Question

What will be the probability of getting $3$ red and $4$ black cards respectively, when seven cards are drawn at random from a pack of $52$ cards?
(a) $\dfrac{1}{6}$
(b) $\dfrac{{27}}{{37}}$
(c) Cannot be determined
(d) None of these

Answer 1

We will use the definition of probability for each iteration/s for a given number of outcomes, that is mathematically indented formulae that to be analyzed with the required outcome or solution. To find the desired outcome/s we will use the concept of combination i.e. $^n{C_r} = \dfrac{{n!}}{{r!(n - r)}}$ , so as to get the desire (or, exact) probability of both the required outcomes that is $3$ red and $4$ black cards respectively and then using the definition of probability i.e. $Probability = \dfrac{{Favorable{\text{ }}outcomes}}{{Total{\text{ }}outcomes}}$ respectively, substituting the respective values using the respective combinational formula, to obtained the required probability.

Complete answer: Since, the pack of card is fair, all the outcomes of the pack are equally probable; the probability of getting $3$ red and $4$ black cards are equally authorized with probability of the outcomes from $26$ red and $26$ black cards that is from out of total $52$ cards respectively.
As a result, to find the total number of possible cases of getting $3$ red and $4$ black cards in the pack, according to the combinational statement in the mathematics, we have
Hence, the event contains number of outcomes. Therefore, using combinational statement/condition can be better to find an appropriate value!
$\Rightarrow {\text{Total outcomes = }}{{\text{ }}^n}{{\text{C}}_r}$
Where, ‘n’ is total outcomes and ‘r’ is required outcome of respective parameter
$\Rightarrow {\text{Total outcomes = }}{{\text{ }}^{52}}{{\text{C}}_7}$ … (i)
Where,
‘$n$’ is the required outcomes i.e. $= 52$,
‘$r$’ is the required outcomes i.e. $3 + 4 = 7$ (including both red and black cards).
Hence, we know that
As a result, using this relation to find the required probability, $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$, we get
Considering the required $3$ red cards out of $26$ red cards, we get
$^{26}{C_3} = \dfrac{{26!}}{{3!(26 - 3)!}}$ … (ii)
Similarly,
Considering the required $4$ black cards out of $26$ red cards, we get
$^{26}{C_4} = \dfrac{{26!}}{{4!(26 - 4)!}}$ … (iii)
Now, since by the definition of probability i.e. $Probability = \dfrac{{n\left( s \right)}}{{n\left( A \right)}}$ where,
$n\left( s \right) =$ Total favourable outcomes (here, we will bifurcate into two given/required outcomes so as to make solution in simplified form), and
$n\left( A \right) =$ Total number of outcomes in the respective (given) event.
Hence, we get
$Probability,p\left( A \right) = \dfrac{{^{26}{C_3}{ \times ^{26}}{C_4}}}{{^{52}{C_7}}}$
Substituting the respective values from the equations (i), (ii) and (ii), we get
$p\left( A \right) = \dfrac{{\dfrac{{26!}}{{3!(26 - 3)!}} \times \dfrac{{26!}}{{4!(26 - 4)!}}}}{{\dfrac{{52!}}{{7!(52 - 7)!}}}}$
Solving the equation predominantly, we get
$p\left( A \right) = \dfrac{{26!}}{{3!(26 - 3)!}} \times \dfrac{{26!}}{{4!(26 - 4)!}} \times \dfrac{{7!(52 - 7)!}}{{52!}}$
$p\left( A \right) = \dfrac{{26!}}{{3! \times 23!}} \times \dfrac{{26!}}{{4! \times 22!}} \times \dfrac{{7! \times 45!}}{{52!}}$
Since, the above equation is solved by the definition of factorial which seems that the multiplication of a respective number preceding to every number in the sequence of it (till one),
Hence, simplifying it
$p\left( A \right) = \dfrac{{26 \times 25 \times 24 \times 23!}}{{3! \times 23!}} \times \dfrac{{26 \times 25 \times 24 \times 23 \times 22!}}{{4! \times 22!}} \times \dfrac{{7! \times 45!}}{{52 \times 51 \times 50 \times 49 \times 48 \times 47 \times 46 \times 45!}}$
$p\left( A \right) = \dfrac{{26 \times 25 \times 24}}{{3 \times 2 \times 1}} \times \dfrac{{26 \times 25 \times 24 \times 23}}{{4 \times 3 \times 2 \times 1}} \times \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{52 \times 51 \times 50 \times 49 \times 48 \times 47 \times 46}}$
As a result, solving the equation mathematically, we get
$p\left( A \right) = 26 \times 25 \times 4 \times 26 \times 25 \times 23 \times \dfrac{{52 \times 51 \times 25 \times 7 \times 47 \times 46}}{3}$
Again, simplifying the equation, we get
$p\left( A \right) = 26 \times 25 \times 26 \times 25 \times 23 \times \dfrac{3}{{13 \times 51 \times 25 \times 7 \times 47 \times 46}}$
$p\left( A \right) = \dfrac{{26 \times 25 \times 3}}{{51 \times 7 \times 47}}$
Hence, the required probability is
$p\left( A \right) = \dfrac{{1950}}{{16779}}$
$p\left( A \right) = \dfrac{{650}}{{5593}}$
Since, from the given options no value exists!
$\therefore$The option (d) is correct.

Note:
The probability of an event A is denoted by ‘P(A)’ and the basic formula of finding probability is $P(A) = \dfrac{{n(s)}}{{n(A)}}$, where n(s) denotes the favorable outcomes and n(A) denotes the total number of outcomes for an respective event. Getting rid of misconceptions one must know the basic formulae of solving such application based problem like $^n{C_r} = \dfrac{{n!}}{{r!(n - r)}}$ (in case of finding the probability of complex events). As a result, we should take care of the calculations so as to be sure of our final answer.