Question

The queen, the king, and the jack of diamonds are removed from a deck of 52 playing cards and the pack is well-shuffled. A card is drawn from the remaining cards. Find the probability that of getting a card of
(i) a diamond.
(ii) a jack.

Answer 1

Hint:
Here, we need to find the probability of getting a diamond, and the probability of getting a jack. First, we will find the total number of cards in the deck. Then, we will find the number of diamonds, and the number of jacks in the deck. Finally, we will use the formula for probability to get the required probabilities.
Formula Used: The probability of an event is given by \[P\left( E \right) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Number of total outcomes}}}}\].

Complete step by step solution:
First, we will find the number of favourable outcomes and total outcomes.
The total number of cards in the deck is 52.
It is given that the queen, king, and jack of diamonds is removed from the deck.
Therefore, 3 of the 52 cards are removed
Thus, the number of cards left in the deck are \[52 - 3 = 49\] cards.
Therefore, the total number of possible outcomes is 49.
Next, we will find the number of favourable outcomes.
There are 13 cards of diamond in a deck of 52 cards.
However, 3 cards of diamond (the queen, king, and jack) are removed from the deck.
Therefore, the remaining number of diamonds in the deck are \[13 - 3 = 10\] cards.
Therefore, the number of favourable outcomes is 10.
Finally, we will use the formula for probability of an event to calculate the probability of getting a card of diamond.
We know that the probability of an event \[E\] is given by \[P\left( E \right) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Number of total outcomes}}}}\].
Let \[E\] be the event that a card of diamond is selected.
Substituting 10 for the number of favourable outcomes, and 49 for the number of total outcomes in the formula, we get
\[ \Rightarrow P\left( E \right) = \dfrac{{10}}{{49}}\]
Therefore, the probability of getting a card of diamond is \[\dfrac{{10}}{{49}}\].
(ii)
First, we will find the number of favourable outcomes and total outcomes.
The number of cards left in the deck are \[52 - 3 = 49\] cards.
Therefore, the total number of possible outcomes is 49.
Next, we will find the number of favourable outcomes.
There are 4 cards of jack (1 each of hearts, diamonds, spade, clubs) in a deck of 52 cards.
However, 1 card of jack (the jack of diamond) is removed from the deck.
Therefore, the remaining number of cards of jack in the deck are \[4 - 1 = 3\] cards.
Therefore, the number of favourable outcomes is 3.
Finally, we will use the formula for probability of an event to calculate the probability of getting a card of diamond.
Let \[F\] be the event that a card of jack is selected.
Substituting 3 for the number of favourable outcomes, and 49 for the number of total outcomes in the formula, we get
\[ \Rightarrow P\left( F \right) = \dfrac{3}{{49}}\]

Therefore, the probability of getting a card of jack is \[\dfrac{3}{{49}}\].

Note:
There are a total of 52 cards in a deck in which 26 cards are red and 26 are blacks. There are 13 spades, 13 diamonds, 13 hearts and 13 clubs. Also there are 4 kings, 4 queens, 4 jacks and 4 aces in a pack of 52 cards. We need to keep in mind that since the three cards are removed, the total number of cards in the deck also decreases. A common mistake is to take the total number of outcomes as 52 in both the cases, and get the answers \[\dfrac{{10}}{{52}} = \dfrac{5}{{26}}\] and \[\dfrac{3}{{52}}\] respectively which is incorrect.