Two cards are lost in a pack of 52 playing cards. If the probability of getting a spade card when a card is drawn from the remaining cards is$\frac{1}{m}$ , then $m$  is

So, in this case there are three possibilities you need to consider.

1. Both removed cards are spades

2. One removed card is a spade

3. Neither of the removed cards is a spade

Assuming there is an even chance that any card from the deck is missing, what is the probability of each of the above events occurring?

1. The probability of both lost cards being spades is  $\frac{13}{52}\times \frac{12}{51}=\frac{1}{17}$

2. The probability of one lost card being a spade is

$\frac{13}{52}\times \frac{39}{51}+\frac{39}{52}\times \frac{13}{51}=\frac{13}{34}$  

3. The probability the neither is a spade is $\frac{39}{52}\times \frac{38}{51}=\frac{19}{34}$

Now in each scenario, you have 50 cards remaining, but 11 spades remain in the first scenario, 12in the second scenario and 13 in the third scenario. The probabilities of picking a spade in each scenario respectively is

$\frac{11}{50},\frac{12}{50},\frac{13}{50}$

So, now all the remains is to apply the law of total probability to consider all 3 cases.

P(Pick a spade) $=$ P(Event) $\times$ P(Pick spade given event 1) $+$ P(Pick spade given event 3)

P(Pick a spade) $=$  $\frac{1}{17}\times \frac{11}{50}+\frac{13}{34}\times \frac{12}{50}+\frac{19}{34}\times \frac{13}{50}=\frac{1}{4}$

$m=4$