91Ó°ÊÓ

Two six-sided fair dice are tossed independently and the random variable\(M\)represent the maximum of the two tosses.

The possible outcomes of a throwing a single die is 6 i.e.\(\left\{ {1,2,3,4,5,6} \right\}\). If two dice are thrown then the possible outcomes is\({6^2} = 36\). The sample space S is given as:

\(S = \left\{ \begin{array}{l}\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right)\\\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right)\\\left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,4} \right),\left( {3,5} \right),\left( {3,6} \right)\\\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right)\\\left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {5,4} \right),\left( {5,5} \right),\left( {5,6} \right)\\\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)\end{array} \right\}\)

Consider M to be the maximum number in pair of two tosses and it can be taken for each outcome of an experiment.

Let\(M\left( {1,2} \right) = 2\), the value of M is 2. Similarly, for\(M\left( {6,3} \right) = 6\)the value is 6. The possible values for each outcome can be written as:

\(M = \left\{ \begin{array}{l}1,2,3,4,5,6\\2,2,3,4,5,6\\3,3,3,4,5,6\\4,4,4,4,5,6\\5,5,5,5,5,6\\6,6,6,6,6,6\end{array} \right\}\)

From the above list of values, the possible value of random variable M is\(\left\{ {1,2,3,4,5,6} \right\}\).

As M is a random variable that can take maximum value among two tosses, when\(M = 1\) it means the maximum number in the pair is 1. Since the dice are fair so each outcome is equally likely to get selected. Then the probability of each outcome is\(\frac{1}{{36}}\).

The possible pair is\(\left\{ {\left( {1,1} \right)} \right\}\). So its probability can be calculated as:

\(\begin{aligned}P\left( {M = 1} \right) &= P\left( {1,1} \right)\\ &= \frac{1}{{36}}\end{aligned}\)

Similarly, when\(M = 2\)it means the maximum number in the pair is 2. The possible pairs are\(\left\{ {\left( {1,2} \right),\left( {2,1} \right),\left( {2,2} \right)} \right\}\). So its probability is calculated as,

\(\begin{aligned}P\left( {M = 2} \right) &= \left( {P\left( {1,2} \right) + P\left( {2,1} \right) + P\left( {2,2} \right)} \right)\\ &= \left( {\frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}}} \right)\\ &= \frac{3}{{36}}\end{aligned}\)

Similarly, when\(M = 3\)it means the maximum number in the pair is 3. The possible pairs are\(\left\{ {\left( {1,3} \right),\left( {2,3} \right),\left( {3,1} \right),\left( {3,3} \right),\left( {3,2} \right)} \right\}\). So its probability is calculated as,

\(\begin{aligned}P\left( {M = 3} \right) &= \left( {P\left( {1,3} \right) + P\left( {2,3} \right) + P\left( {3,1} \right) + P\left( {3,3} \right) + P\left( {3,2} \right)} \right)\\ &= \left( {\frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}}} \right)\\ &= \frac{5}{{36}}\end{aligned}\)

Similarly, when\(M = 4\)it means the maximum number in the pair is 4. The possible pairs are\(\left\{ {\left( {1,4} \right),\left( {2,4} \right),\left( {3,4} \right),\left( {4,3} \right),\left( {4,2} \right),\left( {4,1} \right),\left( {4,4} \right)} \right\}\). So its probability is calculated as,

\(\begin{aligned}P\left( {M = 4} \right) &= \left( {P\left( {1,4} \right) + P\left( {2,4} \right) + P\left( {3,4} \right) + P\left( {4,3} \right) + P\left( {4,2} \right) + P\left( {4,1} \right) + P\left( {4,4} \right)} \right)\\ &= \left( {\frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}}} \right)\\ &= \frac{7}{{36}}\end{aligned}\)

Similarly, when\(M = 5\)it means the maximum number in the pair is 5. The possible pairs are\(\left\{ {\left( {1,5} \right),\left( {2,5} \right),\left( {3,5} \right),\left( {4,5} \right),\left( {5,2} \right),\left( {5,1} \right),\left( {5,4} \right),\left( {5,3} \right),\left( {5,5} \right)} \right\}\). So its probability is calculated as,

\(\begin{aligned}P\left( {M = 5} \right) &= \left( \begin{array}{l}P\left( {1,5} \right) + P\left( {2,5} \right) + P\left( {3,5} \right) + P\left( {4,5} \right) + P\left( {5,2} \right)\\ + P\left( {5,1} \right) + P\left( {5,4} \right) + P\left( {5,3} \right) + P\left( {5,5} \right)\end{array} \right)\\ &= \left( {\frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}}} \right)\\ &= \frac{9}{{36}}\end{aligned}\)

Similarly, when\(M = 6\)it means the maximum number in the pair is 6. The possible pairs are\(\left\{ {\left( {1,6} \right),\left( {2,6} \right),\left( {3,6} \right),\left( {4,6} \right),\left( {5,6} \right),\left( {6,1} \right),\left( {6,2} \right),\left( {6,4} \right),\left( {6,3} \right),\left( {6,5} \right),\left( {6,6} \right)} \right\}\). So its probability is calculated as,

