldots ,n$, $S$ is the sample space and ${E_1} \cup {E_2} \cup \ldots \cup {E_n} = S$.

Complete step by step solution:
Let $A$ be the event of getting exactly one head.
Let ${E_1}$ be the event of getting a 5 or a 6 on the die.
There are 6 outcomes on rolling a die, out of which only 2 outcomes are favourable.
Therefore, we get $P\left( {{E_1}} \right) = \dfrac{2}{6} = \dfrac{1}{3}$.
Let ${E_2}$ be the event of getting a 1, 2, 3, or 4 on the die.
There are 6 outcomes on rolling a die, out of which only 4 outcomes are favourable.
Therefore, we get $P\left( {{E_2}} \right) = \dfrac{4}{6} = \dfrac{2}{3}$.
Now, we will find the probability that $A$ occurs, given that ${E_1}$ occurs.
Since the event ${E_1}$ occurs, the coin is tossed thrice.
The possible outcomes of tossing a coin three times are HHH, HHT, HTH, THH, TTT, TTH, THT, HTT.
Out of these 8 possible outcomes, the outcomes in which we get exactly one head are HTT, THT, and TTH.
Hence, the number of favourable outcomes is 3.
Therefore, we get the probability that $A$ occurs, given that ${E_1}$ occurs as
$P\left( {A|{E_1}} \right) = \dfrac{3}{8}$
Next, we will find the probability that $A$ occurs, given that ${E_2}$ occurs.
Since the event ${E_2}$ occurs, the coin is tossed once.
The possible outcomes on tossing a coin once are H or T.
Out of these 2 possible outcomes, the outcomes in which we get exactly one head is H.
Hence, the number of favourable outcomes is 1.
Therefore, we get the probability that $A$ occurs, given that ${E_2}$ occurs as
$P\left( {A|{E_2}} \right) = \dfrac{1}{2}$
Now, we need to find the probability that the girl threw 1, 2, 3 or 4 with the die, given that she obtained exactly 1 head, that is $P\left( {{E_2}|A} \right)$.
We will use Bayes theorem to get the required probability.
Substitute $i = 2$ and $n = 2$ in Bayes theorem where ${E_1} \cup {E_2} = S$, we get
P\left( {{E_2}|A} \right) = \dfrac{{P\left( {{E_2}} \right)P\left( {A|{E_2}} \right)}}{{\sum\limits_{j = 1}^2 {P\left( {{E_j}} \right)P\left( {A|{E_j}} \right)} }} \\\ \Rightarrow P\left( {{E_2}|A} \right) = \dfrac{{P\left( {{E_2}} \right)P\left( {A|{E_2}} \right)}}{{P\left( {{E_1}} \right)P\left( {A|{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A|{E_2}} \right)}} \\\
Substituting $P\left( {{E_1}} \right) = \dfrac{1}{3}$, $P\left( {{E_2}} \right) = \dfrac{2}{3}$, $P\left( {A|{E_1}} \right) = \dfrac{3}{8}$, and $P\left( {A|{E_2}} \right) = \dfrac{1}{2}$ in the expression, we get
$\Rightarrow P\left( {{E_2}|A} \right) = \dfrac{{\dfrac{2}{3} \cdot \dfrac{1}{2}}}{{\dfrac{1}{3} \cdot \dfrac{3}{8} + \dfrac{2}{3} \cdot \dfrac{1}{2}}}$
Multiplying the terms in the expression, we get
$\Rightarrow P\left( {{E_2}|A} \right) = \dfrac{{\dfrac{1}{3}}}{{\dfrac{1}{8} + \dfrac{1}{3}}}$
Simplifying the expression, we get the required probability as
\Rightarrow P\left( {{E_2}|A} \right) = \dfrac{{\dfrac{1}{3}}}{{\dfrac{{11}}{{24}}}} \\\ \therefore P\left( {{E_2}|A} \right) = \dfrac{8}{{11}} \\\

Therefore, if the girl obtained exactly 1 head, the probability that she threw 1, 2, 3 or 4 with the die is $\dfrac{8}{{11}}$ or approximately $0.72$.

Note:
You need to remember how to use Bayes theorem and where it is applicable. The event of rolling the die occurs before the event of tossing the coin. Since we needed to find the conditional probability of getting a 1, 2, 3, or 4 on the die (event occurring first) given that the girl got exactly one head (event occurring later), we used Bayes theorem.