Question

A deck of $52$ cards is shuffled and dealt.

Find the probabilities of the following events:

The last five cards are spades.

Answer 1

To find the probability that the last five cards are spades:

The total number of cards are $52$.

In the total cards, there are the number of spades are $13$.

Using the concept of the probability is ratio of $\frac{\text{Event occur}}{\text{Total number of outcomes}}$, find the probability of the card is spade :

$\begin{array}{rcl}p& =& \frac{{}^{13}C_1}{{}^{52}{C}_1}\\ & \Rightarrow & p=\frac{13}{52}\\ \therefore p& =& \frac{1}{4}\end{array}$

Using the binomial distribution function $P\left(x\right)={}^{n}C_r\times p^r\times {\left(1-p\right)}^{n-r}$ and calculate the probability that the last five $\left(52-5=47\right)$ cards are spades in below :

$\begin{array}{rcl}P\left(x\right)& =& {}^{47}C_1\times {\left(\frac{1}{4}\right)}^1\times {\left(1-\frac{1}{4}\right)}^{47-1}\left[\text{Substitute all values}\right]\\ & \Rightarrow & P\left(x\right)=47\times \frac{1}{4}\times {\left(\frac{3}{4}\right)}^{46}\left[\text{Simplifying}\right]\\ \therefore P\left(x\right)& =& \frac{47\times 3^{46}}{4^{47}}\end{array}$

Hence, the probability is $\begin{array}{l}\frac{\mathbf{47}\mathbf{\times }{\mathbf{3}}^{\mathbf{46}}}{{\mathbf{4}}^{\mathbf{47}}}\end{array}$.