Question

A card from a pack 52 cards is lost. From the remaining card of the pack, one card is drawn and is found be heart, Find the probability that the lost cards were both hearts.

Answer 1

Consider the problem

We have $52$ cards initially, $13$ hearts and $39$ hearts

Consider the events

$E_1$ be both lost card are hearts

$E_2$ be both lost cards are non-hearts

$E_3$ One lost card is non-heart and One is heart

And, $A$ is the probability that heart is picked from the remaining $50$ cards.

So,

$\begin{array}{c}P\left(E_1\right)=\frac{{}^{13}{C}_2}{{}^{52}{C}_2}\\ =\frac{13\times 12}{52\times 51}\\ =\frac{1}{17}\end{array}$

$P\left(\frac{A}{E_1}\right)=\frac{11}{50}$

And

$\begin{array}{c}P\left(E_2\right)=\frac{{}^{39}{C}_2}{{}^{52}{C}_2}\\ =\frac{39\times 38}{52\times 51}\\ =\frac{19}{34}\end{array}$

$P\left(\frac{A}{E_2}\right)=\frac{13}{50}$

And,

$\begin{array}{c}P\left(E_3\right)=\frac{\left(Oneheart\right)and\left(Onenonheart\right)}{{}^{52}{C}_2}\\ =\frac{13\times 39}{{}^{52}{C}_2}\\ =\frac{13\times 39\times 2}{52\times 51}\\ =\frac{39}{102}\end{array}$

$P\left(\frac{A}{E_3}\right)=\frac{12}{50}$

Now Required probability is

$\begin{array}{c}=\frac{P\left(E_1\right)\times P\left(\frac{A}{E_1}\right)}{P\left(E_1\right)\times P\left(\frac{A}{E_1}\right)+P\left(E_2\right)\times P\left(\frac{A}{E_2}\right)+P\left(E_3\right)\times P\left(\frac{A}{E_3}\right)}\\ =\frac{\frac{1}{17}\times \frac{11}{50}}{\frac{1}{17}\times \frac{11}{50}+\frac{19}{34}\times \frac{13}{50}+\frac{39}{102}\times \frac{12}{50}}\\ =\frac{\frac{1}{17}\times \frac{11}{50}}{\frac{1}{17}\times \frac{11}{50}+\frac{19}{34}\times \frac{13}{50}+\frac{39}{17}\times \frac{2}{50}}\\ =\frac{11}{11+\frac{247}{2}+78}\left(Multiplyingeachtermby17\times 50\right)\\ =\frac{11\times 2}{22+247+156}\\ =\frac{22}{425}\end{array}$