Question

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

Answer 1

Hint: According to the question given in the question we have to determine the probability distribution of the maximum of the two scores obtained when die is thrown twice and also the mean of the distribution. So, first of all we have to let the random variable score obtained when a die is thrown twice.
Now, we have to obtain all the possible outcomes when a die is thrown at random and when we throw twice a dice the possible outcomes are:
$ \Rightarrow X = 1,2,3,4,5,6$
And sample space which mean total number of possible outcomes are:
$ \Rightarrow S = \{ (1,1),(1,2),(2,1),(2,2),(1,3),(2,3),(3,1),(3,2),(3,3),...............(6,6)\} $

Formula used: To obtain the probability first we have to use the formula as mentioned below:
$ \Rightarrow n(P) = \dfrac{{n(E)}}{{n(S)}}........................(A)$
Where, n(P) is the required probability, n(E) is the obtained event and n(S) is the total number of outcomes that are possible.
Now, we have to determine all the probabilities when a dice a thrown twice and if the possible out- comes are $P(X = 1)$
Now, we have to determine all the probabilities when a dice a thrown twice and if the possible out- comes are $P(X = 2)$
Now, we have to determine all the probabilities when a dice a thrown twice and if the possible out- comes are $P(X = 3)$
Now, we have to determine all the probabilities when a dice a thrown twice and if the possible out- comes are $P(X = 4)$
Now, we have to determine all the probabilities when a dice a thrown twice and if the possible out- comes are $P(X = 5)$
Now, we have to determine all the probabilities when a dice a thrown twice and if the possible out- comes are $P(X = 6)$
After obtaining the probability for all the possibilities now, we have to draw a table for all the required distributions.
Now, we have to determine the mean which can be obtained with the help of the formula as mentioned below:
$ \Rightarrow E(X) = \sum {XP(X)......................(B)} $
Where, X is the possible probabilities and P(X) is the obtained probabilities.

Complete step-by-step solution:
Step 1: First of all we have to let the random variable scores obtained when a die is thrown twice as X as mentioned in the solution hint. Hence,
$ \Rightarrow X = 1,2,3,4,5,6$
Step 2: Now, we have to determine all the same spaces which mean all the possible outcomes when a dice is thrown twice.
$ \Rightarrow S = \{ (1,1),(1,2),(2,1),(2,2),(1,3),(2,3),(3,1),(3,2),(3,3),...............(6,6)\} $
Step 3: Now, we have to determine all the probabilities when a dice is thrown twice and if the possible out- comes are $P(X = 1)$ as mentioned in the solution hint. Hence,
$
   \Rightarrow P(X = 1) = \dfrac{1}{6} \times \dfrac{1}{6} \\
   \Rightarrow P(X = 1) = \dfrac{1}{{36}}
 $
Step 4: Now, we have to determine all the probabilities when a dice is thrown twice and if the possible out- comes are $P(X = 2)$ as mentioned in the solution hint. Hence,
$
   \Rightarrow P(X = 2) = \dfrac{1}{6} \times \dfrac{1}{6} + \dfrac{1}{6} \times \dfrac{1}{6} + \dfrac{1}{6} \times \dfrac{1}{6} \\
   \Rightarrow P(X = 2) = \dfrac{1}{{36}} + \dfrac{1}{{36}} + \dfrac{1}{{36}} \\
   \Rightarrow P(X = 2) = \dfrac{3}{{36}}
 $
Step 5: Now, we have to determine all the probabilities when a dice is thrown twice and if the possible out- comes are $P(X = 3)$ as mentioned in the solution hint. Hence,
$
   \Rightarrow P(X = 3) = \dfrac{1}{6} \times \dfrac{1}{6} + \dfrac{1}{6} \times \dfrac{1}{6} + \dfrac{1}{6} \times \dfrac{1}{6} + \dfrac{1}{6} \times \dfrac{1}{6} + \dfrac{1}{6} \times \dfrac{1}{6} \\
   \Rightarrow P(X = 3) = \dfrac{1}{{36}} + \dfrac{1}{{36}} + \dfrac{1}{{36}} + \dfrac{1}{{36}} + \dfrac{1}{{36}} \\
   \Rightarrow P(X = 3) = \dfrac{5}{{36}}
 $
Step 6: Now, we have to determine all the probabilities when a dice is thrown twice and if the possible out- comes are $P(X = 4)$ as mentioned in the solution hint. Hence,
$
   \Rightarrow P(X = 4) = \dfrac{1}{6} \times \dfrac{1}{6} + \dfrac{1}{6} \times \dfrac{1}{6} + \dfrac{1}{6} \times \dfrac{1}{6} + \dfrac{1}{6} \times \dfrac{1}{6} + \dfrac{1}{6} \times \dfrac{1}{6} + \dfrac{1}{6} \times \dfrac{1}{6} + \dfrac{1}{6} \times \dfrac{1}{6} \\
   \Rightarrow P(X = 4) = \dfrac{1}{{36}} + \dfrac{1}{{36}} + \dfrac{1}{{36}} + \dfrac{1}{{36}} + \dfrac{1}{{36}} + \dfrac{1}{{36}} + \dfrac{1}{{36}} \\
   \Rightarrow P(X = 4) = \dfrac{7}{{36}}
 $
Step 7: Now, we have to determine all the probabilities when a dice is thrown twice and if the possible out- comes are $P(X = 5)$ as mentioned in the solution hint. Hence,
$ \Rightarrow P(X = 5) = \dfrac{9}{{36}}$
Step 8: Now, we have to determine all the probabilities when a dice is thrown twice and if the possible out- comes are $P(X = 6)$ as mentioned in the solution hint. Hence,
$ \Rightarrow P(X = 6) = \dfrac{{11}}{{36}}$
Step 9: Now, we can obtained the required table for required probability distribution is:

X	P(X)
1	$\dfrac{1}{{36}}$
2	$\dfrac{3}{{36}}$
3	$\dfrac{5}{{36}}$
4	$\dfrac{7}{{36}}$
5	$\dfrac{9}{{36}}$
6	$\dfrac{{11}}{{36}}$

Step 10: Now, we have to obtain the mean for the distribution which can be determined with the help of the formula (B) as mentioned in the solution hint. Hence,
$ \Rightarrow \sum {XP(X) = \dfrac{1}{{36}} + \dfrac{6}{{36}} + \dfrac{{15}}{{36}} + \dfrac{{28}}{{36}} + \dfrac{{45}}{{36}} + \dfrac{{66}}{{36}}} $
To solve the expression as obtained just above we have to determine the L.C.M,
\[
   \Rightarrow \sum {XP(X) = \dfrac{{1 + 6 + 15 + 28 + 45 + 66}}{{36}}} \\
   \Rightarrow \sum {XP(X) = \dfrac{{161}}{{36}}} \\
 \]

Hence, with the help of formula (A) and formula (B) as mentioned in the solution hint we have obtained the required probability distribution which is:

X	P(X)
1	$\dfrac{1}{{36}}$
2	$\dfrac{3}{{36}}$
3	$\dfrac{5}{{36}}$
4	$\dfrac{7}{{36}}$
5	$\dfrac{9}{{36}}$
6	$\dfrac{{11}}{{36}}$

And mean of the distribution which is \[\sum {XP(X) = \dfrac{{161}}{{36}}} \]

Note: To determine the required probability n(P) we have to determine the required event for the given case and then we have to obtain the sample space which is the total number of outcomes and required probability can be obtained by dividing the required event by the total number of outcomes or sample space.
To determine the mean of the distribution we have to determine the sum of all the required probabilities and apply the formula as $E(X) = \sum {XP(X)} $