Two cards are drawn one by one at random from a pack of 52 cards. The probability that both of them are king, is
A.\[1)\dfrac{2}{13}\]
B.\[2)\dfrac{1}{169}\]
C.\[3)\dfrac{1}{221}\]
D.\[4)\dfrac{30}{221}\]
Answer
573.6k+ views
Hint: Total number of cards in the pack is\[52\]. When the first card is drawn from a pack, the total number of cards will remain as it is i.e. \[52\] and in a pack of cards the total number of kings is\[4\]. But when the next card is drawn the total number of cards is \[51\] because there is no replacement of cards and kings left are \[3\]. Now try it by yourself, you will definitely crack this problem.
Complete step-by-step answer:
Firstly, assume that the pack is well shuffled so that there is no condition of being biased and the cards are selected randomly. After one card is drawn from a pack there will be no replacement of the card allowed.
As we know that a pack contains \[52\] cards, therefore the number of sample spaces for the first card will become\[52\].
For the first card,
Total number of cards in a pack= n(S) = \[52\]
Total number of kings in a pack = \[4\]
Let, \[{{E}_{1}}=\]event of getting the first card as a king.
So, by the formula of probability, we get:
\[P({{E}_{1}})=\dfrac{4}{52}\]
where, \[4\] is the number of favorable outcomes and \[52\] is the total number of outcomes.
For the second card left,
Total number of cards left in a pack = n(S) = \[51\] (because one card is already drawn and as per question there is no replacement of card previously drawn, so\[52-1=51\])
Total number of kings left in a pack= \[3\]
Let, \[{{E}_{2}}=\]event of getting the second card as a king.
So, by the formula of probability, we get:
\[P({{E}_{2}})=\dfrac{3}{51}\]
where, \[3\]is the total number of favorable outcome and \[51\] is the total number of outcome for the second card
Now, the total probability of getting both the cards as king:
Total Probability of two cards drawn = \[P({{E}_{1}})\times P({{E}_{2}})\]
Substituting the values of \[P({{E}_{1}})\] and\[P({{E}_{2}})\], we get:
\[\Rightarrow \] Total probability of two cards drawn at random \[=\dfrac{4}{52}\times \dfrac{3}{51}\]
By simplifying the above expression, we get:
\[\Rightarrow \] Total Probability of two cards drawn at random\[=\dfrac{1}{221}\]
So, the final answer is\[\dfrac{1}{221}\].
So, the correct answer is “\[\dfrac{1}{221}\]”.
Note: Probability means possibility. It is used to predict the likely events to happen. To solve the questions of probability we always consider the unbiased situation or biased situation so that there is no unfair means and you should have a clear picture of card names and their colors.
Complete step-by-step answer:
Firstly, assume that the pack is well shuffled so that there is no condition of being biased and the cards are selected randomly. After one card is drawn from a pack there will be no replacement of the card allowed.
As we know that a pack contains \[52\] cards, therefore the number of sample spaces for the first card will become\[52\].
For the first card,
Total number of cards in a pack= n(S) = \[52\]
Total number of kings in a pack = \[4\]
Let, \[{{E}_{1}}=\]event of getting the first card as a king.
So, by the formula of probability, we get:
\[P({{E}_{1}})=\dfrac{4}{52}\]
where, \[4\] is the number of favorable outcomes and \[52\] is the total number of outcomes.
For the second card left,
Total number of cards left in a pack = n(S) = \[51\] (because one card is already drawn and as per question there is no replacement of card previously drawn, so\[52-1=51\])
Total number of kings left in a pack= \[3\]
Let, \[{{E}_{2}}=\]event of getting the second card as a king.
So, by the formula of probability, we get:
\[P({{E}_{2}})=\dfrac{3}{51}\]
where, \[3\]is the total number of favorable outcome and \[51\] is the total number of outcome for the second card
Now, the total probability of getting both the cards as king:
Total Probability of two cards drawn = \[P({{E}_{1}})\times P({{E}_{2}})\]
Substituting the values of \[P({{E}_{1}})\] and\[P({{E}_{2}})\], we get:
\[\Rightarrow \] Total probability of two cards drawn at random \[=\dfrac{4}{52}\times \dfrac{3}{51}\]
By simplifying the above expression, we get:
\[\Rightarrow \] Total Probability of two cards drawn at random\[=\dfrac{1}{221}\]
So, the final answer is\[\dfrac{1}{221}\].
So, the correct answer is “\[\dfrac{1}{221}\]”.
Note: Probability means possibility. It is used to predict the likely events to happen. To solve the questions of probability we always consider the unbiased situation or biased situation so that there is no unfair means and you should have a clear picture of card names and their colors.
Recently Updated Pages
What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

A Paragraph on Pollution in about 100-150 Words

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

