Question

Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that
I.All the five cards are spade
II.Only 3 cards are spade
III.None is a spade

Answer 1

Hint: In any question, whenever the possibility of yes or no, success or failure has to be dete\textined, the Bernoulli trials are used. The Bernoulli distribution is given by $ {}^n{C_r}{p^r}{q^{n - r}} $ where, $ n $ is the number of trials, $ r = 0,1,2,.. $ , $ p $ is the probability of success and $ q $ is the probability of failure, such that $ p + q = 1 $ .

Complete step-by-step answer:
In any question, whenever the possibility of yes or no, success or failure has to be dete\textined, the Bernoulli trials are used.
Bernoulli trials is given by $ {}^n{C_r}{p^r}{q^{n - r}} $ where, $ n $ is the number of trials, $ r = 0,1,2,.. $ , $ p $ is the probability of success and $ q $ is the probability of failure, such that $ p + q = 1 $ .
Let us first assume that the number of spades present among the 5 cards drawn be $ r $ .
In a deck of cards, 52 cards are present among which 5 cards are drawn successively with replacement.
The four types of cards in any deck of cards are diamond, spade, heart and club.
There are 13 cards of each type. So, there are 13 spades in a deck of well-shuffled cards.
We know that, the probability of an event can be found by the formula
\[{\text{Probability = }}\dfrac{{{\text{Favorable cases}}}}{{{\text{Total number of cases}}}}\].
The probability that the card drawn is spade can be given by
 $
\Rightarrow {\text{Probability = }}\dfrac{{{\text{Number of spade cards}}}}{{{\text{Total number of cards}}}}\\
 = \dfrac{{13}}{{52}}\\
 = \dfrac{1}{4}
 $
Let, $ p $ be the probability that the card drawn is spade, so, \[p = \dfrac{1}{4}\].
Let, $ q $ be the probability that the card drawn is not spade, so
 $ p + q = 1\\
\dfrac{1}{4} + q = 1\\
q = 1 - \dfrac{1}{4}\\
 = \dfrac{3}{4}
 $
Out of the 52 cards, only 5 cards are drawn.
Now, we can apply Bernoulli trial with $ n = 5,{\text{ }}p = \dfrac{1}{4},{\text{ }}q = \dfrac{3}{4} $ .
 $ P\left( {x = r} \right) = {}^n{C_r}{p^r}{q^{n - r}} $ , after substituting $ n = 5,{\text{ }}p = \dfrac{1}{4},{\text{ }}q = \dfrac{3}{4} $ , we get,
 $ P\left( {x = r} \right) = {}^5{C_r}{p^r}{q^{5 - r}} $
Also, we know that $ {}^n{C_r} = \dfrac{{n!}}{{r! \cdot \left( {n - r} \right)!}} $ , after substituting $ r = n $ , we get,
 $
\Rightarrow {}^n{C_n} = \dfrac{{n!}}{{n! \cdot \left( {n - n} \right)!}}\\
 = \dfrac{{n!}}{{n! \cdot 0!}}\\
 = 1
 $
I.We have to find the probability that all the 5 cards are spade.
Now, substitute $ r = 5 $ in $ P\left( {x = r} \right) = {}^5{C_r}{p^r}{q^{5 - r}} $ to find the required probability.
 $ P\left( {x = 5} \right) = {}^5{C_5}{p^5}{q^{5 - 5}} $
We know that $ {}^n{C_n} = 1 $ , so $ {}^5{C_5} = 1 $
So, after substituting $ {}^5{C_5} = 1 $ , we get,
 $
\Rightarrow P\left( {x = 5} \right) = 1 \cdot {\left( {\dfrac{1}{4}} \right)^5}{q^0}\\
 = \dfrac{1}{4} \cdot \dfrac{1}{4} \cdot \dfrac{1}{4} \cdot \dfrac{1}{4} \cdot \dfrac{1}{4}\\
 = \dfrac{1}{{1024}}
 $
Therefore, the probability that all the 5 cards are spade is $ \dfrac{1}{{1024}} $ .
II.We have to find the probability that only 3 cards are spayed.
Now, substitute $ r = 3 $ in $ P\left( {x = r} \right) = {}^5{C_r}{p^r}{q^{5 - r}} $ to find the required probability.
 $ P\left( {x = } \right) = {}^5{C_3}{p^3}{q^{5 - 3}} $
We know that $ {}^n{C_r} = \dfrac{{n!}}{{r! \cdot \left( {n - r} \right)!}} $ , so,
 $
\Rightarrow {}^5{C_3} = \dfrac{{5!}}{{3!\left( {5 - 3} \right)!}}\\
 = \dfrac{{5!}}{{3!2!}}\\
 = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 2 \times 1}}\\
 = 10
 $
After substituting $ {}^5{C_3} = 10 $ , we get,
\[
\Rightarrow P\left( {x = 3} \right) = 10 \cdot {\left( {\dfrac{1}{4}} \right)^3}{\left( {\dfrac{3}{4}} \right)^{5 - 3}}\\
 = 10 \cdot {\left( {\dfrac{1}{4}} \right)^3} \cdot {\left( {\dfrac{3}{4}} \right)^2}\\
 = 10 \cdot \dfrac{1}{4} \cdot \dfrac{1}{4} \cdot \dfrac{1}{4} \cdot \dfrac{3}{4} \cdot \dfrac{3}{4}\\
 = \dfrac{{45}}{{512}}
\]
Therefore, the probability that only 3 cards are spade is \[\dfrac{{45}}{{512}}\].
III.We have to find the probability that none is a spade.
Now, substitute $ r = 0 $ in $ P\left( {x = r} \right) = {}^5{C_r}{p^r}{q^{5 - r}} $ to find the required probability.
 $ P\left( {x = } \right) = {}^5{C_0}{p^0}{q^{5 - 0}} $
We know that $ {}^n{C_0} = 1 $ , so, $ {}^5{C_0} = 1 $
After substituting $ {}^5{C_0} = 1 $ , we get,
\[
\Rightarrow P\left( {x = 0} \right) = 1 \cdot {\left( {\dfrac{1}{4}} \right)^0}{\left( {\dfrac{3}{4}} \right)^{5 - 0}}\\
 = 1 \cdot {\left( {\dfrac{1}{4}} \right)^0} \cdot {\left( {\dfrac{3}{4}} \right)^5}\\
 = 1 \cdot \dfrac{3}{4} \cdot \dfrac{3}{4} \cdot \dfrac{3}{4} \cdot \dfrac{3}{4} \cdot \dfrac{3}{4}\\
 = \dfrac{{243}}{{1024}}
\]
Therefore, the probability that none is a spade is \[\dfrac{{243}}{{1024}}\].

Note: Students should take care of while substituting the values in Bernoulli distribution. The language of the question should be clearly understood by the students. Read every piece of info\taxation, think and analyse properly before applying Bernoulli trial. Also, students often get confused whether $ 0! = 1 $ or not. Please note that the value of zero factorial is 1. Also, $ {}^n{C_0} = 1 $ and the value of $ {}^n{C_0} $ is not equal to zero.