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

Last updated date: 19th Sep 2024
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Hint: Here it is given that the three cards i.e. king, queen and jack of diamonds are removed from a pack of playing cards. We will calculate the remaining cards by subtracting the number of removed cards from the total number of cards and then we will calculate the number of diamond cards left and the number of jack cards left, then we will calculate the required probability.

Formula used:
We will use the formula ${\rm{Probability}} = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$, where $n\left( E \right)$ is the number of favorable outcomes and $n\left( S \right)$ is the total number of outcomes.

As it is given that the king, queen and jack of Diamonds are removed from a deck of 52 playing cards.
Total 3 cards are removed from a deck of 52 playing cards.
So remaining cards in a deck $= 52 - 3 = 49$
Total number of outcomes $= 49$
(i) A card is drawn from the remaining cards and we need to find the probability that the card drawn is a card of diamond.
We know that there are a total 13 cards of diamond and after removing three cards i.e. king, queen and jack of diamonds only ten diamond cards are left in a Deck.
Number of favorable outcomes $= 10$
We know that;
${\rm{Probability}} = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$
On substituting $n\left( E \right) = 10$ and $n\left( S \right) = 49$ in the formula ${\rm{Probability}} = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$, we get $\Rightarrow {\rm{Probability}} = \dfrac{{10}}{{49}}$
Hence, the probability of getting a diamond card from the remaining cards $= \dfrac{{10}}{{49}}$
(ii) There are four Jacks in a deck. After removing one jack of diamond, only 3 jack cards are left.
Number of favorable outcomes $= 3$
On substituting $n\left( E \right) = 3$ and $n\left( S \right) = 49$ in the formula ${\rm{Probability}} = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$, we get $\Rightarrow {\rm{Probability}} = \dfrac{3}{{49}}$
Hence, the probability of getting a jack card from the remaining cards $= \dfrac{3}{{49}}$

Note: To solve this question, we need to know about probability and its properties. Probability is defined as the ratio of number of desired or favorable outcomes to the total number of possible outcomes. We need to keep in mind that the value of probability cannot be greater than 1 and the value of probability cannot be negative. Also, the probability of a sure event is always one.