What is the probability that four S’s come consecutively if all the letters of the word “MISSISSIPPI” are rearranged randomly?

Hint: Take all S’s as one block and the rest of the letters separately. The next step is to find the total number of ways in arranging all the letters with the above condition and the total ways of arranging the letters without condition and finding the probability using these two values.

Complete step-by-step answer:
MISSISSIPPI $ \Rightarrow 11$letters $\left( {4 - S,4 - I,2 - P,1 - M} \right)$.
As we know the number of ways of selecting n objects of which a are alike, b are alike and c are alike is $\dfrac{{n!}}{{a! \times b! \times c!}}$.
So, Total number of ways of arranging the word “MISSISSIPPI” without any condition
 $ \Rightarrow \dfrac{{11!}}{{\left( {4!} \right)\left( {4!} \right)\left( {2!} \right)}}.....(1)$
Now, the condition given to us is all the S’s should come together,
So to do that, we are going take all the S’s as one block, therefore,
 Total number of letters will be \[8\]( \[4 - I\], \[2 - P\], \[1 - M\], \[1\] block of S)
Therefore,
Total number of ways of arranging the above letters are
\[\dfrac{{8!}}{{\left( {4!} \right)\left( {2!} \right)}} = \dfrac{{8!}}{{\left( {4!} \right)2}}\] ….. (2)
We know probability is the ratio of number of favourable outcomes to the total number of outcomes.
Now, we will calculate the probability that all 4 S’s are together will be equation (2) divided by equation (1),
$ \Rightarrow \dfrac{{8!}}{{\left( {4!} \right)2}} \times \dfrac{{4!4!2!}}{{11!}}$
$ \Rightarrow \dfrac{{8!}}{{11!}} \times 4!$
$ \Rightarrow \dfrac{{\left( {4 \times 3 \times 2 \times 1} \right) \times 8!}}{{11 \times 10 \times 9 \times 8!}}$
Probability that all S’s will be together are$\dfrac{4}{{165}}$.

Note: Make sure to start by finding the number of ways of arrangement of the words without any condition and then find the number of ways of arrangement with any condition in these types of problems. Then find the probability.