
All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is drawn at random from the remaining pack. Find the probability of getting
(i) a black face card
(ii) a queen
Answer
522.3k+ views
Hint: To solve this question we will use the basic concept of probability. First we will find the number of total possible outcomes then we will find the number of outcomes that the card drawn is a black face card and a queen. Then we will substitute the values in the basic formula of probability which is given as
$P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
Where, A is an event which we wanted
\[n\left( E \right)=\] Number of favorable outcomes and $n\left( S \right)=$ number of total possible outcomes
Complete step by step solution:
We have been given that all the three face cards of spades are removed from a well-shuffled pack of 52 cards and a card is drawn at random from the remaining pack.
We have to find the probability of getting a black face card and a queen individually.
Now, as given that all three faces of spades are removed from the pack of 52. So the number of total possible outcomes will be
$\Rightarrow n\left( S \right)=49$
Now, a card is drawn at random from the total 49 cards.
(i) a black face card
Now, we know that there are total 3 black face cards so the number of favorable outcomes will be
\[\Rightarrow n\left( E \right)=3\]
So the probability of getting a black face card will be
$\begin{align}
& \Rightarrow P\left( \text{black face card} \right)=\dfrac{n\left( E \right)}{n\left( S \right)} \\
& \Rightarrow P\left( \text{black face card} \right)=\dfrac{3}{49} \\
\end{align}$
Hence we get the probability of getting a black face card as $\dfrac{3}{49}$.
(ii) a queen
Now, we know that there are a total 4 queen cards but a queen of spades is removed from the cards so we will have only 3 queen cards left in the total pack. So the number of favorable outcomes will be
\[\Rightarrow n\left( E \right)=3\]
So the probability of getting a queen will be
$\begin{align}
& \Rightarrow P\left( queen \right)=\dfrac{n\left( E \right)}{n\left( S \right)} \\
& \Rightarrow P\left( queen \right)=\dfrac{3}{49} \\
\end{align}$
Hence we get the probability of getting a queen as $\dfrac{3}{49}$.
Note: The possibility of mistake is while considering the total number of possible outcomes. Students may consider the total number of possible outcomes as 52 and solve accordingly which leads to incorrect solutions. So carefully read the conditions given in the question and then solve further.
$P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
Where, A is an event which we wanted
\[n\left( E \right)=\] Number of favorable outcomes and $n\left( S \right)=$ number of total possible outcomes
Complete step by step solution:
We have been given that all the three face cards of spades are removed from a well-shuffled pack of 52 cards and a card is drawn at random from the remaining pack.
We have to find the probability of getting a black face card and a queen individually.
Now, as given that all three faces of spades are removed from the pack of 52. So the number of total possible outcomes will be
$\Rightarrow n\left( S \right)=49$
Now, a card is drawn at random from the total 49 cards.
(i) a black face card
Now, we know that there are total 3 black face cards so the number of favorable outcomes will be
\[\Rightarrow n\left( E \right)=3\]
So the probability of getting a black face card will be
$\begin{align}
& \Rightarrow P\left( \text{black face card} \right)=\dfrac{n\left( E \right)}{n\left( S \right)} \\
& \Rightarrow P\left( \text{black face card} \right)=\dfrac{3}{49} \\
\end{align}$
Hence we get the probability of getting a black face card as $\dfrac{3}{49}$.
(ii) a queen
Now, we know that there are a total 4 queen cards but a queen of spades is removed from the cards so we will have only 3 queen cards left in the total pack. So the number of favorable outcomes will be
\[\Rightarrow n\left( E \right)=3\]
So the probability of getting a queen will be
$\begin{align}
& \Rightarrow P\left( queen \right)=\dfrac{n\left( E \right)}{n\left( S \right)} \\
& \Rightarrow P\left( queen \right)=\dfrac{3}{49} \\
\end{align}$
Hence we get the probability of getting a queen as $\dfrac{3}{49}$.
Note: The possibility of mistake is while considering the total number of possible outcomes. Students may consider the total number of possible outcomes as 52 and solve accordingly which leads to incorrect solutions. So carefully read the conditions given in the question and then solve further.
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