Question

# Suppose a girl throws a die. If she gets a 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 tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?(a) $\dfrac{1}{9}$(b) $\dfrac{5}{16}$(c) $\dfrac{2}{13}$(d) $\dfrac{8}{11}$

Hint: First, we need to find the probabilities of the events when she gets 1, 2, 3 or 4 and when she gets 5 or 6 after throwing the die. Now, we need to find the probability when exactly one head is obtained so for that we will get 2 cases, first i.e. probability of getting exactly one head if she gets 5 or 6 on the die and second as event A, probability of getting one head when 1, 2, 3 or 4 is obtained on the die as event B. Then event C will be the event when she get exactly one head. Now, by using Bayes Theorem we will get the probability that she threw 1, 2, 3 or 4 with the die, when exactly one head is obtained as $P\left( B|C \right)=\dfrac{P\left( B \right).P\left( C|B \right)}{P\left( A \right).P\left( C|A \right)+P\left( B \right).P\left( C|B \right)}$.

First we will have to find the probability that when she throws the dice whether she gets (5 or 6), or (1, 2, 3 or 4) on the die, so
Let A be the event when she gets 5 or 6 on the die, and
Let B be the event when she gets 1, 2, 3 or 4 on the die.
We know that there total are 6 possibilities i.e. {1,2,3,4,5,6} that she can get when die is thrown, so we get
P(A) = $\dfrac{2}{6}=\dfrac{1}{3}$, and
P(B) = $\dfrac{4}{6}=\dfrac{2}{3}$
We are given that she gets 1, 2, 3 or 4 with the die and we need to find when she gets exactly one head. So, for that we need to find the probability when she gets exactly one head, there will be two cases i.e. first probability of getting exactly one head if she gets 5 or 6 on the die and she tosses the coin for 3 times and second, probability of getting one head when 1, 2, 3 or 4 is obtained on the die and she tosses the coin once.
Let C be the event when she gets exactly one head.
So probability of getting exactly one head when coin is tossed 3 times $P\left( C|A \right)$ i.e. she gets 5 or 6 on the die, we get as
So sample space when coin is tossed three times = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}
Hence we get
$P\left( C|A \right)=\dfrac{3}{8}$
And probability of getting exactly one head when coin is tossed once $P\left( C|B \right)$ i.e. she gets 5 or 6 on the die, we get as
Sample space = {H, T}
Hence we get,
$P\left( C|B \right)=\dfrac{1}{2}$
Now we have to find $P\left( B|C \right)$ and it can be found by using the Bayes Theorem according to which,
$P\left( B|C \right)=\dfrac{P\left( B \right).P\left( C|B \right)}{P\left( A \right).P\left( C|A \right)+P\left( B \right).P\left( C|B \right)}$
So, we get
$P\left( B|C \right)=\dfrac{\dfrac{2}{3}\times \dfrac{1}{2}}{\dfrac{2}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{3}{8}}$
After further solving the above expression, we get
$P\left( B|C \right)=\dfrac{8}{11}$
Hence our answer matches with option (d), so it is the correct answer.

Note:
Try to solve this problem step by step and don’t try any shortcuts. And whenever we have to find the partial probability i.e. an event occurred and we have to find the probability of sub-event within that event then mostly Bayes theorem is applied. In every case try to make a sample space of the event first it’ll help to solve the problem more easily.