Question

Suppose a girl throws a die. If she gets 1 or 2 , she tosses a coin three times and notes the number of tails. If she gets 3,4,5 or 6 , She tosses a coin once and notes whether a ‘Head’ or ‘Tail’ is obtained. If she obtained exactly one ’Tail’, What is the probability that she threw 3,4,5 or 6 with the die?

Answer 1

Start by finding the probability of getting 1 or 2 and 3,4,5 or 6 separately. After that bring in the condition of obtaining a tail in both of the above two cases and apply Bayes theorem to get the required solution.

Complete step-by-step answer :
Let ${{\text{E}}_1}$be the event of the girl getting 1 or 2.
Now, It’s probability will be given by the relation
${\text{P(E) = }}\dfrac{{n(E)}}{{n(S)}} = \dfrac{{{\text{No}}{\text{. of favourable outcomes}}}}{{{\text{Total number of observation made(sample space)}}}} \\\ \therefore {\text{P(}}{{\text{E}}_1}) = \dfrac{2}{6} = \dfrac{1}{3} \\\$
Let ${{\text{E}}_2}$be the event of the girl getting 3,4,5 or 6
Now, It’s probability will be
${\text{P(}}{{\text{E}}_2}) = \dfrac{4}{6} = \dfrac{2}{3}$
Let X be the event of the girl obtaining exactly one ‘Tail’.
Let us see all the possibilities of getting a tail when the coin is tossed 3 times.
The possible outcomes are ={THH , HTH ,HHT} = 3
Probability of the event X when event ${{\text{E}}_1}$ has already occurred = ${\text{P(}}\dfrac{{\text{X}}}{{{{\text{E}}_1}}}) = \dfrac{3}{8}$
Now, if the coin is tossed only once and gets a Tail,
Then, The total number of favourable outcomes = 1
Probability of the event X when event ${{\text{E}}_1}$ has already occurred = ${\text{P(}}\dfrac{{\text{X}}}{{{{\text{E}}_2}}}) = \dfrac{1}{2}$
Now , the probability of getting 3,4,5 or 6 with die , when she got exactly a tail will be as follows
∴ It will be given by ${\text{P(}}\dfrac{{{{\text{E}}_2}}}{{\text{X}}})$
Now , We will use Bayes theorem. Which gives the relation
${\text{P(}}\dfrac{{{{\text{E}}_2}}}{{\text{X}}}) = \dfrac{{{\text{P(}}{{\text{E}}_2}) \cdot {\text{P}}(\dfrac{{\text{X}}}{{{{\text{E}}_2}}})}}{{{\text{P(}}{{\text{E}}_1}) \cdot {\text{P}}(\dfrac{{\text{X}}}{{{{\text{E}}_1}}}) + {\text{P(}}{{\text{E}}_2}) \cdot {\text{P}}(\dfrac{{\text{X}}}{{{{\text{E}}_2}}})}} \\\ \therefore {\text{P(}}\dfrac{{{{\text{E}}_2}}}{{\text{X}}}) = \dfrac{{\dfrac{1}{2} \cdot \dfrac{2}{3}}}{{\dfrac{3}{8} \cdot \dfrac{1}{3} + \dfrac{1}{2} \cdot \dfrac{2}{3}}} = \dfrac{{\dfrac{1}{3}}}{{\dfrac{1}{3} + \dfrac{1}{8}}} = \dfrac{8}{{11}} \\\$
Therefore, the probability of getting 3,4,5 or 6 with die , when she got exactly a tail is $\dfrac{8}{{11}}$

Note :Writing of the sample space gives a clear idea of the outcomes possible , makes it easier to find out the required data , This might not be possible for all the questions. Attention must be given while representing the conditional probability and Bayes theorem.