Question

Two cards are drawn one by one from a pack of cards. Find the probability of getting the first card an ace and the second an honor card (before drawing the second card first card is not placed again in the pack).

Answer 1

Here we will be using the concept of probability which is defined as the possibility of an event to happen is equal to the ratio of the number of favorable outcomes and the total number of outcomes. The formula of probability is shown below:
$P\left( E \right) = \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}}$

Complete step-by-step solution:
Step 1: Now as per the given information in the question, two cards are drawn from a pack of cards. As we know the total number of cards in a pack is equals to
$52$.
As we know the number of ace cards in a pack is equals to
$4$.So, for calculating the probability of an ace card, we will use the formula of probability as below:
Let, ${E_1} = {\text{Getting first card as an ace}}$is an event.
So, by using the formula, we get:
$\Rightarrow P\left( {{E_1}} \right) = \dfrac{{\text{4}}}{{{\text{52}}}}$ , where the Number of favorable outcomes is equaled to $4$and the total number of outcomes is equaled to $52$.
By solving the expression $P\left( {{E_1}} \right) = \dfrac{{\text{4}}}{{{\text{52}}}}$, we get:
$\Rightarrow P\left( {{E_1}} \right) = \dfrac{1}{{13}}$ …………………….. (1)
Step 2: Now we know that an honor card is a card including all cards of ace, king, queen, and jack.
So, we also know that the number of ace/king/queen/jack cards in a pack of cards is always equal to $4$. From this information we can say that:
$\Rightarrow {\text{Number of favorable outcomes = 4 of king + 4 of queen + 4 of jack + 3 of ace}}$, (because as per the question after drawing the first card as an ace, it is not placed back in the pack.)
After adding the number of cards in the above expression we get:
$\Rightarrow {\text{Number of favorable outcomes = 15}}$
$\Rightarrow {\text{Number of total outcomes = 51}}$ (because one card is already drawn so, $52 - 1 = 51$)
Let, ${E_2}$ is an event of drawing an honor card, so by using the formula of calculating probability, we get:
$\Rightarrow P\left( {{E_2}} \right) = \dfrac{{15}}{{{\text{51}}}}$ ……………………………. (2)
Step 3: Now the total probability of two cards in which the first card we get as an ace and the second one is honor is shown as below:
$\Rightarrow {\text{Total probability of two cards drawn}} = P\left( {{E_1}} \right) \times P\left( {{E_2}} \right)$
By substituting the value from expression (1) and (2) in the above one, we get:
$\Rightarrow {\text{Total probability of two cards drawn}} = \dfrac{1}{{13}} \times \dfrac{{15}}{{51}}$
By simplifying the above expression, we get:
$\Rightarrow {\text{Total probability of two cards drawn}} = \dfrac{5}{{221}}$

The required probability is $\dfrac{5}{{221}}$.

Note: Students should remember the formula for calculating the probability for any event. Also, you should know how many total cards are there in a pack or how many ace/king/queen/jack/honor cards are there in a jack because any error in this can lead to an incorrect answer.