
Probability of getting an ace from a well shuffled deck of 52 playing cards is \[\dfrac{1}{m}\]. Value of \[m\] is
Answer
578.1k+ views
Hint:
We will find the number of aces. We know that there are 52 cards in a deck in total. We will use the formula to find the probability. We will equate the calculated probability with \[\dfrac{1}{m}\]. We will find the reciprocal of the probability to get the value of \[m\],
Formulas used: The probability (P) is given by \[{\rm{P}} = \dfrac{{\rm{F}}}{{\rm{T}}}\] where F is the number of favourable outcomes and T is the number of total outcomes.
Complete step by step solution:
A standard deck of 52 cards has 4 aces; 1 of hearts, 1 of clubs, 1 of diamond and 1 of spades. There are 52 cards in total.
We will substitute 4 for F and 52 for T in the formula of probability.
\[{\Rightarrow \rm{P}} = \dfrac{4}{{52}} = \dfrac{4}{{13 \times 4}}\]
We can see that the numerator and denominator have 4 as a common factor. We will cancel the common factor and find the probability.
\[{\Rightarrow \rm{P}} = \dfrac{{\rm{1}}}{{13}}\]
We will equate the probability with \[\dfrac{1}{m}\] :
\[\Rightarrow \dfrac{1}{4} = \dfrac{1}{m}\]
We will find the reciprocal of both sides to find value of \[m\]:
\[\begin{array}{l} \Rightarrow \dfrac{{13}}{1} = \dfrac{m}{1}\\ \Rightarrow 13 = m\end{array}\]
The value of \[m\] is 13.
Note:
A standard deck of cards has 4 suits – Hearts, Clubs, Diamonds and Spades. Each suite has 13 cards. They are Ace, 2, 3, 4, 5, 6, 7, 8, 9, Jack, Queen and King. There are 4 cards of each value in a standard deck. We can also obtain the value of \[m\] by cross multiplying the numerator of the first fraction with the denominator of the second fraction and the denominator of the first fraction with the numerator of the second fraction in the following step:
\[\begin{array}{l}\dfrac{1}{{13}} = \dfrac{1}{m}\\1 \cdot m = 1 \cdot 13\\m = 13\end{array}\]
We get the value of \[m\] as 13 by using this method as well.
We will find the number of aces. We know that there are 52 cards in a deck in total. We will use the formula to find the probability. We will equate the calculated probability with \[\dfrac{1}{m}\]. We will find the reciprocal of the probability to get the value of \[m\],
Formulas used: The probability (P) is given by \[{\rm{P}} = \dfrac{{\rm{F}}}{{\rm{T}}}\] where F is the number of favourable outcomes and T is the number of total outcomes.
Complete step by step solution:
A standard deck of 52 cards has 4 aces; 1 of hearts, 1 of clubs, 1 of diamond and 1 of spades. There are 52 cards in total.
We will substitute 4 for F and 52 for T in the formula of probability.
\[{\Rightarrow \rm{P}} = \dfrac{4}{{52}} = \dfrac{4}{{13 \times 4}}\]
We can see that the numerator and denominator have 4 as a common factor. We will cancel the common factor and find the probability.
\[{\Rightarrow \rm{P}} = \dfrac{{\rm{1}}}{{13}}\]
We will equate the probability with \[\dfrac{1}{m}\] :
\[\Rightarrow \dfrac{1}{4} = \dfrac{1}{m}\]
We will find the reciprocal of both sides to find value of \[m\]:
\[\begin{array}{l} \Rightarrow \dfrac{{13}}{1} = \dfrac{m}{1}\\ \Rightarrow 13 = m\end{array}\]
The value of \[m\] is 13.
Note:
A standard deck of cards has 4 suits – Hearts, Clubs, Diamonds and Spades. Each suite has 13 cards. They are Ace, 2, 3, 4, 5, 6, 7, 8, 9, Jack, Queen and King. There are 4 cards of each value in a standard deck. We can also obtain the value of \[m\] by cross multiplying the numerator of the first fraction with the denominator of the second fraction and the denominator of the first fraction with the numerator of the second fraction in the following step:
\[\begin{array}{l}\dfrac{1}{{13}} = \dfrac{1}{m}\\1 \cdot m = 1 \cdot 13\\m = 13\end{array}\]
We get the value of \[m\] as 13 by using this method as well.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Why cannot DNA pass through cell membranes class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

In a human foetus the limbs and digits develop after class 12 biology CBSE

AABbCc genotype forms how many types of gametes a 4 class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

The correct structure of ethylenediaminetetraacetic class 12 chemistry CBSE

