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# 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.${\text{a}}{\text{. }}\dfrac{1}{{34}} \\ {\text{b}}{\text{. }}\dfrac{8}{{221}} \\ {\text{c}}{\text{. }}\dfrac{1}{{26}} \\ {\text{d}}{\text{. }}\dfrac{2}{{51}} \\$

Last updated date: 24th Mar 2023
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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
$P = \dfrac{{{}^{13}{C_1} \times {}^4{C_1}}}{{{}^{52}{C_2}}} = \dfrac{{\dfrac{{13!}}{{12!\left( {1!} \right)}} \times \dfrac{{4!}}{{3!\left( {1!} \right)}}}}{{\dfrac{{52!}}{{50!\left( {2!} \right)}}}} = \dfrac{{13! \times 4! \times 50! \times 2!}}{{52! \times 12! \times 3!}} = \dfrac{{13 \times 12! \times 4 \times 3! \times 50! \times 2 \times 1}}{{52 \times 51 \times 50! \times 12! \times 3!}} \\ P = \dfrac{{13 \times 4 \times 2}}{{52 \times 51}} = \dfrac{2}{{51}} \\$