Solveeit Logo
Login

Question

Question: \[5\] cards are drawn one after another successively with replacement from a well-shuffled pack of \...

55 cards are drawn one after another successively with replacement from a well-shuffled pack of 5252 cards. The probability that all the 55 cards are spades-
A.(34)5{\left( {\dfrac{3}{4}} \right)^5}
B.1−(34)51 - {\left( {\dfrac{3}{4}} \right)^5}
C.(14)5{\left( {\dfrac{1}{4}} \right)^5}
D.1−(14)51 - {\left( {\dfrac{1}{4}} \right)^5}

Explanation

Solution

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-
⇒P(X=x)=nCxqn−xpx\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−p1 - p and x represents the number of success. Find p by using the formula- the probability of getting spade cards=no. of spade cardsTotal number of 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 5252 cards. We know that there are 1313 cards of each suit- club, diamond, heart, and spade.
We have to find the probability that all the 55 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-
⇒P(X=x)=nCxqn−xpx\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−p1 - p
Here we know that n=55,
The probability of getting spade cards=no. of spade cardsTotal number of cards.\dfrac{{{\text{no}}{\text{. of spade cards}}}}{{{\text{Total number of cards}}{\text{.}}}}
On putting value we get,
⇒\Rightarrow p=1352=14\dfrac{{13}}{{52}} = \dfrac{1}{4}
Then q=1−p=1−141 - p = 1 - \dfrac{1}{4}
On solving we get,
⇒\Rightarrow q=4−14=34\dfrac{{4 - 1}}{4} = \dfrac{3}{4}
Now putting all these values in eq. (i), we get-
⇒P(X=x)=5Cx(34)5−x(14)x\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=55
On putting this value in the formula, we get-
⇒P(X=5)=5C5(34)5−5(14)5\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,
⇒P(X=5)=5C5(34)0(14)5\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 nCr=n!n−r!r!{}^n{C_r} = \dfrac{{n!}}{{n - r!r!}} where n!=n(n−1)...3,2,1n! = n\left( {n - 1} \right)...3,2,1
So on applying this formula we get,
⇒P(X=5)=5!5!0!×1×(14)5\Rightarrow P\left( {X = 5} \right) = \dfrac{{5!}}{{5!0!}} \times 1 \times {\left( {\dfrac{1}{4}} \right)^5}
And we know that 0!=10! = 1 so-
⇒P(X=5)=(14)5\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 5252 cards=1352=14\dfrac{{13}}{{52}} = \dfrac{1}{4}
Then the probability of drawing 55 cards successively with replacement and all of them to be a card of spade=14×14×14×14×14=(14)5\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}