Question

Three cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of spades. Hence find the mean of the distribution.

Answer 1

Hint: In this question, we have been given a situation for which we need to calculate the probability by the binomial distribution. So, for that firstly we will be calculating the probability of getting a spade in a draw after that the probability of not getting a spade in a draw by the formula:
$\text{Probability} = \dfrac{\text{Favorable outcomes}}{\text{Total number of outcomes}}$
So according to the question number of trials would be equal to 3. In the next step we will be using the formula of binomial distribution $P(x = r)\Rightarrow { ^n}{C_r}{p^r}{q^{n - r}}$. So, as we need to calculate the probability distribution of the number of spades, so we will be calculating the probability for r equal to 0,1,2 and 3 and add them which will be our desired solution.

Complete step-by-step solution:
We have been provided with a condition in this question, so we have a pack of 52 cards.
Now we will be calculating the probability of getting spade in a draw and the probability of not getting spade in a draw by using the formula: Probability = favorable outcomes ÷ total number of outcomes
Using the above formula
Probability of getting spade in a draw = $\dfrac{{13}}{{52}} = \dfrac{1}{4}$
Probability of not getting spade in a draw = $1 - \dfrac{1}{4} = \dfrac{3}{4}$
Now according to the question, three cards were thrown with replacement so the number of trials is equal to 3
Number of trials n=3
Using the formula of the binomial distribution
 $P(x = r) => { ^n}{C_r}{p^r}{q^{n - r}}$
Now as in this question we need to calculate the probability for we can take the value of r=0,1,2,3respectively.
For r=0, $^3{C_0}{\dfrac{1}{4}^0}{\dfrac{3}{4}^3}$
Now we will be simplifying it and finding its value which comes out to be $\dfrac{{27}}{{64}}$
Now we will be calculating for r=1, $^3{C_1}{\dfrac{1}{4}^1}{\dfrac{3}{4}^2}$
Now we will be simplifying it and finding its value which comes out to be $\dfrac{{27}}{{64}}$
Now we will be calculating for r=2, $^3{C_2}{\dfrac{1}{4}^2}{\dfrac{3}{4}^1}$
Now we will be simplifying it and finding its value which comes out to be $\dfrac{9}{{64}}$
Now we will be calculating for r=3, $^3{C_3}{\dfrac{1}{4}^3}{\dfrac{3}{4}^0}$
Now we will be simplifying it and finding its value which comes out to be $\dfrac{1}{{64}}$
Now after finding all the probabilities we will be finding the mean of the distribution:
For that firstly we will write all the probability distributions for r=0,1,2,3.

X	P(X)
0	$\dfrac{{27}}{{64}}$
1	$\dfrac{{27}}{{64}}$
2	$\dfrac{9}{{64}}$
3	$\dfrac{1}{{64}}$

Now we will be finding the mean using the formula:$\sum {X.P(X)} $
Now we will be putting the values: $\left[ {\left( {0.\dfrac{{27}}{{64}}} \right) + \left( {1.\dfrac{{27}}{{64}}} \right) + \left( {2.\dfrac{9}{{64}}} \right) + \left( {3.\dfrac{1}{{64}}} \right)} \right]$
Now we will be simplifying it further: $\dfrac{{27}}{{64}} + \dfrac{{18}}{{64}} + \dfrac{3}{{64}} = \dfrac{{49}}{{64}}$
So, the mean comes out to be $\dfrac{{48}}{{64}}$.

Note: In this question, be careful while finding the number of trials as we need to include 0 also along with 1,2, and 3. Do draw a table mentioning the probability distribution to avoid any kind of mistakes and use the formulas appropriately while calculating probability distributions.