\(\begin{aligned}P\left( {M = 6} \right) &= \left( \begin{array}{l}P\left( {1,6} \right) + P\left( {2,6} \right) + P\left( {3,6} \right) + P\left( {4,6} \right) + P\left( {5,6} \right) + P\left( {6,1} \right)\\ + P\left( {6,2} \right) + P\left( {6,4} \right) + P\left( {6,3} \right) + P\left( {6,5} \right) + P\left( {6,6} \right)\end{array} \right)\\ &= \left( {\frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}} + \frac{1}{{36}}} \right)\\ &= \frac{{11}}{{36}}\end{aligned}\)

The probability distribution table is given as:

The probability mass function for M is,

\(P\left( M \right) = \left\{ \begin{array}{l}\frac{1}{{36}},\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when}}\,M = 1\\\frac{3}{{36}},\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when}}\,M = 2\\\frac{5}{{36}},\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when}}\,M = 3\\\frac{7}{{36}},\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when}}\,M = 4\\\frac{9}{{36}},\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when}}\,M = 5\\\frac{{11}}{{36}},\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when}}\,M = 6\end{array} \right.\)

From the problem, the cdf\(F\left( {M = m} \right)\) can be defined as,

\(F\left( {M = m} \right) = P\left( {M \le m} \right)\)

First, compute\(F\left( 1 \right)\) using the above formula:

\(\begin{aligned}F\left( {M = 1} \right) &= P\left( {M \le 1} \right)\\ &= P\left( {M = 1} \right)\\ &= \frac{1}{{36}}\end{aligned}\)

Then, compute\(F\left( 2 \right)\) using the above formula:

\(\begin{aligned}F\left( {M = 2} \right) &= P\left( {M \le 2} \right)\\ &= \left( {P\left( {M = 1} \right) + P\left( {M = 2} \right)} \right)\\ &= \left( {\frac{1}{{36}} + \frac{3}{{36}}} \right)\\ &= \frac{4}{{36}}\end{aligned}\)

Then, compute\(F\left( 3 \right)\) using the above formula:

\(\begin{aligned}F\left( {M = 3} \right) &= P\left( {M \le 3} \right)\\ &= \left( {P\left( {M = 1} \right) + P\left( {M = 2} \right) + P\left( {M = 3} \right)} \right)\\ &= \left( {\frac{1}{{36}} + \frac{3}{{36}} + \frac{5}{{36}}} \right)\\ &= \frac{9}{{36}}\end{aligned}\)

Similarly, compute\(F\left( 4 \right)\) using the above formula:

\(\begin{aligned}F\left( {M = 4} \right) &= P\left( {M \le 4} \right)\\ &= \left( {P\left( {M = 1} \right) + P\left( {M = 2} \right) + P\left( {M = 3} \right) + P\left( {M = 4} \right)} \right)\\ &= \left( {\frac{1}{{36}} + \frac{3}{{36}} + \frac{5}{{36}} + \frac{7}{{36}}} \right)\\ &= \frac{{16}}{{36}}\end{aligned}\)

Similarly, compute\(F\left( 5 \right)\) using the above formula:

\(\begin{aligned}F\left( {M = 5} \right) &= P\left( {M \le 5} \right)\\ &= \left( {P\left( {M = 1} \right) + P\left( {M = 2} \right) + P\left( {M = 3} \right) + P\left( {M = 4} \right) + P\left( {M = 5} \right)} \right)\\ &= \left( {\frac{1}{{36}} + \frac{3}{{36}} + \frac{5}{{36}} + \frac{7}{{36}} + \frac{9}{{36}}} \right)\\ &= \frac{{25}}{{36}}\end{aligned}\)

Finally, compute\(F\left( 6 \right)\) using the above formula:

\(\begin{aligned}F\left( {M = 6} \right) &= P\left( {M \le 6} \right)\\ &= \left( {P\left( {M = 1} \right) + P\left( {M = 2} \right) + P\left( {M = 3} \right) + P\left( {M = 4} \right) + P\left( {M = 5} \right) + P\left( {M = 6} \right)} \right)\\ &= \left( {\frac{1}{{36}} + \frac{3}{{36}} + \frac{5}{{36}} + \frac{7}{{36}} + \frac{9}{{36}} + \frac{{11}}{{36}}} \right)\\ &= \frac{{36}}{{36}}\\ &= 1\end{aligned}\)

The cumulative distribution function of M is given as,

\(F\left( M \right) = \left\{ \begin{array}{l}0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when}}\,M < 1\\\frac{1}{{36}},\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when 1}} \le M < 2\\\frac{4}{{36}},\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when}}\,2 \le M < 3\\\frac{9}{{36}},\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when}}\,3 \le M < 4\\\frac{{16}}{{36}},\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when}}\,4 \le M < 5\\\frac{{25}}{{36}},\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when}}\,5 \le M < 6\\1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when}}\,M \ge 6\end{array} \right.\)

Following are the steps to construct a line plot of cdf for M:

1. Draw two axes in x-y Cartesian plane.
2. The scale of X-axis represents the values ofMand the scale of Y-axis represents the cdf of each value.
3. Label the marks and give names to the horizontal and vertical axes.
4. Create dot for each value of cdf w.r.t to the value of M. Join all the data points to make a single line.