probabilty problem how to solve

Question

Six cards are drawn with replacement form on ordinary deck. What is the probabilty that each of four suits will be represented at least once among the six cards?

Answer 1

Consider the different cases:

1 suit $$ \begin{align*} P(\text{Only } \clubsuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\\ P(\text{Only } \heartsuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\\ P(\text{Only } \spadesuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\\ P(\text{Only } \diamondsuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6 \end{align*} $$ So 4 cases.

2 suits $$ \begin{align*} P(\text{Only } \clubsuit \text { AND } \heartsuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{2}{4}\end{pmatrix}^6 - 2\times \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\\ &.\\&.\\&.\\ \end{align*} $$ The idea here is you EXCLUDE the cases where only $\clubsuit$ or $\heartsuit$ are chosen. The residual probability is the probability of a combination of both $\clubsuit$ and $\heartsuit$
You will realize that there are ${4 \choose 2} = 6$ cases ie

$$ (\clubsuit,\heartsuit) (\clubsuit,\spadesuit) (\clubsuit,\diamondsuit) \\ (\heartsuit,\spadesuit) (\heartsuit,\diamondsuit) (\spadesuit,\diamondsuit) $$

So do your multiplication accordingly.

3 suits $$ \begin{align*} P(\text{Only } \clubsuit \text { AND } \heartsuit { AND } \spadesuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{3}{4}\end{pmatrix}^6 - 3\times\begin{pmatrix}\frac{2}{4}\end{pmatrix}^6 - 3\times \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\\ &.\\&.\\&.\\ \end{align*} $$

Similarly to 2 suits, the idea is to exclude the case where only 2 suits are selected, and only 1 suit is selected. There are $3$ ways that only 2 suits are selected and $3$ ways that only 1 suit can be selected.

There are 4 cases, namely $$ (\clubsuit,\heartsuit, \spadesuit) (\heartsuit, \spadesuit, \diamondsuit) \\ (\spadesuit,\diamondsuit, \clubsuit) (\diamondsuit,\clubsuit, \heartsuit) $$

So do the addition, and take compliment. $$ \begin{align*} \text{Req. Probability} &= 1\\&- 4\begin{bmatrix}\begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\end{bmatrix}\\&- 6\begin{bmatrix}\begin{pmatrix}\frac{2}{4}\end{pmatrix}^6 - 2\times \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\end{bmatrix}\\&- 4\begin{bmatrix}\begin{pmatrix}\frac{3}{4}\end{pmatrix}^6 - 3\times\begin{pmatrix}\frac{2}{4}\end{pmatrix}^6 + 3\times \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\end{bmatrix} \end{align*} $$

Answer 2

This can be done by inclusion-exclusion.

The probability that at most three suits are represented is $$4\times\left({3\over4}\right)^6$$ because there are $4$ ways to pick three suits and then $3/4$ chance any given card is from one of the three suits.

Probability of at most two suits is $$6\times\left({1\over2}\right)^6$$ and probability of at most one suit is $$4\times\left({1\over4}\right)^6$$

Now, do you know how to combine these to get the answer you are after?

Answer 3

The probability that at most 3 suits are represented, is paradoxically given in such a situation by applying inclusion-exclusion, viz.

4*0.75^6 - 6*0.5^6 + 4*0.25^6 = 0.6191

and the indicated probability is thus

1 - 0.6191 = 0.3809

See the question "non-routine application of inclusion-exclusion" for a full discussion of this apparent paradox