Question

A fair coin is tossed four times. Let X denote the longest string of heads occurring. Find the probability distribution, mean and variance of X

Answer 1

Hint: In this question, we need to determine the probability distribution, mean and variance of X where X denotes the longest string of heads occurring when a coin is tossed four times. For this, we will evaluate the total number of possible outcomes and total number of outcomes on tossing four fair coins.

Complete step-by-step answer:
The possible outcome when four coins are tossed together:
(HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT)
So the total number of outcome\[ = 16\]
Now for the longest string of heads, X can be 0, 1, 2, 3 and 4.
Now,
When X=0, no head
\[P\left( {X = 0} \right) = \dfrac{1}{{16}}\]
When X=1, one head
\[P\left( {X = 1} \right) = \dfrac{7}{{16}}\]
When X=2, two head
\[P\left( {X = 2} \right) = \dfrac{5}{{16}}\]
When X=3, three head
\[P\left( {X = 3} \right) = \dfrac{2}{{16}}\]
When X=4, three head
\[P\left( {X = 4} \right) = \dfrac{1}{{16}}\]
Hence we can write the probability distribution as

x	P(x)
0	\[\dfrac{1}{{16}}\]
1	\[\dfrac{7}{{16}}\]
2	\[\dfrac{5}{{16}}\]
3	\[\dfrac{2}{{16}}\]
4	\[\dfrac{1}{{16}}\]

Now to compute the mean and variance, we can write the obtained data as

\[{x_i}\]	\[{p_i}\]	\[{p_i}{x_i}\]	\[{p_i}x_i^2\]
0	\[\dfrac{1}{{16}}\]	0	0
1	\[\dfrac{7}{{16}}\]	\[\dfrac{7}{{16}}\]	\[\dfrac{7}{{16}}\]
2	\[\dfrac{5}{{16}}\]	\[\dfrac{{10}}{{16}}\]	\[\dfrac{{20}}{{16}}\]
3	\[\dfrac{2}{{16}}\]	\[\dfrac{6}{{16}}\]	\[\dfrac{{18}}{{16}}\]
4	\[\dfrac{1}{{16}}\]	\[\dfrac{4}{{16}}\]	1
		\[\sum {{p_i}{x_i}} = \dfrac{{27}}{{16}}\]	\[\sum {{p_i}x_i^2} = \dfrac{{61}}{{16}}\]

We know mean of given data can be found by using the formula\[\sum {{p_i}{x_i}} \], hence from the obtained table we can write
$\Rightarrow$ Mean \[\sum {{p_i}{x_i}} = \dfrac{{27}}{{16}} = 1.7\]
Hence the mean of \[X = 1.7\]
Now also we know the variance of certain observation is given by the formula
\[\Rightarrow Varience = \sum {{p_i}x_i^2 - {{\left( {Mean} \right)}^2}} \]
Hence by substituting the values from the above table, we can write
\[
\Rightarrow Varience = \sum {{p_i}x_i^2 - {{\left( {Mean} \right)}^2}} \\
   = \dfrac{{61}}{{16}} - \dfrac{{729}}{{256}} \\
   = \dfrac{{247}}{{256}} \\
   = 0.9 \;
 \]
Variance of \[X = 0.9\]
Note: Probability distribution is the tabulated form of the probabilities of the different outcomes of an event. Mean is the average of the outcomes in the probability distribution while variance is the difference of the particular outcome from the mean of the probability distribution.