Question

\[5\] cards are drawn one after another successively with replacement from a well-shuffled pack of $52$ cards. The probability that all the $5$ cards are spades-
A.${\left( {\dfrac{3}{4}} \right)^5}$
B.$1 - {\left( {\dfrac{3}{4}} \right)^5}$
C.${\left( {\dfrac{1}{4}} \right)^5}$
D.$1 - {\left( {\dfrac{1}{4}} \right)^5}$

Answer 1

Hint: Bernoulli trial represents only two possible outcomes- success or failure, of any random experiment. The number of success and failure is given by binomial distribution formula-
$ \Rightarrow P\left( {X = x} \right) = {}^n{C_x}{q^{n - x}}{p^x}$ Where n is the number of cards drawn, p is the probability of getting spade cards (success) and q (probability of failure)=$1 - p$ and x represents the number of success. Find p by using the formula- the probability of getting spade cards=$\dfrac{{{\text{no}}{\text{. of spade cards}}}}{{{\text{Total number of cards}}{\text{.}}}}$
Then find q using the value of p. Put all the values in the formula and solve.

Complete step-by-step answer:
Given, five cards are drawn one after another with replacement from a pack of $52$ cards. We know that there are $13$ cards of each suit- club, diamond, heart, and spade.
We have to find the probability that all the $5$ cards are spades.
Let the number of spade cards be X.
So drawing a card is a Bernoulli trial and the binomial distribution is given as-
$ \Rightarrow P\left( {X = x} \right) = {}^n{C_x}{q^{n - x}}{p^x}$ --- (i)
Where n is the number of cards drawn, p is the probability of getting spade cards and q=$1 - p$
Here we know that n=$5$,
The probability of getting spade cards=$\dfrac{{{\text{no}}{\text{. of spade cards}}}}{{{\text{Total number of cards}}{\text{.}}}}$
On putting value we get,
$ \Rightarrow $ p=$\dfrac{{13}}{{52}} = \dfrac{1}{4}$
Then q=$1 - p = 1 - \dfrac{1}{4}$
On solving we get,
$ \Rightarrow $ q=$\dfrac{{4 - 1}}{4} = \dfrac{3}{4}$
Now putting all these values in eq. (i), we get-
$ \Rightarrow P\left( {X = x} \right) = {}^5{C_x}{\left( {\dfrac{3}{4}} \right)^{5 - x}}{\left( {\dfrac{1}{4}} \right)^x}$
Now we have to find the probability that all the cards are spade so x=$5$
On putting this value in the formula, we get-
$ \Rightarrow P\left( {X = 5} \right) = {}^5{C_5}{\left( {\dfrac{3}{4}} \right)^{5 - 5}}{\left( {\dfrac{1}{4}} \right)^5}$
On solving we get,
$ \Rightarrow P\left( {X = 5} \right) = {}^5{C_5}{\left( {\dfrac{3}{4}} \right)^0}{\left( {\dfrac{1}{4}} \right)^5}$
Now we know that ${}^n{C_r} = \dfrac{{n!}}{{n - r!r!}}$ where $n! = n\left( {n - 1} \right)...3,2,1$
So on applying this formula we get,
$ \Rightarrow P\left( {X = 5} \right) = \dfrac{{5!}}{{5!0!}} \times 1 \times {\left( {\dfrac{1}{4}} \right)^5}$
And we know that $0! = 1$ so-
$ \Rightarrow P\left( {X = 5} \right) = {\left( {\dfrac{1}{4}} \right)^5}$
Hence the correct answer is C.

Note: Here we can also solve this question by this method-
Since we know that the probability of drawing one card of spade from a pack of $52$ cards=\[\dfrac{{13}}{{52}} = \dfrac{1}{4}\]
Then the probability of drawing $5$ cards successively with replacement and all of them to be a card of spade=$\dfrac{1}{4} \times \dfrac{1}{4} \times \dfrac{1}{4} \times \dfrac{1}{4} \times \dfrac{1}{4} = {\left( {\dfrac{1}{4}} \right)^5}$