Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Three cards are drawn at random (without replacement) from a well shuffled pack of 52 cards. Find the probability distribution of the number of red cards. Hence find the mean of the distribution.

Last updated date: 15th Jun 2024
Total views: 52.2k
Views today: 1.52k
Verified
52.2k+ views
Hint: In this question we use the theory of permutation and combination. So, before solving this question you need to first recall the basics of this chapter. For example, if we need to select two cards out of four cards. In this case, this can be done in ${}^{\text{4}}{{\text{C}}_2}$ =6 ways.
Formula used- ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$

Three Cards are drawn from a pack of 52 cards.
Let X be the number of red cards drawn.
In a pack of 52 cards there are 26 red cards and 26 black cards.
We can draw 3 cards out of 52 in${}^{52}{{\text{C}}_3}$ ways.
P(X=0) = P(No red cards drawn)
= $\dfrac{{{}^{26}{{\text{C}}_3} \times {}^{26}{{\text{C}}_0}}}{{{}^{52}{{\text{C}}_3}}}$
= $\dfrac{{26 \times 25 \times 24}}{{52 \times 51 \times 50}}$
= $\dfrac{2}{{17}}$
P(X=1) = P(One red card drawn )
= $\dfrac{{{}^{26}{{\text{C}}_1} \times {}^{26}{{\text{C}}_2}}}{{{}^{52}{{\text{C}}_3}}}$
= $\dfrac{{26 \times \dfrac{{26 \times 25}}{{2 \times 1}}}}{{\dfrac{{52 \times 51 \times 50}}{{3 \times 2 \times 1}}}}$
= $\dfrac {{13}}{{34}}$
P(X=2) = P(two red card drawn )
= $\dfrac{{{}^{26}{{\text{C}}_2} \times {}^{26}{{\text{C}}_1}}}{{{}^{52}{{\text{C}}_3}}}$
= $\dfrac{{\dfrac{{26 \times 25}}{{2 \times 1}} \times \dfrac{{26}}{1}}}{{\dfrac{{52 \times 51 \times 50}}{{3 \times 2 \times 1}}}}$
= $\dfrac {{13}}{{34}}$
P(X=3) = P(three red card drawn )
= $\dfrac{{{}^{26}{{\text{C}}_3} \times {}^{26}{{\text{C}}_0}}}{{{}^{52}{{\text{C}}_3}}}$
= $\dfrac{{26 \times 25 \times 24}}{{52 \times 51 \times 50}}$
= $\dfrac{2}{{17}}$
Therefore, as we know mean is defined as-
$\sum\limits_0^n {{x_i}.P({x_i})}$= $\sum\limits_0^3 {{x_i}.P({x_i})}$
= ${x_0}.P({x_0})$+ ${x_1}.P({x_1})$+${x_2}.P({x_2})$+${x_3}.P({x_3})$
= $0 \times \dfrac{2}{{17}} + 1 \times \dfrac{{13}}{{34}} + 2 \times \dfrac{{13}}{{34}} + 3 \times \dfrac{2}{{17}}$
= $\dfrac{3}{2}$
= 1.5
Thus, the mean of the distribution is 1.5.

Note: Here we need to keep in mind the difference between terms permutation and combination. A permutation is an act of arranging the objects or numbers so as . Combinations are the way of selecting the objects or numbers from a group of objects or collections, in such a way that the order of the objects does not matter. For example, suppose we've a group of three letters: A, B, and C. Each possible selection would be an example of a mixture .