Answer

Verified

70.5k+ views

Hint:Probability$\left( P \right)$$ = \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}}$

Total number of cards$ = 52$

The total number of ways of choosing two cards at random$ = {}^{52}{C_2}$

Total no of spades in a deck of cards is$ = 13$

So, number of ways choosing a spade from 13 spades${}^{13}{C_1}$

And the number of aces in a deck of cards is 4, so, in a set of spades there is only one ace.

So, number of ways of choosing an ace from four aces${}^4{C_1}$

The number of ways choosing at least a spade and an ace$ = {}^{13}{C_1} \times {}^4{C_1}$

So, required probability$\left( P \right)$$ = \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}}$

$ = \dfrac{{{}^{13}{C_1} \times {}^4{C_1}}}{{{}^{52}{C_2}}}$

Now, we know${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$, so, using this formula

$

P = \dfrac{{{}^{13}{C_1} \times {}^4{C_1}}}{{{}^{52}{C_2}}} = \dfrac{{\dfrac{{13!}}{{12!\left( {1!} \right)}} \times \dfrac{{4!}}{{3!\left( {1!} \right)}}}}{{\dfrac{{52!}}{{50!\left( {2!} \right)}}}} = \dfrac{{13! \times 4! \times 50! \times 2!}}{{52! \times 12! \times 3!}} = \dfrac{{13 \times 12! \times 4 \times 3! \times 50! \times 2 \times 1}}{{52 \times 51 \times 50! \times 12! \times 3!}} \\

P = \dfrac{{13 \times 4 \times 2}}{{52 \times 51}} = \dfrac{2}{{51}} \\

$

Hence option d is correct.

Note: - In such types of questions first find out the total number of outcomes, then find out the number of possible outcomes, then divide them using the formula which is stated above, we will get the required probability.

Total number of cards$ = 52$

The total number of ways of choosing two cards at random$ = {}^{52}{C_2}$

Total no of spades in a deck of cards is$ = 13$

So, number of ways choosing a spade from 13 spades${}^{13}{C_1}$

And the number of aces in a deck of cards is 4, so, in a set of spades there is only one ace.

So, number of ways of choosing an ace from four aces${}^4{C_1}$

The number of ways choosing at least a spade and an ace$ = {}^{13}{C_1} \times {}^4{C_1}$

So, required probability$\left( P \right)$$ = \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}}$

$ = \dfrac{{{}^{13}{C_1} \times {}^4{C_1}}}{{{}^{52}{C_2}}}$

Now, we know${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$, so, using this formula

$

P = \dfrac{{{}^{13}{C_1} \times {}^4{C_1}}}{{{}^{52}{C_2}}} = \dfrac{{\dfrac{{13!}}{{12!\left( {1!} \right)}} \times \dfrac{{4!}}{{3!\left( {1!} \right)}}}}{{\dfrac{{52!}}{{50!\left( {2!} \right)}}}} = \dfrac{{13! \times 4! \times 50! \times 2!}}{{52! \times 12! \times 3!}} = \dfrac{{13 \times 12! \times 4 \times 3! \times 50! \times 2 \times 1}}{{52 \times 51 \times 50! \times 12! \times 3!}} \\

P = \dfrac{{13 \times 4 \times 2}}{{52 \times 51}} = \dfrac{2}{{51}} \\

$

Hence option d is correct.

Note: - In such types of questions first find out the total number of outcomes, then find out the number of possible outcomes, then divide them using the formula which is stated above, we will get the required probability.

Recently Updated Pages

Write an article on the need and importance of sports class 10 english JEE_Main

Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main

Choose the one which best expresses the meaning of class 9 english JEE_Main

What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main

A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main

A man stands at a distance of 250m from a wall He shoots class 9 physics JEE_Main

Other Pages

The reaction of Zinc with dilute and concentrated nitric class 12 chemistry JEE_Main

The positive temperature coefficient of resistance class 11 physics JEE_Main

Which is the correct statement As the scharacter of class 11 chemistry JEE_Main

A shell explodes and many pieces fly in different directions class 11 physics JEE_Main

The bond order of ClO bond in ClO4 ion and the effective class 11 chemistry JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main