Question

Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice. Determine also the mean of the distribution.

Answer 1

To find the probability we have to find the sample space according to the problem. Then check the probability distribution for each $x = 1, 2, 3, 4, 5, 6\;$ and use the formula of the mean to find our needed answer.

Complete step by step solution:
Let X be the random variable of the maximum score obtained when a die is thrown twice.
Here X can take the values, $x = 1,2,3,4,5,6\;$
Now, if we have the total number of results that can take place is, S,

\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) \\\ } \right\\}$$ So, number of elements of S, $$n(S) = 36$$ , We know that probability of a function is defined as the number of favorable outcomes divided by the total number of outcomes. So, we can find the probabilities, We get the maximum value as 1 only when both dies show 1, i.e. $$(1,1)$$ . As there is only one favorable outcome out of 36 possible outcomes in the sample space, its probability is given by $$P\left( {X = 1} \right) = \dfrac{1}{{36}}$$ We get maximum value as 2 when we get $$(1,2),(2,1),(2,2)$$ in the two dice. As there are 3 favorable outcomes out of 36 possible outcomes in the sample space, its probability is given by $$P\left( {X = 2} \right) = \dfrac{3}{{36}}\;$$ We get maximum value as 3 when we get $$(1,3),(3,1),(3,2),(2,3),(3,3)$$ in the two dice. As there are 5 favorable outcomes out of 36 possible outcomes in the sample space, its probability is given by $$P\left( {X = 3} \right) = \dfrac{5}{{36}}$$ We get maximum value as 3 when we get $$(1,4),(4,1),(4,2),(2,4),(3,4),(4,3),(4,4)$$ in the two dice. As there are 7 favorable outcomes out of 36 possible outcomes in the sample space, its probability is given by $$P\left( {X = 4} \right) = \dfrac{7}{{36}}$$ We get maximum value as 3 when we get $$(1,5),(5,1),(5,2),(2,5),(3,5),(5,3),(4,5),(5,4),(5,5)$$ in the two dice. As there are 9 favorable outcomes out of 36 possible outcomes in the sample space, its probability is given by $$P\left( {X = 5} \right) = \dfrac{9}{{36}}$$ We get maximum value as 3 when we get $$(1,6),(6,1),(6,2),(2,6),(3,6),(6,3),(4,6),(6,4),(5,6),(6,5),(6,6)$$ in the two dice. As there are 11 favorable outcomes out of 36 possible outcomes in the sample space, its probability is given by $$P\left( {X = 6} \right) = \dfrac{{11}}{{36}}$$ So, the required distribution is, $$X$$| 1| 2| 3| 4| 5| 6 ---|---|---|---|---|---|--- $$P(X)$$ | $$\dfrac{1}{{36}}$$ | $$\dfrac{3}{{36}}$$ | $$\dfrac{5}{{36}}$$ | $$\dfrac{7}{{36}}$$ | $$\dfrac{9}{{36}}$$ | $$\dfrac{{11}}{{36}}$$ Also, we know that, from the formula of the mean of the distribution, $$E(X) = \sum {XP(X)} $$ $$ = 1.\dfrac{1}{{36}} + 2.\dfrac{3}{{36}} + 3.\dfrac{5}{{36}} + 4.\dfrac{7}{{36}} + 5.\dfrac{9}{{36}} + 6.\dfrac{{11}}{{36}} = \dfrac{1}{{36}} + \dfrac{6}{{36}} + \dfrac{{15}}{{36}} + \dfrac{{28}}{{36}} + \dfrac{{45}}{{36}} + \dfrac{{66}}{{36}} = \dfrac{{161}}{{36}}$$ **Note:** A probability distribution is a statistical function that describes all the possible values and likelihoods that a random variable can take within a given range. This range will be bounded between the minimum and maximum possible values, but precisely where the possible value is likely to be plotted on the probability distribution depends on a number of factors. These factors include the distribution's mean (average), standard deviation, skewness, and kurtosis.