Question

A card from a pack of 52 playing cards is lost. From the remaining cards of the pack, three cards are drawn at random (without replacement) and are found to be all spades. Find the probability of the lost card being a spade.

Answer 1

Hint: Here, we will proceed by assuming event ${{\text{E}}_1}$ as the lost card is a spade and event ${{\text{E}}_2}$ as the lost card is not a spade (i.e., the lost can be a heart or a club or a diamond) and event A as the three randomly drawn cards (without replacement) are all spades. Then, using the formula for Bayes theorem i.e., ${\text{P}}\left( {\dfrac{{{{\text{E}}_1}}}{{\text{A}}}} \right) = \dfrac{{{\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_1}}}} \right){\text{P}}\left( {{{\text{E}}_1}} \right)}}{{{\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_1}}}} \right){\text{P}}\left( {{{\text{E}}_1}} \right) + {\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_2}}}} \right){\text{P}}\left( {{{\text{E}}_2}} \right)}}$.

Complete step-by-step answer:
If ${\text{P}}\left( {\dfrac{{{{\text{E}}_1}}}{{\text{A}}}} \right)$ is the probability of occurrence of event ${{\text{E}}_1}$ when event A had already occurred.
If \[{\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_1}}}} \right)\] is the probability of occurrence of event A when event ${{\text{E}}_1}$ had already occurred.
If \[{\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_{\text{2}}}}}} \right)\] is the probability of occurrence of event A when event ${{\text{E}}_2}$ had already occurred.
If ${\text{P}}\left( {{{\text{E}}_1}} \right)$ is the probability of occurrence of event ${{\text{E}}_1}$ and if ${\text{P}}\left( {{{\text{E}}_2}} \right)$ is the probability of occurrence of event ${{\text{E}}_2}$.
According to Bayes theorem, we can write
${\text{P}}\left( {\dfrac{{{{\text{E}}_1}}}{{\text{A}}}} \right) = \dfrac{{{\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_1}}}} \right){\text{P}}\left( {{{\text{E}}_1}} \right)}}{{{\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_1}}}} \right){\text{P}}\left( {{{\text{E}}_1}} \right) + {\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_2}}}} \right){\text{P}}\left( {{{\text{E}}_2}} \right)}}{\text{ }} \to {\text{(1)}}$

Also we know that the general formula for the probability of occurrence of an event is given by
Probability of occurrence of an event$ = \dfrac{{{\text{Number of favourable cases}}}}{{{\text{Total number of possible cases}}}}$
Let ${{\text{E}}_1}$ be the event that the lost card is a spade and ${{\text{E}}_2}$ be the event that the lost card is not a spade (i.e., the lost can be a heart or a club or a diamond) and A be the event that the three randomly drawn cards (without replacement) are all spades.
Probability that the lost card is a spade, ${\text{P}}\left( {{{\text{E}}_1}} \right) = \dfrac{{{\text{Number of spades}}}}{{{\text{Total number of cards}}}} = \dfrac{{13}}{{52}} = \dfrac{1}{4}$
Probability that the lost card is not a spade, ${\text{P}}\left( {{{\text{E}}_2}} \right) = 1 - {\text{P}}\left( {{{\text{E}}_1}} \right) = 1 - \dfrac{1}{4} = \dfrac{{4 - 1}}{4} = \dfrac{3}{4}$
Probability that the three randomly drawn cards (without replacement) are all spades given that the lost card is a spade, \[{\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_1}}}} \right) = \left( {\dfrac{{12}}{{51}}} \right) \times \left( {\dfrac{{11}}{{50}}} \right) \times \left( {\dfrac{{10}}{{49}}} \right) = \dfrac{{44}}{{4165}}\]
Probability that the three randomly drawn cards (without replacement) are all spades given that the lost card is not a spade, \[{\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_2}}}} \right) = \left( {\dfrac{{13}}{{51}}} \right) \times \left( {\dfrac{{12}}{{50}}} \right) \times \left( {\dfrac{{11}}{{49}}} \right) = \dfrac{{286}}{{20825}}\]
Using the formula given by equation (1), we will get the required probability of the lost card being a spade given that the three randomly drawn cards (without replacement) are all spades.
 $
  {\text{P}}\left( {\dfrac{{{{\text{E}}_1}}}{{\text{A}}}} \right) = \dfrac{{\left( {\dfrac{{44}}{{4165}}} \right)\left( {\dfrac{1}{4}} \right)}}{{\left( {\dfrac{{44}}{{4165}}} \right)\left( {\dfrac{1}{4}} \right) + \left( {\dfrac{{286}}{{20825}}} \right)\left( {\dfrac{3}{4}} \right)}} \\
   \Rightarrow {\text{P}}\left( {\dfrac{{{{\text{E}}_1}}}{{\text{A}}}} \right) = \dfrac{{\left( {\dfrac{{11}}{{4165}}} \right)}}{{\left( {\dfrac{{11}}{{4165}} + \dfrac{{429}}{{41650}}} \right)}} \\
   \Rightarrow {\text{P}}\left( {\dfrac{{{{\text{E}}_1}}}{{\text{A}}}} \right) = \dfrac{{\left( {\dfrac{{11}}{{4165}}} \right)}}{{\left( {\dfrac{{110 + 429}}{{41650}}} \right)}} \\
   \Rightarrow {\text{P}}\left( {\dfrac{{{{\text{E}}_1}}}{{\text{A}}}} \right) = \dfrac{{\left( {\dfrac{{11}}{{4165}}} \right)}}{{\left( {\dfrac{{539}}{{41650}}} \right)}} \\
   \Rightarrow {\text{P}}\left( {\dfrac{{{{\text{E}}_1}}}{{\text{A}}}} \right) = \dfrac{{110}}{{539}} \\
   \Rightarrow {\text{P}}\left( {\dfrac{{{{\text{E}}_1}}}{{\text{A}}}} \right) = \dfrac{{10}}{{49}} \\
 $
Therefore, the required probability of the lost card being a spade given that the three randomly drawn cards (without replacement) are all spades is $\dfrac{{10}}{{49}}$.

Note: In a deck of 52 cards, there are 13 cards each of spade, club, heart and diamond where cards of spade and club are black in colour and cards of heart and diamond are red in colour. In each of these 13 cards, there are three face cards including king, queen and jack. Here, either the lost card is a spade or it is not a spade i.e., ${\text{P}}\left( {{{\text{E}}_1}} \right) + {\text{P}}\left( {{{\text{E}}_2}} \right) = 1$.