Question

From a pack of cards, \[2\] cards are chosen randomly. Find the probability of the event of one card is $10$ which are not hearts and another hearts card.
$1)$$\dfrac{1}{{34}}$
$2)$$\dfrac{1}{{102}}$
$3)$$\dfrac{8}{{663}}$
$4)$$\dfrac{{33}}{{34}}$

Answer 1

Hint: First we have to define what the terms we need to solve the problem are.
Since a pack of cards means it will contain $52$ cards, in four different shapes they are $13$cards each thus a total of fifty-two cards in a pack of cards, and the shapes are heart, spade, diamond and clubs.
Each will contain thirteen cards.

Complete answer:
Since in a pack of cards there are fifty-two cards, also in each of the four shapes there are thirteen cards like $1,2,3,...,10$ and Jake king and queen.
Thus, in the given question \[2\] cards are chosen random and now we need to find the probability of the event of one card is ten and which is not a heart
So, in $52$ cards, $13$ cards are hearts and \[4\] cards are number \[10\]
To find the probability we need the formula for the probability is $\dfrac{{\text{Number of favourable outcomes}}}{{\text{total number of outcomes}}}$ which is the number of favorable events divided by the number of total outcomes of the given event.
First, we find the favorable outcomes which is $13$ cards of heart is $13{c_1}$(in combination)
And four cards of number \[10\] but it is not a heart thus four minus one is three
Hence four cards of number \[10\] are $3{c_1}$(number of combination) now the product yields the required favorable outcomes which is $ = 13{c_1} \times 3{c_1}$(c one is combination with base one factorial)
Total number of outcomes is $52{c_2}$(combination with two cards outcome)
Total number = $52{c_2}$= $\dfrac{{52!}}{{2!}} = 26 \times 51$(since the two will cancel to $52$ factorials elsewhere)
Hence probability = $\dfrac{{3{c_1} \times 13{c_1}}}{{26 \times 51}} = \dfrac{3}{{102}} = \dfrac{1}{{34}}$which is the required probability of another heart card.

So, the correct answer is “Option 1”.

Note: Since there is no possible of getting other options such as$\dfrac{1}{{102}}$,$\dfrac{8}{{663}}$,$\dfrac{{33}}{{34}}$because of another heart card will be possible in lesser probability of chance and count the favorable events carefully so
there will be any calculation mistakes and yields the correct option.