Question

The Queen, King and Jack of diamonds are removed from a deck of 52 playing cards. One card is drawn. Find the probability of getting a card of
(i). A diamond (ii). A Jack

Answer 1

Hint: Now in the question it is given that three cards have been removed from the deck of 52 playing cards which are King, Queen and Jack of diamonds. So now total outcomes will be 49 instead of 52. We will use the formula of probability which is, \[P\left( X \right)=\dfrac{\rm{Number \space of \space favourable \space outcomes}}{\rm{Number \space of \space Total \space outcomes}}\].

Complete step-by-step answer:
Now we have two parts in the question to solve and in the first we have to find the probability of picking a card and it comes out to be diamond card, which will come out to be \[P\left( A \right)=\dfrac{10}{49}\] where event A is picking a diamond card when 3 diamond cards out of 13 are removed from the deck and in the second part we have to find the probability of picking a card and it comes out to be a Jack, due to the removal of one Jack there are three Jacks left in the deck so the probability comes out to be \[P\left( B \right)=\dfrac{3}{49}\] where event B is picking a Jack when three diamond cards are removed and out of which one is Jack.

Now in the question it is given that three diamond cards have been removed from the full deck of 52 playing cards and the deck involves 13 spades, 13 diamonds, 13 clubs and 13 hearts of cards, so there is no need to construct the sample space on the paper because it is well understood by everyone. The cards in all the four types are the same which are one King, one Queen , one Jack, one Ace and the natural numbers from 2 to 10 as there are four types 13 in each type so the number of cards becomes 52.
The cards that are removed from the deck are one King, one Queen and one Jack and all belong to diamonds type. So as diamonds are 13 in the whole deck now the number reduces to \[13-3=10\] diamond cards, also as there are 4 Kings, 4 Queens and 4 Jacks in the full deck as of now it remains only 3 Kings, 3 Queens and 3 Jacks. The total number of cards in the full deck is 52 and after removing the three mentioned cards the number reduces to \[52-3=49\].
Now the formula for calculating the probability of an event is,
\[P\left( X \right)=\dfrac{\rm{Number \space of \space favourable \space outcomes}}{\rm{Number \space of \space Total \space outcomes}}\]
(i). Now in the first part of the question it is given that we have to find the probability of picking a card from the left off deck and it comes out to be a diamond card,
So now the number of total outcomes become 49 and the number of favourable outcomes become 10 for the event A which is to pick a card from the left off deck and it comes out to be a diamond.
Now applying the formula to find the probability, we get,
\[P\left( A \right)=\dfrac{\rm{Number \space of \space favourable \space outcomes}}{\rm{Number \space of \space Total \space outcomes}}\]
Putting the known values in the formula, we get,
\[P\left( A \right)=\dfrac{13-3}{52-3}\]
\[P\left( A \right)=\dfrac{10}{49}\]
Hence it is the probability of picking a card from the left off deck and it comes out to be a diamond card.
(ii). Now in the second part of the question it is given that we have to find the probability of picking a card from the left off deck and it comes out to be a Jack.
So now the number of total outcomes becomes 49 and the number of favourable outcomes becomes 3 for the event B which is to pick a card from the left off deck and it comes out to be a Jack.
Now applying the formula to find the probability, we get,
\[P\left( B \right)=\dfrac{\rm{Number \space of \space favourable \space outcomes}}{\rm{Number \space of \space Total \space outcomes}}\]
Putting the known values in the formula, we get,
\[P\left( B \right)=\dfrac{4-1}{52-3}\]
\[P\left( B \right)=\dfrac{3}{49}\]
Hence it is the probability of picking a card from the left off deck and it comes out to be a Jack.

Note:The formula for evaluating the probability must be remembered thoroughly and before putting the values they must be analyzed properly. In these questions there is no need to construct the sample space but always evaluate the number of total outcomes and the number of favourable outcomes of an event by thinking of the sample space in your mind which can be done in the case of card problems, but in case of problems involving tossing a coin three times or tossing a coin and throwing a dice at a same time then you need to construct a sample space. Remember we are interested in the number of outcomes after applying the conditions given in the question and a mistake of taking the total number of outcomes as 52 shall be avoided.