Question

A card from a pack of 52 cards is lost. From the remaining card of the pack, one card is drawn and is found to be heart, find the probability of the missing card to be (i) heart, (ii) club.

Answer 1

Hint: In this particular question use the concept of events assume three events, in first event assume that the missing card is heart, in second event that the missing card is not heart and in third event that one card drawn is heart now assume two other events that drawn card is of heart and missing card is also of heart, and that drawn card is of heart and missing card is not heart, so calculate the probabilities of these events, now use Bayes' theorem based on these events so use these concepts to reach the solution of the question.

Complete step-by-step answer:
$\left( i \right)$ Probability of the missing card to be heart.
As we all know there are 52 cards in a deck of cards, with four suits 13 cards each.
Spade, heart, diamond and club.
As we know that the probability is the ratio of the favorable number of outcomes to the total number of outcomes.
Let A be the event that the missing card is heart.
Therefore, P (A) = $\dfrac{{13}}{{52}}$ = $\dfrac{1}{4}$.
Let B be the event that the missing card is not heart.
Therefore, P (B) = $\dfrac{{39}}{{52}}$ = $\dfrac{3}{4}$
Let, C be the event that one card drawn is of heart. As one card is lost so the remaining cards in the deck are 51.
Let $\dfrac{C}{A}$ be the event that the drawn card is of heart and the missing card is also of heart, so there are 12 cards of heart remaining in the deck.
Therefore, $P\left( {\dfrac{C}{A}} \right)$ = \[\left( {\dfrac{{12}}{{51}}} \right)\]
Let $\dfrac{C}{B}$ be the event that the drawn card is of heart and the missing card is not heart, so there are 13 cards of heart intact in the deck.
And $P\left( {\dfrac{C}{B}} \right) = \left( {\dfrac{{13}}{{51}}} \right)$
Now according to Bayes' theorem if one card is drawn and is found to be heart so the probability that the missing card is heart is,
$P\left( {\dfrac{A}{C}} \right)$ = $\dfrac{{P\left( A \right)P\left( {\dfrac{C}{A}} \right)}}{{P\left( A \right)P\left( {\dfrac{C}{A}} \right) + P\left( B \right)P\left( {\dfrac{C}{B}} \right)}}$
Now substitute the values we have,
$P\left( {\dfrac{A}{C}} \right)$ = \[\dfrac{{\dfrac{1}{4}\left( {\dfrac{{12}}{{51}}} \right)}}{{\dfrac{1}{4}\left( {\dfrac{{12}}{{51}}} \right) + \dfrac{3}{4}\left( {\dfrac{{13}}{{51}}} \right)}}\]
Now simplify we have,
$P\left( {\dfrac{A}{C}} \right)$ = \[\dfrac{{1\left( {12} \right)}}{{1\left( {12} \right) + 3\left( {13} \right)}} = \dfrac{{12}}{{12 + 39}} = \dfrac{{12}}{{51}}\]
$\left( {ii} \right)$ Probability of the missing card to be club.
As we all know there are 52 cards in a deck of cards, with four suits 13 cards each.
Spade, heart, diamond and club.
As we know that the probability is the ratio of the favorable number of outcomes to the total number of outcomes.
Let A be the event that the missing card is club.
Therefore, P (A) = $\dfrac{{13}}{{52}}$ = $\dfrac{1}{4}$.
Let B be the event that the missing card is not a club.
Therefore, P (B) = $\dfrac{{39}}{{52}}$ = $\dfrac{3}{4}$
Let, C be the event that one card drawn is of heart. As one card is lost so the remaining cards in the deck is 51.
Let $\dfrac{C}{A}$ be the event that the drawn card is of heart and the missing card is of club.
Therefore, $P\left( {\dfrac{C}{A}} \right)$ = \[\left( {\dfrac{{13}}{{51}}} \right)\]
Let $\dfrac{C}{B}$ be the event that the drawn card is of heart and the missing card is not club.
So the missing card can be a heart and cannot be a heart.
So if missing card is heart so probability of drawing heart \[\left( {\dfrac{{12}}{{51}}} \right)\] and if missing card is not heart so probability of drawing heart \[\left( {\dfrac{{13}}{{51}}} \right)\]
Therefore, $P\left( {\dfrac{C}{B}} \right)$ = $\dfrac{{12}}{{51}} \times \dfrac{{13}}{{51}}$
Now according to Bayes' theorem if one card is drawn and is found to be heart so the probability that the missing card is club is,
$P\left( {\dfrac{A}{C}} \right)$ = $\dfrac{{P\left( A \right)P\left( {\dfrac{C}{A}} \right)}}{{P\left( A \right)P\left( {\dfrac{C}{A}} \right) + P\left( B \right)P\left( {\dfrac{C}{B}} \right)}}$
Now substitute the values we have,
$P\left( {\dfrac{A}{C}} \right)$ = \[\dfrac{{\dfrac{1}{4}\left( {\dfrac{{13}}{{51}}} \right)}}{{\dfrac{1}{4}\left( {\dfrac{{13}}{{51}}} \right) + \dfrac{3}{4}\left( {\dfrac{{12}}{{51}} \times \dfrac{{13}}{{51}}} \right)}}\]
Now simplify we have,
$P\left( {\dfrac{A}{C}} \right)$ = \[\dfrac{{1\left( {13} \right)}}{{1\left( {13} \right) + 3\left( {\dfrac{{12 \times 13}}{{51}}} \right)}} = \dfrac{{13\left( {51} \right)}}{{13\left( {51} \right) + 39\left( {12} \right)}} = \dfrac{{13\left( {17} \right)}}{{13\left( {17} \right) + 13\left( {12} \right)}} = \dfrac{{17}}{{17 + 12}} = \dfrac{{17}}{{29}}\]
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall that the probability is the ratio of the favorable number of outcomes to the total number of outcomes and always recall the Bayes' theorem which is stated above this theorem is the basis of the question.