Question

The two cards are drawn at random from a pack of 52 cards. The probability of getting at least a spade and an ace, is.
$
  {\text{a}}{\text{. }}\dfrac{1}{{34}} \\
  {\text{b}}{\text{. }}\dfrac{8}{{221}} \\
  {\text{c}}{\text{. }}\dfrac{1}{{26}} \\
  {\text{d}}{\text{. }}\dfrac{2}{{51}} \\
$

Answer 1

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.